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GCSE Science/Safety in Mains circuits

GCSE Science/Electricity

Mains electricity is potentially dangerous. There are, however, safety features included in plugs. This module looks at how to wire a plug correctly.

The three pin plug
Nowadays most appliances are sold with moulded plugs already fitted. Nevertheless, it is still important to understand the correct wiring of a plug because enough of the old plugs still exist. It is also the case when you bring in equipments overseas. British Standard compliant adaptors are not always available for such non-UK plugs. You are very likely to need to change a plug at some time in your life. In the UK mains electricity is 230 V. (In Hong Kong, it is 220 V.) If you were to touch a live wire a current would flow through your body to the ground. This current may be enough to kill you.

The cable from the appliance usually consist of three wires, an earth and two other wires, live and neutral, which carry the current to and from the power station (live is from the power station and neutral is back to the power station). The wires are made of copper surrounded by an insulation casing. The casing is made of plastic and is coloured:

 * The live wire is brown
 * The neutral wire is blue
 * The earth wire is green and yellow

The three wires are covered by an outer sheath made of plastic.

Q1) Use your knowledge of insulators and conductors to explain
 * Why the wires are insulated.
 * Why the sheaths are made of plastic.

The plug has the following features:
 * A cable grip, to grip the outer sheath of cable and prevent it being pulled out of the plug
 * Three pins made of brass, one of which is the earth pin.
 * A fuse.
 * A case made of plastic.

Q2) Why are the pins made of brass and why is the case plastic?

The live and neutral wires
The live and neutral wires carry the current around the circuit. Mains current is A.C. (alternating current); this means that it is going backward and forwards in cycles (clockwise and anticlockwise around the circuit). The frequency of the cycle is 50 hertz (50 times per second). This cycling of current is achieved by varying the voltage on the line wire from about +325V to – you to the earth. This is where the earth wire is included, for your safety. The earth wire connects the case of the appliance out down the flex to the earth pin on its plug. This connection goes into the socket, then inside the wiring of your house down to the earth through the earthing system (not necessarily plumbing).

If the live wire were to touch the case a huge current would flow through the earth wire. This would probably blow the fuse and break the circuit (see next section). However even if the fuse doesn't blow the current would still prefer to flow through a wire with low resistance than a human body with relatively higher resistance.

Thus the earth wire helps protect you if you touch the case of an appliance that is "live".

The earth pin on a plug is longer than the live and neutral pins. This ensures that the earth pin always connects with the socket first. All sockets have shutters which prevent access to the live contacts when there is no plug in the socket. On some sockets these shutters are operated by the earth pin pushing the shutter mechanism down to uncover the line and neutral socket contacts, on other sockets there is a mechanism which opens the shutters when the two live pins are inserted simultaneously, and on other sockets when all three pins are inserted simultaneously.

The fuse
A fuse is simply a very thin wire. The wire has quite a low melting point. As current flows through the wire it heats up. If too large a current flows, it melts, breaking the circuit. Fuses are used to protect the flexible lead between the plug and the appliance. If too large a current flows through a lead it may overheat or catch fire. Fuses are unlikely to act quickly enough to prevent human electrocution – their main purpose is to prevent fires due to large currents.

Fuses are rated according to how much current they can carry before melting. In plugs fuses are usually 3A (red), 5A (black), or 13A (brown). The correct fuse is the one that matches the current rating of the lead.

All plug fuses must comply to British Standard BS1362. The rating and "BS1362" should be explicitly marked on such fuses.

Q4) A table lamp usually carries a current of 0.5A. What fuse should be put in the plug: 3A, 5A, or 13A?

Q5) An iron usually carries a current of 5.2A. What fuse should be put in the plug: 3A, 5A, or 13A?

Q6) A kettle is protected by an earth wire and a 13A fuse. The line wire comes loose and touches the side of the kettle. The fuse blows. Explain why.

Q7) Explain why the fuse is always located on the line wire and not the neutral wire?

Q8) Describe and Explain what happens in the following scenarios:

 * The earth and live wire switch terminals on the plug
 * The live and neutral wire switch terminals on the plug
 * The neutral and earth wire switch terminals on the plug
 * When the earth wire is removed

Answers | <>

title: GCSE Science/Safety in Mains circuits


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Botany/Plant structure laboratory

I'm experimenting here with the practicality of having laboratory excercises as part of the Botany Study Guide Comments invited.

Leaves
Ludwigia palustris (Onagraceae)([[media:leaves_opposite.jpg|2x Enlarge]]) Examine this 'marsh purslane' plant and determine the following about its leaves:

3-1. Leaf arrangement is:
 * a) opposite
 * b) alternate
 * c) whorled
 * d) entire
 * e) smooth

3-2. Which of the following describes the leaves:
 * a) sessile, margin toothed
 * b) peltate, margin entire
 * c) sessile, margin entire
 * d) petiolate, margin crenate
 * e) petiolate and hirsute

3-3. Only one of the following statements is true:
 * a) A monocot with clasping leaves
 * b) A dicot with hairy, waxy leaves to reduce water loss
 * c) A monocot growing in relatively wet areas
 * d) A dicot with leaves opposite, leaf bases attenuate
 * e) A wetland plant with compound leaves

<< Chapter 3

Answers to Laboratory Questions:
 * 3-1 ~ a
 * 3-2 ~ d (but tricky, as petiole is about 1 mm long and margins weakly crenate).
 * 3-3 ~ d

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Botany/Plant structure laboratory


text:
Discrete Mathematics/Number theory

Introduction
'Number theory' is a large encompassing subject in its own right. Here we will examine the key concepts of number theory.

Unlike real analysis and calculus which deals with the dense set of real numbers, number theory examines mathematics in discrete sets, such as N or Z. If you are unsure about sets, you may wish to revisit ../Set theory/.

Number Theory, the study of the integers, is one of the oldest and richest branches of mathematics. Its basic concepts are those of divisibility, prime numbers, and integer solutions to equations -- all very simple to understand, but immediately giving rise to some of the best known theorems and biggest unsolved problems in mathematics. The Theory of Numbers is also a very interdisciplinary subject. Ideas from combinatorics (the study of counting), algebra, and complex analysis all find their way in, and eventually become essential for understanding parts of number theory. Indeed, the greatest open problem in all mathematics, the Riemann Hypothesis, is deeply tied into Complex Analysis. But never fear, just start right into Elementary Number Theory, one of the warmest invitations to pure mathematics, and one of the most surprising areas of applied mathematics!

Divisibility
Note that in R, Q, and C, we can divide freely, except by zero. This property is often known as closure -- the quotient of two rationals is again a rational, etc.. However, if we move to performing mathematics purely in a set such as Z, we come into difficulty. This is because, in the integers, the result of a division of two integers might not be another integer. For example, we can of course divide 6 by 2 to get 3, but we cannot divide 6 by 5, because the fraction 6/5 is not in the set of integers.

However we can introduce a new relation where division is defined. We call this relation divisibility, and if $$\frac{b}{a}$$ is an integer, we say:
 * $$a$$ divides $$b$$
 * $$a$$ is a factor of $$b$$
 * $$b$$ is a multiple of $$a$$
 * $$b$$ is divisible by $$a$$

Formally, if there exists an integer $$q$$ such that $$b = qa$$ then we say that $$a$$ divides $$b$$ and write $$a \mid b$$. If $$a$$ does not divide $$b$$ then we write $$a \nmid b$$:

Proposition. The following are basic consequences of this definition. Let a, b, and c be integers:
 * (a) If a|b then a|(bc).
 * (b) If a|b and b|c, then a|c.
 * (c) If a|b and a|c, then for any integers x and y, a|(xb+yc) -- in other words a divides any linear combination of its multiples.
 * (d) If both a|b and b|a, then a = b or a = -b.
 * (e) If c is not 0, then a|b is equivalent to ca|cb.

Quotient and divisor theorem
For any integer n and any k > 0, there is a unique q and r such that:

 * n = qk + r (with 0 ≤ r < k)

Here n is known as dividend.

We call q the quotient, r the remainder, and k the divisor.

It is probably easier to recognize this as division by the algebraic re-arrangement:

 * n/k = q + r/k (0 ≤ r/k < 1)

Modular arithmetic
What can we say about the numbers that divide another? Pick the number 8 for example. What is the remainder on dividing 1 by 8? Using the division theorem above

 * 0 = 8*0 + 0
 * 1 = 8*0 + 1
 * 2 = 8*0 + 2
 * 8 = 8*1 + 0
 * 9 = 8*1 + 1
 * 10 = 8 * 1 + 2
 * and so on
 * and so on
 * and so on

We have a notation for the remainders, and can write the above equations as

 * 0 mod 8 = 0
 * 1 mod 8 = 1
 * 2 mod 8 = 2
 * 3 mod 8 = 3
 * 4 mod 8 = 4
 * 5 mod 8 = 5
 * 6 mod 8 = 6
 * 7 mod 8 = 7
 * 8 mod 8 = 0
 * 9 mod 8 = 1
 * 10 mod 8 = 2

We can also write
 * 1 ≡ 1 (mod 8)
 * 2 ≡ 2 (mod 8)
 * 3 ≡ 3 (mod 8)
 * 4 ≡ 4 (mod 8)
 * 5 ≡ 5 (mod 8)
 * 6 ≡ 6 (mod 8)
 * 7 ≡ 7 (mod 8)
 * 8 ≡ 0 (mod 8)
 * 9 ≡ 1 (mod 8)
 * 10 ≡ 2 (mod 8)

These notations are all short for

 * a = 8k+r for some integer k.

So x ≡ 1 (mod 8), for example, is the same as saying

 * x = 8k+1

Observe that the remainder here, in comparing it with the division algorithm is 1. x ≡ 1 (mod 8) asks what numbers have the remainder 1 on division by 8? Clearly the solutions are x=8×0+1, 8×1+1,... = 1, 9, ...

Often the entire set of remainders on dividing by n - which we say modulo n - are interesting to look at. We write this set Zn. Note that this set is finite. The remainder on dividing 9 by 8 is 1 - the same as dividing 1 by 8. So in a sense 9 is really "the same as" 1. In fact, the relation "≡"

 * x ≡ y iff x mod n = y mod n.

is an equivalence relation. We call this relation congruence. Note that the equivalence classes defined by congruence are precisely the elements of Zn.

We can find some number a modulo n (or we say a congruent to n) by finding its decomposition using the division algorithm.

Addition, subtraction, and multiplication work in Zn - for example 3 + 6 (mod 8) = 9 (mod 8) = 1 (mod 8). The numbers do look strange but they follow many normal properties such as commutativity and associativity.

If we have a number greater than n we often reduce it modulo n first - again using the division algorithm. For example if we want to find 11+3 mod 8, its often easier to calculate 3 + 3 (mod 8) rather than reducing 14 mod 8. A trick that's often used is that, say, if we have 6 + 7 (mod 8) we can use negative numbers instead so the problem becomes -2 + -1 = -3 = 5 (mod 8).

We often use the second notation when we want to look at equations involving numbers modulo some n. For example, we may want to find a number x such that
 * 3x ≡ 5 (mod 8)

We can find solutions by trial substitution (going through all the numbers 0 through 7), but what if the moduli are very large? We will look at a more systematic solution later.

Note: we often say that we are working in Zn and use equals signs throughout. Familiarize yourself with the three ways of writing modular equations and expressions.

The Consistency of Modular Arithmetic
Let $$n \geq 2$$ denote an arbitrary base. Given an arbitrary integer $$x$$, the sequence of integers $$..., x-2n, x-n, x, x+n, x+2n, ...$$ are all congruent to each other modulo $$n$$:

$$ ... \equiv x-2n \equiv x-n \equiv x \equiv x+n \equiv x+2n \equiv ... \;\;(\text{mod}\; n) $$

In modular arithmetic, two integers $$x$$ and $$y$$ that are congruent modulo $$n$$ ($$x \equiv y \;\;(\text{mod}\; n)$$) both "represent" the same quantity from $$\Z_n$$. It should be possible to substitute an arbitrary integer $$x$$ in place of integer $$y$$ provided that $$x \equiv y \;\;(\text{mod}\; n)$$.

This means that:
 * Given arbitrary integers $$x_1, x_2$$ and $$y_1, y_2$$, if $$x_1 \equiv x_2 \;\;(\text{mod}\; n)$$ and $$y_1 \equiv y_2 \;\;(\text{mod}\; n)$$, then $$x_1 + y_1 \equiv x_2 + y_2 \;\;(\text{mod}\; n)$$.

 * Given arbitrary integers $$x_1, x_2$$ and $$y_1, y_2$$, if $$x_1 \equiv x_2 \;\;(\text{mod}\; n)$$ and $$y_1 \equiv y_2 \;\;(\text{mod}\; n)$$, then $$x_1y_1 \equiv x_2y_2 \;\;(\text{mod}\; n)$$.

Number Bases
Converting between various number bases is one of the most tedious processes in mathematics.

The numbers that are generally used in transactions are all in base-10. This means that there are 10 digits that are used to describe a number. These ten digits are {0,1,2,3,4,5,6,7,8,9}.

Similarly, base-4 has 4 digits {0,1,2,3} and base-2 has two digits {0,1}. Base two is sometimes referred to as Binary.

There are also bases greater than 10. For these bases, it is customary to use letters to represent digits greater than 10. An example is Base-16 (Hexadecimal). The digits used in this base are {0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F}.

In order to convert between number bases, it is critical that one knows how to divide numbers and find remainders.

To convert from decimal to another base one must simply start dividing by the value of the other base, then dividing the result of the first division and overlooking the remainder, and so on until the base is larger than the result (so the result of the division would be a zero). Then the number in the desired base is the remainders read from end to start.

The following shows how to convert a number (105) which is in base-10 into base-2.

Answer : 1101001

After finishing this process, the remainders are taken and placed in a row (from bottom to top) after the final quotient (1101001, in this example) is shown as the base-2 equivalent of the number 105.

To sum up the process, simply take the original number in base 10, and divide that number repeatedly, keeping track of the remainders, until the quotient becomes less than the numerical value of the base.

This works when converting any number from base-10 to any base. If there are any letters in the base digits, then use the letters to replace any remainder greater than 9. For example, writing 11(of base-10) in base 14.

Answer: B

As 11 is a single remainder, it is written as a single digit. Following the pattern {10=A, 11=B, 12=C...35=Z}, write it as B. If you were to write "11" as the answer, it would be wrong, as "11" Base-14 is equal to 15 in base-10!

In order to convert from a number in any base back to base ten, the following process should be used:

Take the number 3210 (in base-10). In the units place (100), there is a 0. In the tens place (101), there is a 1. In the hundreds place (102), there is a 2. In the thousands place (103), there is a 3.

The formula to find the value of the above number is:

3×103 + 2×102 + 1×101 + 0×100 = 3000 + 200 + 10 + 0 = 3210.

The process is similar when converting from any base to base-10. For example, take the number 3452 (in base-6). In the units place (60), there is a 2. In the sixths place (61) there is a 5. In the thirty-sixths place (62), there is a 4. In the 216th place (63), there is a 3.

The formula to find the value of the above number (in base-10) is:

3×63 + 4×62 + 5×61 + 2×60 = 648 + 144 + 30 + 2 = 824.

The value of 3452 (base-6) is 824 in base-10.

A more efficient algorithm is to add the left-most digit and multiply by the base, and repeat with the next digit and so on.

((3 * 6 + 4) * 6 + 5) * 6 + 2 = 824

The processes to convert between number bases may seem difficult at first, but become easy if one practices often.

Prime numbers
Prime numbers are the building blocks of the integers. A prime number is a positive integer greater than one that has only two divisors: 1, and the number itself. For example, 17 is prime because the only positive integers that divide evenly into it are 1 and 17. The number 6 is not a prime since more than two divisors 1, 2, 3, 6 divide 6. Also, note that 1 is not a prime since 1 has only one divisor.

Some prime numbers
The prime numbers as a sequence begin
 * 2, 3, 5, 7, 11, 13, 17, 19, 23, ...

Euclid's Proof that There are Infinitely Many Primes
The Greek mathematician Euclid gave the following elegant proof that there are an infinite number of primes. It relies on the fact that all non-prime numbers --- composites --- have a unique factorization into primes.

Euclid's proof works by contradiction: we will assume that there are a finite number of primes, and show that we can derive a logically contradictory fact.

So here we go. First, we assume that that there are a finite number of primes:

 * p1, p2, ..., pn

Now consider the number M defined as follows:

 * M = 1 + p1 *  p2 * ... *  pn

There are two important --- and ultimately contradictory --- facts about the number M:

 * 1) It cannot be prime because pn is the biggest prime (by our initial assumption), and M is clearly bigger than pn. Thus, there must be some prime p that divides M.
 * 2) It is not divisible by any of the numbers p1, p2, ..., pn.   Consider what would happen if you tried to divide M by any of the primes in the list p1, p2, ..., pn.  From the definition of M, you can tell that you would end up with a remainder of 1. That means that p --- the prime that divides M --- must be bigger than any of p1, ..., pn.

Thus, we have shown that M is divisible by a prime p that is not on the finite list of all prime. And so there must be an infinite number of primes.

These two facts imply that M must be divisible by a prime number bigger than pn. Thus, there cannot be a biggest prime.

Note that this proof does not provide us with a direct way to generate arbitrarily large primes, although it always generates a number which is divisible by a new prime. Suppose we know only one prime: 2. So, our list of primes is simply p1=2. Then, in the notation of the proof, M=1+2=3. We note that M is prime, so we add 3 to the list. Now, M = 1 +2 *3 = 7. Again, 7 is prime. So we add it to the list. Now, M = 1+2*3*7 = 43: again prime. Continuing in this way one more time, we calculate M = 1+2*3*7*43 = 1807 =13*139. So we see that M is not prime.

Viewed another way: note that while 1+2, 1+2*3, 1+2*3*5, 1+2*3*5*7, and 1+2*3*5*7*11 are prime, 1+2*3*5*7*11*13=30031=59*509 is not.

Testing for primality
There are a number of simple and sophisticated primality tests. We will consider some simple tests here. In upper-level courses we will consider some faster and more sophisticated methods to test whether a number is prime.

Inspection
The most immediate and simple test to eliminate a number n as a prime is to inspect the units digit or the last digit of a number.

If the number n ends in an even number 0, 2, 4, 6, 8 we can show that number n cannot be a prime. For example, take n = 87536 = 8753(10) + 6. Since 10 is divisible by 2 and 6 is divisible by 2 then 87536 must be divisible by 2. In general, any even number can be expressed in the form n = a*10 + b, where b = 0, 2, 4, 6, 8. Since 10 is divisible by 2 and b is divisible by 2 then n = a*10 + b is divisible by 2. Consequently, any number n which ends in an even number such as 7777732 or 8896 is divisible by 2 so n is not a prime.

In a similar type of argument, if a number n ends in a 5 we can show the number n cannot be a prime. If the last digit of n, call it b, is a 5 we can express n in the form n = a*10 + b, where b = 5. Since 10 is divisible by 5 and b = 5 is divisible by 5 then n = a*10 + b is divisible by 5. Hence, any number n which ends in a 5 such as 93475 is divisible by 5 so n is not a prime.

Thus, if a number greater than 5 is a prime it must end with either a 1, 3, 7, or 9. Note that this does not mean all numbers that end in a 1, 3, 7, or 9 are primes. For example, while the numbers 11, 23, 37, 59 are primes, the numbers 21 = 3*7, 33 = 3*11, 27 = 3*9, 39 = 3*13 are not primes. Consequently, if a number ends in a 1, 3, 7, or 9 we have to test further.

Trial Division Method
To test if a number n that ends in a 1, 3, 7, or 9 is prime, we could simply try the smallest prime number and try to divide it in n. If that doesn't divide, we would take the next largest prime number and try again etc. Certainly, if we took all primes numbers in this manner that were less than n and we could not divide n then we would be justified in saying n is prime. However, it can be shown that you don't have to take all primes smaller than n to test if n is prime. We can stop earlier by using the Trial Division Method.

The justification of the Trial Division Method is if a number n has no divisors less than or equal to $$\sqrt{n}$$ then n must be a prime. We can show this by contradiction. Let us assume n has no divisors less than or equal to $$\sqrt{n}$$. If n is not a prime, there must be two numbers a and b such that $$a*b = n$$. In particular, by our assumption $$\sqrt{n} < a $$ and $$\sqrt{n} < b $$. But then $$n = \sqrt{n}\sqrt{n} < a*b = n$$. Since a number can not be greater than itself the number n must be a prime.

Trial Division Method is a method of primality testing that involves taking a number n and then sequentially dividing it by primes up to $$\sqrt{n}$$.

For example, is 113 prime? $$\sqrt{113}$$ is approximately 10.63... We only need to test whether 2, 3, 5, 7 divide 113 cleanly (leave no remainder, i.e., the quotient is an integer).
 * 113/2 is not an integer since the last digit is not even.
 * 113/3 (=37.666...) is not an integer.
 * 113/5 is not an integer since the last digit does not end in a 0 or 5.
 * 113/7 (=16.142857...) is not an integer.

So we need not look at any more primes such as 11, 13, 17 etc. less than 113 to test, since 2, 3, 5, 7 does not divide 113 cleanly, 113 is prime.

Notice that after rejecting 2 and 3 as a divisor, we next considered the next prime number 5 and not the next number 4. We know not to consider 4 because we know 2 does not divide 113. If 2 cannot divide 113 then certainly 4 cannot because if 4 divided 113 and since 2 divides 4 then 2 would divide 113. So we only use the next cheapest available prime to test not the next consecutive number.

If we test 91 we get,

 * 91/2 is not an integer since that last digit is not even.
 * 91/3= (30.333) is not an integer.
 * 91/5= is not an integer since the last digit does not end in a 0 or 5.
 * 91/7=13 is an integer

So we know since 7 divides 91, 91 is not a prime.

Trial division is normally only used for relatively small numbers due to its inefficiency. However this technique has the two advantages that firstly once we have tested a number we know for sure that it is prime and secondly if a number is not prime it also gives us the number's factors.

To obtain a few small primes, it may be best to use the Sieve of Eratosthenes than to test each number sequentially using trial division. The Sieve of Eratosthenes method is basically a process of finding primes by elimination. We start by taking a list of consecutive numbers say 1 to 100. Cross out the number 1 because the number is not prime. Take the next least uncrossed off number which is 2 and circle it. Now cross out all multiples of 2 on the list. Next take the next least uncircled number which is 3. Circle the number 3 and cross out all multiples of 3. The next least uncircled number should be 5 since 4 is a multiple of 2 and should have been crossed off. Circle the number 5 and cross out all multiples of 5. The next least uncircled number should be a 7 since 6 is a multiple of 2. Circle the 7 and mark off all multiples of 7. Now the next uncrossed off number should be 11 since 8,9,10 is a multiple of 2, 3, and 2. If we continue in this manner what is left is the circled numbers which are primes. But notice we can actually stop now and circle all the unmarked numbers after crossing off multiples of 7 because of the result that since $$\sqrt{100} = 10$$ any number less than 100 which is not prime must be divisible by 2, 3, 5, or 7.

The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic is an important theorem relating to the factorization of numbers. The theorem basically states that every positive integer can be written as the product of prime numbers in a unique way (ignoring reordering the product of prime numbers).

In particular, The Fundamental Theorem of Arithmetic means any number such as 1,943,032,663 is either a prime or can be factored into a product of primes. If a number such as 1,943,032,663 can be factored into primes such as 11×13×17×19×23×31×59 it is futile to try to find another different combination of prime numbers that will also give you the same number.

To make the theorem work even for the number 1, we think of 1 as being the product of zero prime numbers.

More formally,
 * For all n∈N
 * n=p1p2p3...
 * where the pi are all prime numbers, and can be repeated.

Here are some examples.
 * 4 = 2 × 2 = 22
 * 12 = 2 × 2 × 3 = 22 × 3
 * 11 = 11.

A proof of the Fundamental Theorem of Arithmetic will be given after Bezout's identity has been established.

LCM and GCD
Two characteristics we can determine between two numbers based on their factorizations are the lowest common multiple, the LCM and greatest common divisor, the GCD (also greatest common factor, GCF)

LCM
The lowest common multiple, or the least common multiple, for two numbers a and b is the smallest number designated by LCM(a,b) that is divisible by both the number a and the number b. We can find LCM(a,b) by finding the prime factorization of a and b and choosing the maximum power for each prime factor.

In another words, if the number a factors to $$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$, and the number b factors to $$p_1^{\beta_1}p_2^{\beta_2}\cdots p_n^{\beta_n}$$, then LCM(a,b) = $$p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$ where $$\gamma_i = Maximum (\alpha_i, \beta_i)$$ for i = 1 to n.

An example, let us see the process on how we find lowest common multiple for 5500 and 450 which happens to be 49500. First, we find the prime factorization for 5500 and 450 which is
 * 5500=22 53 11
 * 450=2 32 52

Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power.
 * 5500=22 53 11 = 22 30 53 111
 * 450=2 32 52 = 21 32 52 110

The LCM(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what the powers of each individual prime will be.

So now we compare the power of each prime for 5500 and 450. Let us consider the powers of the first prime 2. In the number 5500, the prime 2 is raised to the second power and in the number 450, prime 2 is raised to the first power. Since the maximum between 2 and 1 for the power of the prime 2 is 2, we use 2 for the power of the prime 2.

Now let us consider the powers of the prime 3. In the number 5500, the prime 3 is raised to the zero power and in the number 450 the prime 3 is raised to the second power. Since the maximum between 0 and 2 for the power of the prime 3 is 2, we use 2 for the power of the prime 3.

Similarly, let us consider the powers of the next prime 5. In the number 5500, the prime 5 is raised to the third power and in the number 450 the prime 5 is raised to the second power. Since the maximum between 3 and 2 for the power of the prime 5 is 3, we use 3 for the power of the prime 5.

Finally, let us consider the powers of the prime 11, the last prime on our list. In the number 5500, the prime 11 is raised to the first power and in the number 450 the prime 11 is raised to the zero power. Since the maximum between 1 and 0 for the power of the prime 11 is 1, we use 1 for the power of the last prime 11.

Consequently, the product of our results is LCM(5500,450)=22 32 53 111 = 49500.

GCD
The greatest common divisor for two numbers a and b is the biggest number designated by GCD(a,b) that divides both the number a and the number b. In a similar process to finding LCM(a,b), we can find GCD(a,b) by finding the prime factorization of a and b but choosing the minimum power for each prime factor instead.

In other words, if the number a factors to $$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$, and the number b factors to $$p_1^{\beta_1}p_2^{\beta_2}\cdots p_n^{\beta_n}$$, then GCD(a,b) = $$p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$ where $$\gamma_i = Minimum (\alpha_i, \beta_i)$$ for i = 1 to n.

An example, let us see the process on how we find the greatest common divisor for 5500 and 450 which happens to be 50. First, we find the prime factorization for 5500 and 450 which is
 * 5500=22 53 11
 * 450=2 32 52

Notice the different primes we came up for both the number 5500 and the number 450 are 2, 3, 5, and 11. Now let us express 5500 and 450 completely in a product of these primes raised to the appropriate power.
 * 5500=22 53 11 = 22 30 53 111
 * 450=2 32 52 = 21 32 52 110

The GCD(5500,450) is going to be in the form 2? 3? 5? 11?. All we now have to do is find what the powers of each individual prime will be.

So now we compare the power of each prime for 5500 and 450. Let us consider the powers of the first prime 2. In the number 5500, the prime 2 is raised to the second power and in the number 4 50, prime 2 is raised to the first power. Since the minimum between 2 and 1 for the power of the prime 2 is 1, we use 1 for the power of the prime 2.

Now let us consider the powers of the prime 3. In the number 5500, the prime 3 is raised to the zero power and in the number 450 the prime 3 is raised to the second power. Since the minimum between 0 and 2 for the power of the prime 3 is 0, we use 0 for the power of the prime 3.

Similarly, let us consider the powers of the next prime 5. In the number 5500, the prime 5 is raised to the third power and in the number 450 the prime 5 is raised to the second power. Since the minimum between 3 and 2 for the power of the prime 5 is 2, we use 2 for the power of the prime 5.

Finally, let us consider the powers of the prime 11, the last prime on our list. In the number 5500, the prime 11 is raised to the first power and in the number 450 the prime 11 is raised to the zero power. Since the minimum between 1 and 0 for the power of the prime 11 is 0, we use 0 for the power of the last prime 11.

Consequently, the product of our results is GCD(5500,450)=21 30 52 110 = 50.

Properties

 * gcd(a,b)=gcd(b,a)
 * gcd(a,b) = gcd(b,q), where q is the remainder of a divided by b
 * gcd(a/d, b/d)=1, where d is gcd(a,b)

The Euclidean algorithm
The Euclidean algorithm is such that we can find the gcd of two numbers without finding the factorization*. The Euclidean algorithm consists of only addition and multiplication, and uses the above properties of gcd as its basis.

 * Factorization is a "hard" process, that is, to factor a number takes a long time depending on the length of the number. This is why later, when you need to find the gcd of a pair of numbers, you will most likely never factorize the numbers and use the properties of the primes but will use the Euclidean algorithm instead.

An example
We will see how this works by calculating gcd(458,44)

First, divide 458 by 44 and obtain the remainder:
 * 458 = 44 × 10 + 18

Now suppose that a number is a common divisor of 458 and 44. Then it must also be a divisor of 18. To see this, rearrange the above equation to:
 * 458 - 44×10 =18

When this equation is divided by a common divisor of 44 and 458, an integer is obtained on the left, and so must also be obtained on the right. This, by definition, means that the number is also a divisor of 18. By the same reasoning, any common divisor of 18 and 44 is also a divisor of 458. Since all of the common divisors of 458 and 44 are equal to common divisors of 44 and 18, then in particular the greatest common divisors are equal. So we have gcd(458,44)=gcd(44,18)

The next step in the algorithm is to divide 44 by 18 and find the remainder.
 * 44 = 18 × k + r
 * 44 = 18 × 2 + 8

Repeat this process; keep dividing the previous divisor by the previous remainder:
 * 18 = 8 × 2 + 2
 * 8 = 2 × 4 + 0

Our gcd is the last remainder before the zero, in this case, 2. This is because the reasoning that proved gcd(458,44)=gcd(44,18) applies at every step, so gcd(458,44)=gcd(44,18)=gcd(18,8)=gcd(8,2)=gcd(2,0)=2.

The Matrix Method
We can construct a matrix that provides an alternative method for calculating the greatest common divisor. In its general form, the matrix is

$$ \begin{bmatrix} 1 & 0 & b \\ 0 & 1 & a \\ \end{bmatrix} $$

Recall that one way to write the gcd of two numbers is as an integral linear combination. If we are finding the gcd, for example, we could represent it as as + bt, where a and b are the two numbers we are comparing, and s and t are integers. We also know that b = aq + r where r is the remainder upon division of b by a. After we subtract row 2 from row 1, we get

$$ \begin{bmatrix} 1 & -q_{1} & r_{1} \\ 0 & 1 & a \\ \end{bmatrix} $$

If r_2 is nonzero, we must continue the process; this time, subtracting row 1 from row 2. We continue this process until one of the  r's  has been reduced as far as possible. We now have our gcd. The numbers that are in that row, where the 1 and the 0 used to be, represent t and s, respectively.

Let us now look at a computational example.

$$ \begin{bmatrix} 1 & 0 & 99 \\  0 & 1 & 7 \\ \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & 0 & 99 \\  0 & 14 & 98 \\   \end{bmatrix} \rightsquigarrow \begin{bmatrix} 1 & -14 & 1 \\  0 & 14 & 98 \\   \end{bmatrix} $$

We see that it would be trivial at this point to go any further; we would just end up with row-2 containing a zero where a used to be. So we look at row-1 and remember that the 1 represents our remainder, 1(=t) multiplies b and -14(=s) multiplies a such that

$$1 = 99 \times 1 + -14 \times 7 $$

This can be checked by the Euclidean algorithm that gcd(7,99)=1.

The extended Euclidean algorithm
What happens if we try and reverse the process of the Euclidean algorithm by substituting back? Back-substitution is rather tedious and prone to error, so here is a faster method.

Draw up a table with four columns, label these from left to right q, r, u, v. For convenience label a column i representing the step we're currently up to. Place a and b with the greater of these on top in the column r, and place 1s and 0s accordingly:

$$ \begin{matrix} i & q & r & u & v \\ -1 & . & b & 0 & 1\\ 0 & . & a & 1 & 0 \\ \end{matrix} $$

Now iterate downwards by taking the quotient of b/a and putting it in the next space in the q column, then of b-aq in the r column.

To update u and v, take
 * ui = ui-2-ui-1qi
 * vi = vi-2-vi-1qi

Indeed, you are looking for u and v such that au + bv = gcd (a,b). At some point, gcd (a,b) is in fact the remainder at the ith stage, so you might as well compute ui and vi such that aui + bvi = ri, at EACH stage.

Deriving the recurrences found above results from these three equations (the second equation is Euclid's algorithm's basic property, the other two are constraints we set to attain our desired goal):

 * aui-1 + bvi-1 = ri-1
 * ri-2 = qiri-1 + ri
 * aui + bvi = ri

The trick is to then appropriately express ri-2.

Stop writing when you obtain a 0 in the r column.

Finding then, gcd(450,44) (this is the same as gcd(44,450) )

$$ \begin{matrix} i & q  & r   & u  & v \\ -1 & . & 450 & 0  & 1 \\ 0  & .  & 44  & 1  & 0 \\ 1  & 10 & 10  & -10  & 1 \\ 2  & 4  & 4   & 41 & -4 \\ 3  & 2  & \mathbf{2}   & -92  & 9 \\ 4 & 2  & 0   & -  & - \\ \end{matrix} $$

The bold number is the gcd. Observe (9)×450+(-92)×44=2 Clearly these u and v are very special. What can we say about the general case?

Bezout's identity
In the above case we have 9×450+(-92)×44=gcd(450,44). So the greatest common divisor of 450 and 44 can be written as a linear combination of 450 and 44 with integer coefficients. This turns out to be true of any pair of integers. This result is known as "Bezout's Identity" and can be stated as follows:
 * For any pair of nonzero integers, a and b, there exists integers u and v such that
 * au+bv=gcd(a,b)

Proof
 * If a and b are any pair of integers, not both 0, then clearly there exist integers u and v such that au+bv is positive (just match the signs of u and v to those of a and b, respectively, for instance), and for all integer u and v, au+bv is also an integer (because the integers are closed under addition and multiplication). So there is a non-empty set of positive integers consisting of numbers of the form au+bv; call it S.  Since S is a non-empty set of positive integers, it must have a smallest element (this is the so called  Well-ordering principle).  Call this number d.  The claim is that $$d=gcd(a,b)$$.  Since d belongs to S, then

 * (1)$$d=ua + vb$$

 * for suitable u and v. Let n be any positive common divisor of a and b.  Since n divides the right side of (1), it also divides d.  So $$d=qn$$ for some integer $$q\geq 1$$.  So any common divisor of a and b is less than or equal to d.  Therefore, if d is a common divisor of a and b, it is the greatest one.  It only remains to show that d is in fact a common divisor.

 * To prove this, it will be shown that any element of S is divisible by d. This will prove that d is a common divisor of a and b because a and b are both elements of S  (because a = 1×a + 0×b and b = 0×a + 1×b). Let t be any element of S.  Then, by the division algorithm
 * $$t=qd+r$$

 * for some $$0\leq r<d$$.  If r is not 0, then it is in S.  This is because d and t are in S, so
 * $$r=t-qd=(u_1a+v_1b)-q(u_2a+v_2b)=(u_1-qu_2)a+(v_1-qv_2)b=u'a+v'b$$
 * But, since $$r<d$$ and d is the least element of S, this is impossible. The only other possibility is that r=0.  Therefore any element, t, of S is divisible by d.  Since this includes both a and b, d is a common divisor.  Combining this with the previous result establishes Bezout's Identity.

The numbers u and v can either be obtained using the tabular methods or back-substitution in the Euclidean Algorithm.

Proof of the Fundamental Theorem of Arithmetic
One use of Bezout's identity is in a proof of the Fundamental Theorem of Arithmetic. Before this is proven, two other results are needed: Lemma 1: If a prime number, p, divides a product of two integers, $$ab$$, then it must divide a or b (or both).
 * Proof: If p divides both a and b, there is nothing to prove. Assume p does not divide a.  If it can be proven under that assumption that p does divide b, the lemma will be proven.
 * Since p does not divide a, then gcd(a,p)=1 (because the only divisors of p are 1 and p, but p is not a common divisor). Therefore, by Bezout's Identity, there exist integers u and v such that
 * $$1=ua+vp$$
 * Multiply this equation by b to obtain:
 * $$b=uab+vbp$$
 * p divides both terms on the right hand side and, therefore, divides the left hand side. Hence, p divides b, as was to be shown.

Lemma 2: If a prime number, p, divides a product of integers, $$a_1a_2...a_n$$, then it must divide at least one of the factors.
 * Proof: The proof is by induction on n, the number of factors. The statement is true for n=2, by Lemma 1.  Assume the statement is true for n=k and consider a product of k+1 factors.  If p divides more than one of the factors, once again there is nothing to prove.  Assume that p does not divide any of the factors $$a_1, a_2, ... a_k$$.  It will be shown that p must divide $$a_{k+1}$$.  Since the statement is true for n=k, then since p does not divide any of the factors in $$a_1a_2...a_k$$, it must not divide the product (by Contrapositive).  Let $$a_1a_2...a_k=b$$.  Then $$a_1a_2...a_ka_{k+1}=ba_{k+1}$$.  The conclusion then follows by Lemma 1.

Fundamental Theorem of Arithmetic: Any positive integer, n, can be expressed as a product of primes. This product is unique up to the order in which the terms appear.
 * Proof: The proof of the first part of the theorem is by induction on n.  If n=1, it is the product of 0 primes.  Assume all positive integers less than n can be expressed as a product of primes.  If n is prime, then it is the product of 1 prime.  Assume n is not prime or 1.  Then $$n=ab$$,for some positive integers a and b both less than n.  Since a and b are both less than n, they can be written as a product of primes by the induciton hypothesis.  Combining these products gives n as a product of primes as required.
 * Now to prove the second part. Assume there are two prime factorizations of n,
 * $$n=p_1p_2...p_k=q_1q_2...q_s$$
 * $$p_1$$ divides the left side and so must also divide the right side. By Lemma 2, this means that $$p_1$$ must divide one of the $$q_i$$.  But these are all prime, so the only way $$p_1$$ can divide $$q_i$$ is if $$p_1=q_i$$ for some i.  Canceling $$p_1$$ from both sides of the equation forms another equation of the same form.  So it can likewise be proven that $$p_2=q_i$$ for some other i, and so on until all the factors on the left are exhausted.  After this, there must not be any factors remaining on the right side since it must equal 1.  This proves that any two prime factorizations consist of the same prime factors, as was to be shown.

Partitioning the Divisors of Products
The Fundamental Theorem of Arithmetic can also be derived from the following lemma:

Lemma: Given integers $$a$$, $$b$$, and $$c$$, if $$c$$ divides $$ab$$ (denoted by $$c|ab$$), then there exist integers $$m$$ and $$n$$ such that $$c = mn$$ and $$m|a$$ and $$n|b$$.

In other words, an integer that divides a product can itself be factored into a product where each factor divides the corresponding factor from $$ab$$. This means that no new primes are "created" when $$a$$ and $$b$$ are multiplied together.

This Lemma follows from the Fundamental Theorem of Arithmetic and Bezout's identity, but here a more direct proof will be given.

Proof: If any of $$a$$, $$b$$, or $$c$$ is $$0$$, then the Lemma is trivial. In addition, if any of $$a$$, $$b$$, or $$c$$ is negative, then if the Lemma holds for the absolute values $$|a|$$, $$|b|$$, and $$|c|$$, then it is trivial to show that the Lemma holds for $$a$$, $$b$$, and $$c$$. It will now be assumed that $$a$$, $$b$$, and $$c$$ are all strictly positive integers.

Form an $$a \times b$$ array $$S$$ of integers that has $$a$$ columns and $$b$$ rows. $$s_{(x,y)}$$ will denote the integer at column $$x$$ and row $$y$$. Fill the array by sweeping the array row by row starting with row 1, with each row swept starting from column 1. During this "raster" sweep of $$S$$, assign values to $$s_{(x,y)}$$ using the following cyclical pattern: $$1, 2, ..., c$$. In essence, $$s_{(x,y)}$$ is the unique integer from the range $$\{1, 2, \dots, c\}$$ such that $$s_{(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$. Since $$c|ab$$, it is the case that $$s_{(1,1)} = 1$$ and $$s_{(a,b)} = c$$.

As previously indicated, the "raster sweep" through array $$S$$ is a cyclical progression through the entries of $$S$$ where column index $$x$$ cycles around $$1, 2, ..., a$$, and every time $$x$$ transitions from $$a$$ to $$1$$, row index $$y$$ advances by one step around the cycle $$1, 2, ..., b$$.

In the image below, the grid $$S$$ where $$a = 14$$; $$b = 15$$; and $$c = 6$$ is depicted both explicitly and using a brickwork pattern.

Array $$S$$ can be endlessly replicated and used to form the infinite array $$S_\infty$$. For arbitrary integers $$u$$ and $$v$$, the block of entries in $$S_\infty$$ formed by columns $$a(u-1)+1$$ to $$a(u-1)+a$$ and rows $$(b(v-1)-(u-1))+1$$ to $$(b(v-1)-(u-1))+b$$ is a copy of $$S$$. For arbitrary integers $$x$$ and $$y$$, the entry $$s_{\infty,(x,y)}$$ of $$S_\infty$$ is the unique integer from the range $$\{1, 2, \dots, c\}$$ such that $$s_{\infty,(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$.

Given any column $$x$$ and row $$y$$, the entry $$s_{\infty,(x,y)}$$ of $$S_\infty$$ located at $$(x,y)$$ is the unique integer from $$\{1, 2, ..., c\}$$ such that $$s_{\infty,(x,y)} \equiv x + a(y-1) \;\;(\text{mod}\; c)$$. Given an arbitrary displacement column displacement $$\Delta x$$ and row displacement $$\Delta y$$, the difference $$s_{\infty,(x+\Delta x, y+\Delta y)}-s_{\infty,(x,y)}$$ is separated from $$\Delta x + a\Delta y$$ by a multiple of $$c$$. This gives the entries of $$S_\infty$$ the following symmetries:
 * Given column displacement $$\Delta x$$ and columns $$x_1, x_2$$, and row displacement $$\Delta y$$ and rows $$y_1, y_2$$, then $$s_{\infty,(x_1+\Delta x, y_1+\Delta y)}-s_{\infty,(x_1,y_1)} \equiv s_{\infty,(x_2+\Delta x, y_2+\Delta y)}-s_{\infty,(x_2,y_2)} \;\;(\text{mod}\; c)$$. Specifically if $$s_{\infty,(x_1+\Delta x, y_1+\Delta y)}-s_{\infty,(x_1,y_1)} = 0$$ then $$s_{\infty,(x_2+\Delta x, y_2+\Delta y)}-s_{\infty,(x_2,y_2)} = 0$$ since the difference between any two entries of $$S_\infty$$ is confined to the set $$\{-(c-1), -(c-2), ..., -1, 0, 1, ..., c-1\}$$.
 * Given columns $$x_1, x_2$$, and rows $$y_1, y_2$$, if $$s_{\infty,(x_1,y_1)} = s_{\infty,(x_2,y_2)}$$, then shifting $$S_\infty$$ by $$x_2-x_1$$ columns and $$y_2-y_1$$ rows does not change the entries of $$S_\infty$$.

The columns of $$S_\infty$$ that contain $$1$$ are spaced evenly due to the aforementioned symmetry. Let $$m$$ denote the smallest positive integer such that every $$m^\text{th}$$ column contains $$1$$.
 * The entries $$s_{\infty,(1,1)} = 1$$ and $$s_{\infty,(a+1,b)} = 1$$ so it must be the case that $$a$$ is an integer multiple of the period $$m$$: $$m|a$$.
 * The entries $$s_{\infty,(1,1)} = 1$$ and $$s_{\infty,(c+1,1)} = 1$$ so it must be the case that $$c$$ is an integer multiple of the period $$m$$: $$m|c$$.

The rows of $$S_\infty$$ eventually repeat (not allowing any column shifts) with a period of $$n$$. A row does not appear twice in a single cycle due to the symmetry of $$S_\infty$$. Row $$1$$ is identical to row $$b+1$$ so it must be the case that $$b$$ is an integer multiple of the period $$n$$: $$n|b$$.

It will now be proven that $$c = mn$$ by showing that a sub-block of $$S_\infty$$ that consists of $$m$$ columns and $$n$$ rows contains every integer from $$\{1, 2, ..., c\}$$ exactly once.

To clarify notation, given the column indices $$x_1, x_2$$ where $$x_1 \leq x_2$$, and the row indices $$y_1, y_2$$ where $$y_1 \leq y_2$$, then $$S_\infty([x_1,x_2]\times[y_1,y_2])$$ will denote the sub-block of $$S_\infty$$ consisting of columns $$x_1$$ through to $$x_2$$, and rows $$y_1$$ through to $$y_2$$.

Since a row can be uniquely determined from a single cell and the rows only repeat with a period of $$n$$, any block $$S_\infty([x,x]\times[y,y+(n-1)])$$ will contain exactly $$n$$ unique entries.

Columns that contain $$1$$ occur with a period of $$m$$. Due to the symmetry of $$S_\infty$$, given any integer $$s \in \{1, 2, ..., c\}$$, columns that contain $$s$$ occur with a period of $$m$$. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain exactly $$mn$$ unique entries.

Given any integer $$s \in \{1, 2, ..., c\}$$, integer $$s$$ will occur in every $$m^\text{th}$$ column, and within that column, in every $$n^\text{th}$$ cell. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain $$s$$. Any block $$S_\infty([x,x+(m-1)]\times[y,y+(n-1)])$$ will contain every integer from $$\{1, 2, ..., c\}$$ exactly once. So therefore $$c = mn$$. $$\Box$$.

Solving linear modular equations - back to Bezout
Bezout's identity above provides us with the key to solving equations in the form
 * ax ≡ b (mod m)

Coprime case - gcd(a, m) is 1
Consider the case where
 * ax ≡ b (mod m)

but with gcd(a, m)=1

Because of Bezout's identity
 * 1 = au+mv

When we calculate u, this number is special.

Say if we have the equation
 * 4x=11 (mod 21)

4 and 21 are coprime since gcd(4,21)=1. Now 1=4*16+(-3)*(21). Our u in this case is 16. Observe now that 4*16=64. 64 (mod 21) = 1. This number u is very special - it is known as the multiplicative inverse. It is the number u on multiplication by a gives 1 mod m. Bezout's identity on calculating gcd(a, m) will always give you the multiplicative inverse of a modulo m. The multiplicative inverse of a is often written a-1 but note that this does not mean 1/a since we have seen in the first sections that we can not always divide in the integers.

Note that in Zp there is one number without a multiplicative inverse - 0. It may be useful to exclude 0 when considering modular arithmetic, so instead of having to say Zp\{0} all the time, we merely write Zp*.

Now since we have the magic multiplicative inverse, our problem becomes relatively easy to solve. 4-1=16 in Z21 and now, on multiplying throughout by 16
 * x = 11 × 16 (mod 21)

(since 4×16=1 because 16 is 4's multiplicative inverse mod 21). 11×16=176 and using a calculator or using the division theorem we obtain
 * x = 8 (mod 21)

which is our solution! Verify - 8×4 = 32 = 11 (mod 21).

The general case
Consider the general case where
 * ax ≡ b (mod m)

with no restrictions on a, b and m.

Firstly we calculate gcd(a, m) again to obtain d. Now d is a divisor since the d in gcd means greatest common divisor. So we can now divide a and m - but what about b? Since we have calculated the gcd of a and m but not b we have no guarantees that d will divide b. This then becomes a condition that the equation has no solution.

Now we have reduced the problem to the previous coprime case because gcd(a/d, m/d)=1 with d as above. However we do not have 1 solution any more - this is true because we have reduced the solution to being x = c (mod m/d) and we must bring the solution back mod m. This will be come clearer in the examples.

Let's work through some examples.

Example 1. Solve 4x ≡ 3 (mod 20). Firstly, gcd(4, 20) = 4. 4 does not divide 3 and we have no solution.

Example 2. Solve 9x ≡ 6 (mod 15). gcd(9, 15) = 3 and 3 does divide 6 and we have 3 solutions.

Now, divide through by 3 to obtain
 * 3x ≡ 2 (mod 5)

gcd(3, 5) = 1 = 3 × 2 + -1 × 5 So the inverse of 3 mod 5 is 2. Now we obtain the solution
 * x ≡ 4 (mod 5)

Now in Z15 we must obtain the two extra solutions 9 and 14 mod 15 - 9 mod 5 = 4 and 14 mod 5 = 4.

Generally we can say that if we have the solution to the reduced equation x, the general solution is x+(m/d)k for k={0, 1, .., d-1}.

Chinese Remainder Theorem
Very often congruence relations are required to hold simultaneously. Given positive integer bases $$m_1$$ and $$m_2$$ and arbitrary integers $$a$$ and $$b$$ where $$0 \leq a < m_1$$ and $$0 \leq b < m_2$$, a common question is what integers $$x$$ satisfy the following congruence relations simultaneously:

$$\left\{\begin{array}{c} x \equiv a \;\;(\text{mod}\; m_1) \\ x \equiv b \;\;(\text{mod}\; m_2) \end{array}\right.$$

The Chinese Remainder Theorem dictates that when $$\text{gcd}(m_1, m_2) = 1$$, for any choice of $$a \in \Z_{m_1} = \{0, 1, ..., m_1-1\}$$ and $$b \in \Z_{m_2} = \{0, 1, ..., m_2-1\}$$ there exists a unique integer $$c \in \Z_{m_1m_2} = \{0, 1, ..., m_1m_2-1\}$$ such that:
 * the only integers $$x$$ that satisfy $$\left\{\begin{array}{c} x \equiv a \;\;(\text{mod}\; m_1) \\ x \equiv b \;\;(\text{mod}\; m_2) \end{array}\right.$$ simultaneously are those that satisfy $$x \equiv c \;\;(\text{mod}\; m_1m_2)$$

In essence, when $$m_1$$ and $$m_2$$ are coprime, there is a 1-to-1 correspondence between ordered pairs from $$\Z_{m_1} \times \Z_{m_2}$$ and the set $$\Z_{m_1m_2}$$.

Proof 1
Proof: To begin, observe that $$|\Z_{m_1m_2}| = m_1m_2$$ and $$|\Z_{m_1} \times \Z_{m_2}| = |\Z_{m_1}||\Z_{m_2}| = m_1m_2$$ so it is possible to pair each $$c \in \Z_{m_1m_2}$$ with a unique ordered pair $$\langle a, b \rangle \in \Z_{m_1} \times \Z_{m_2}$$ and vice-versa.

Given any $$c \in \Z_{m_1m_2}$$, integer $$c$$ can be reduced modulo $$m_1$$ to get an integer $$a \in \Z_{m_1}$$, and can be reduced modulo $$b$$ to get an integer $$b \in \Z_{m_2}$$. Integers $$a$$ and $$b$$ satisfy: $$\forall x \in \Z : (x \equiv c \;\;(\text{mod}\; m_1m_2)) \implies ((x \equiv a \;\;(\text{mod}\; m_1)) \;\text{and}\; (x \equiv b \;\;(\text{mod}\; m_2)))$$

It is not obvious that for any choice of $$\langle a, b\rangle \in \Z_{m_1} \times \Z_{m_2}$$ that there exists a unique $$c \in \Z_{m_1m_2}$$ such that $$\forall x \in \Z : ((x \equiv a \;\;(\text{mod}\; m_1)) \;\text{and}\; (x \equiv b \;\;(\text{mod}\; m_2))) \iff (x \equiv c \;\;(\text{mod}\; m_1m_2))$$

Let $$S$$ denote an infinite array with two rows indexed by $$y = 1, 2$$, and an infinite number of columns indexed by $$x \in \Z$$. $$s_{(x,y)}$$ will denote the entry of $$S$$ at column $$x$$ and row $$y$$. $$s_{(x,y)}$$ is the unique integer from $$\Z_y = \{0, 1, ..., m_y-1\}$$ where $$s_{(x,y)} \equiv x \;\;(\text{mod}\; m_y)$$.

Given column indices $$x_1, x_2$$ where $$x_1 \leq x_2$$, then $$S_1[x_1,x_2]; S_2[x_1,x_2]; S_{1,2}[x_1,x_2]$$ will respectively denote the sub-blocks formed by columns $$x_1$$ through to $$x_2$$ and row sets $$\{1\}; \{2\}; \{1,2\}$$.

Partition $$S$$ into the series of $$m_1 \times 2$$ blocks $$..., S_{1,2}[-2m_1,-m_1-1], S_{1,2}[-m_1,-1], S_{1,2}[0,m_1-1], S_{1,2}[m_1,2m_1-1], S_{1,2}[2m_1,3m_1-1], ...$$ where block $$k \in \Z$$ is $$S_{1,2}[km_1, (k+1)m_1-1]$$.

Row 1 of block $$k$$ will always be the sequence $$0, 1, ..., m_1-1$$. Row 2 of block $$k$$ can be uniquely determined by its first entry, $$s_{(km_1, 2)}$$ since $$s_{(km_1+r, 2)} \equiv s_{(km_1, 2)} + r \;\;(\text{mod}\; m_2)$$. The blocks only differ by row 2, and row 2 for each block is uniquely determined by its first entry $$s_{(km_1, 2)}$$.

$$s_{(km_1, 2)} \equiv km_1 \;\;(\text{mod}\; m_2)$$ and $$s_{((k+1)m_1, 2)} \equiv (k+1)m_1 \;\;(\text{mod}\; m_2)$$ so $$s_{((k+1)m_1, 2)} \equiv s_{(km_1, 2)} + m_1 \;\;(\text{mod}\; m_2)$$. This implies that the first entry of row 2 for the next block can be uniquely determined from the first entry of row 2 for the current block.

Since each block is uniquely determined from the previous block, the row 2 pattern for each block will repeat after a regular period of $$p$$ blocks: $$S_{1,2}[km_1, (k+1)m_1-1] = S_{1,2}[(k+p)m_1, (k+p+1)m_1-1]$$. Let set $$U \subseteq \Z_{m_2}$$ denote the total range of values attained by $$s_{(km_1,2)}$$. It is the case that $$p = |U|$$. Let $$d$$ be the minimum positive difference between any two elements from $$U$$. The cyclical nature of the elements from $$U$$ makes it easy to show that any element $$t \in U$$ is congruent to a multiple of $$d$$ modulo $$m_2$$. In essence: $$U = \{t| t \in \Z_{m_2} \;\text{and}\; \exists u \in \Z: t \equiv ud \;\;(\text{mod}\; m_2)\}$$. Since $$d$$ is the minimum positive difference between any two elements of $$U$$, both $$m_1$$ and $$m_2$$ are multiples of $$d$$ (in fact, $$m_2 = pd$$). Since $$\text{gcd}(m_1,m_2) = 1$$, it must be the case that $$d = 1$$. This implies that $$U = \Z_{m_2}$$ and that $$p = m_2$$. A total of $$m_2$$ blocks are encountered before any repetition occurs in array $$S$$, and therefore all $$m_1m_2$$ possible columns occur exactly once in a column period of $$m_1m_2$$ in array $$S$$. This establishes the Chinese Remainder Theorem. $$\Box$$

Proof 2
A second (more intuitive) proof can be derived by constructing a mesh to depict the space $$\mathbb{Z}_{m_1} \times \mathbb{Z}_{m_2}$$. This mesh is a rectangular array of points with $$m_1$$ columns and $$m_2$$ rows. The columns are indexed from left to right by $$0, 1, 2, ..., m_1-1$$, and the rows are indexed from bottom to top by $$0, 1, 2, ..., m_2-1$$. Most importantly, the mesh will "wrap around" in the horizontal and vertical dimensions. This means that moving to the right from column $$m_1-1$$ will return you to column $$0$$; moving to the left from column $$0$$ will send you to column $$m_1-1$$; moving up from row $$m_2-1$$ will return you to row $$0$$; and moving down from row $$0$$ will send you to row $$m_2-1$$.

The mesh has $$m_1m_2$$ points, and the horizontal and vertical coordinates of each point are the remainders from dividing an integer $$x$$ by $$m_1$$ and $$m_2$$ respectively. For convenience, given an arbitrary dividend $$x$$, $$x \;\text{mod}\; m_1$$ and $$x \;\text{mod}\; m_2$$ will denote the remainders after $$x$$ is divided by $$m_1$$ and $$m_2$$ respectively. If the dividend $$x$$ is incremented by $$1$$, then the coordinate formed by the remainders moves to the right by one step, and up by one step, wrapping around if necessary. $$x = 0$$ corresponds to the coordinate $$(0,0)$$. Increasing $$x$$ in steps of $$1$$ will trace a ray that originates from $$(0,0)$$ and moves one step to the right and one step up each interation, wrapping around if necessary. The images below give examples of this ray for $$m_1 = 6; m_2 = 5$$ and for $$m_1 = 6; m_2 = 4$$. In the images below, a copy of column 0 and row 0 appears at the right and top of the mesh respectively to illustrate the wrap around property. When $$m_1 = 6; m_2 = 4$$, $$m_1$$ and $$m_2$$ are not coprime and fail to satisfy the conditions of the Chinese remainder theorem.

The ray forms diagonal "stripes" in the mesh, and if these stripes are all equally spaced by $$1$$, then the ray passes through every point in the mesh exactly once proving that every remainder pair is possible and hence the Chinese remainder theorem. When $$m_1 = 6; m_2 = 4$$, $$m_1$$ and $$m_2$$ are not coprime and fail to satisfy the conditions of the Chinese remainder theorem, hence the ray does not hit every mesh point. The wrap around property of the mesh makes the mesh "symmetric" in both the horizontal and vertical dimensions, which means that if the wrap around "seams" were moved to any column and row where the ray passes through the lower left corner, then the ray is completely unchanged. This requires that the stripes be equally spaced. Let $$d$$ denote this equal spacing, and it will be shown that $$d|m_1$$ and $$d|m_2$$. The ray passes through row $$0$$ every $$d$$ steps to the right. The ray passes through $$(0,0)$$, and the wrap around property implies that moving $$m_1$$ steps to the right returns to this same intersection point. This can only occur if $$d|m_1$$. By a similar argument $$d|m_2$$. If $$m_1$$ and $$m_2$$ are coprime, then $$d = 1$$, the stripes are evenly spaced by $$1$$, every remainder pair is possible, and the Chinese remainder theorem is therefore true. $$\Box$$

title: Discrete Mathematics/Number theory


text:
Talk:Botany/Plant structure laboratory

Marshman this image is beautiful ! I want to make it into a big poster and hang it over my bed. Do you have more like it ? --Karl Wick

title: Talk:Botany/Plant structure laboratory


text:
Talk:Algebra/Complex Numbers

Are we planning to add complex vector notation (polar notation) here or add that on a separate page?

(Descriptions provided for people inexperienced with complex math)

Essentially: Re^(θi), where R = the absolute value of the complex number and theta is the angle from the (1,0i) in the counterclockwise direction.

To convert Re^(θi) to rectangular notation or complex number format, multiply R times sin(θ) for the imaginary and R times cos(θ) for the real dimension. Ergo, from this method of calculation we can determine that e^(πri) = -1.

Additionally, what about hypercomplex numbers, such as quaternions and octonions?

Quaternion = a + bi + cj + dk where i = (-1)^(1/2); j = (-1)^(1/2); k = (-1)^(1/2) and ijk = -1; ij = k; ji = -k; et al...

-- Emperorbma 23:34, 20 Sep 2003 (UTC)

Yes yes, you'll get polar notation and a whole lot more, just wait till I add it! I'm trying to expound this in a logical manner, don't worry :) Quaternions and octonions won't be in complex numbers because they are hypercomplex. Dysprosia 01:19, 21 Sep 2003 (UTC)

 * Just seeing if you needed help, that's all. BTW, should we start a hypercomplex number article, then? -- Emperorbma 20:47, 21 Sep 2003 (UTC)
 * Sorry about the belatedness...sure, why not? :) Dysprosia 11:58, 29 Sep 2003 (UTC)
 * It looks like hypercomplex numbers would best go in Discrete mathematics so I think I'll add it there... -- Emperorbma 17:38, 29 Sep 2003 (UTC)

Introduction Section Editing
Guys, I just edited the introduction section by adding the last para which makes the concept of complex numbers more coherent. - HellKing

i
It is a common mistake to define
 * $$i=\sqrt{-1}$$.

The right definition is to introduce a new number i such that:
 * $$\,i^2=-1$$.

which is not equivalent to the first!

Complex numbers
Better than considering complex numbers as matrices, it is more didactical considering them as pairs of real numbers.

Engineering notation
A suggestion: maybe it would be interesting to add the fact that in engineering notation the imaginary part of a complex number is usually referred to as j instead of $$i$$ (because electric current is also referred to as I). Of course, I don't know about American engineering standards, but as a (European) engineering student I was told to adapt to this convention, even though I of course surely also do understand what is meant by using $$i$$ as a means to indicate the complex part of a complex number.

There's also a convention (in some old engineering texts), to use cis θ instead of (cosθ + isinθ). --Fephisto 19:31, 17 April 2006 (UTC)

Euler's formula
How come one prove that
 * reiθ = r (cos θ + i sin θ)

if reiθ is not even defined?

i is defined, and so is ez via Taylor series. If you want, you can define reiθ through a Taylor series, and see that this has the same radius of convergence of ez and equate them. --Fephisto 19:34, 17 April 2006 (UTC)

Cauchy
Is stuff about Cauchy's Line integral theorem &c. with line integrals under the Argand Plane, going to be included here? --Fephisto 19:39, 17 April 2006 (UTC)

Wow
I think this page goes way too far. Half this stuff deserves to be in Calculus, not Algebra. We all learned primary ideas about Complex Numbers in Algebra, but this definitely will confuse most readers

Forced and explained through psychology
At the opening to this material the phrases 'forced' in relation to i being equal to square root of minus 1. Also the statement suggesting explanations and ideas around root of minus 1 can be explained through psychology. Is it possible for some direction, link, or further explanation on this.

title: Talk:Algebra/Complex Numbers


text:
Spanish/Exercises/Questions

^Lesson 5^

Yes/No Questions
Form the corresponding yes/no question:

1. Ellos tienen hambre.

2. Nosotros estudiamos español.

3. Fernando es alto.

Specific Questions
Put in the correct question word:

4. ¿_________ estás? - Estoy bien.

5. ¿__________ es el hombre alto?

6. ¿_________ hermanos tienes?

7. ¿__________ está España? - Está en Europa.

8. ¿_________ es tu cumpleaños?

9. ¿_________estudias español? - Porque es muy interesante.

Soluciones a los ejercicios Solutions to exercices

^Lesson 5^

title: Spanish/Exercises/Questions


text:
XHTML

Contents

 * /What is XHTML/ —
 * /History of XHTML/ —
 * /Media Types/ —
 * /A Basic XHTML Document/ —
 * /DTD (Document Type Definition)/ —
 * /XHTML Grammar/ —
 * /XHTML Documents/ —
 * /XHTML Attributes/ —
 * /XHTML Objects/ —
 * /XHTML Validating/ —
 * Appendices:
 * /HTML vs. XHTML/ —

Other Resources

 * XHTML Tutorial from HTML Source, detailing the differences in writing XHTML from HTML.
 * XHTML Tutorial from W3Schools Introducing comprehensively XHTML.
 * "HTML Dog" HTML and CSS Tutorials.
 * XForms - XForms Tutorial and Cookbook.
 * Common questions on HTML and XHTML are listed at the W3C FAQ page on XHTML and HTML 4.01.
 * XHTML lessons
 * XHTML Books Collection
 * xHTML Tutorial for Beginners

de:Websiteentwicklung: XHTML XHTML XHTML

title: XHTML


text:
GCSE Science/Electrical Power

GCSE Science/Electricity

The reason electricity has become so popular over the last 100 years or so is because it is a very good way of moving energy from one place to another. When you buy a light bulb is usually rated in watts. The watt is a unit of power. In this unit you are going to lean some equations rating current voltage, energy and power and apply the equations.

Definition of power
Power is defined as the rate of energy flow. Its unit is the watt — one watt is one joule per second.

In electrical circuits the power can be found by multiplying the current and voltage together. One way of remembering this is to use an equation triangle like the one below. Memorize the positions of the symbols. P is power, V is voltage and I is current. If you want to know the power, cover its symbol up and what you have left is I×V, or current times voltage. Want to know the voltage? Cover it up and what you have left is P÷I, or power divided by current.

Where does the equation P equals VI come from?
From the definition above power is energy E per unit time t. ( how many joules per second)
 * P = E/t

So where does P = IV come from?

To find out we look at what I and V are defined as. I is electrical current. It is the amount of charge Q flowing past any point every sec. The voltage also determines the current, so the power is directly proportional to the voltage and the current. We know that from the definition of the volt, if 1 amp flows for 1 second, with a 1 volt potential difference, 1 joule of energy is used up per second, and therefore the power is 1 Watt.

Examples
Michael plugs a 100 W light bulb into the mains, what is the resistance of the bulb? To find the resistance we need to know the current, to find the current we use the equation for power.Covering up the I in the triangle gives:

 * I = P/V = (100 W)/(230 V)
 * = 0.435 A

Now we can use this in the equation for resistance.

 * R = V/I
 * = (230 V)/(0.435 A)
 * = 529 Ω

How much charge flows through this bulb in 15 seconds?

We use the definition of current, to work this out.

I = C/t

So C =It = (0.435 A)(15 s)  = 6.53 coulombs (C)

How much electrical energy is converted into heat and light energy in 15 s?

For this we use the equation for power in its energy per second form.

P = E/t

So E = Pt = (100 W)(15 s) = 1500 joules (J)

Q1) Put Ohm's law into a triangle like the one above.

Q2) If a current of 3 A flows through a 12 V heater, how much energy will it transfer in half an hour? How much charge will have flowed through the heater in the same time?

Q3) In the USA mains voltage is only 120 V. If a 60 W light bulb is connected to the mains in the USA, what current flows through it ?

Q4)What current would flow through the same bulb if plugged into the mains in the UK (230 V) and what would be the power?

Before we go on to looking at power transmission in the power lines of the national grid, we need to look at how energy is generated. We'll come back to this topic later on, after we look at the next module, which is on the magnetic effects of current.

Summary
 * Power is the rate of flow of energy. Power = energy/time
 * The unit of power is the watt, where 1 watt is 1 joule per second
 * Electrical power can be found by multiplying current and voltage. (P = IV)

Answers |<>

title: GCSE Science/Electrical Power


text:
Cell Biology/Endoplasmic Reticulum

Cell Biology | ../Parts of the cell/ | ../Organelles/

Endoplasmic Reticulum | Mitochondria and Chloroplasts >>

The endoplasmic reticulum (endoplasmic="within the cytoplasm", reticulum="little net"; short : ER) is an important organelle in all eukaryotic cells. Prokaryotic organisms do not have organelles and thus do not have an ER. It's base structure and composition is similar to the plasma membrane, though it is an extension of the nuclear membrane. The ER is the site of the translation and folding of and transport of proteins that are to become part of the cell membrane (e.g., transmembrane receptors and other integral membrane proteins) as well as proteins that are to be secreted or "exocytosed" from the cell (e.g., digestive enzymes).

The ER consists of an extensive membrane network of tubes and cisternae (sac-like structures). The membrane encloses a space, the cisternal space (or internal lumen) from the cytosol. This space is acting as a gateway. Parts of the ER membrane are continuous with the outer membrane of the nuclear envelope, and the cisternal space of the ER is continuous with the space in between the two layers of the nuclear envelope.

Parts of the ER are covered with ribosomes (which assemble amino acids into proteins based on instructions from the nucleus). Their rough appearance under electron microscopy led to their being called rough ER (RER), other parts are free of ribosomes and are called smooth ER (SER). The ribosomes on the surface of the rough ER insert the freshly produced proteins directly into the ER, which processes them and then passes them on to the Golgi apparatus (Fig. 1).

Rough and smooth ER differ not only in appearance, but also in function. While the rough ER manufactures and transports proteins destined for membranes and secretion, the smooth ER has functions in several metabolic processes. It takes part in the synthesis of various lipids (e.g., for building membranes) and steroids (e.g., hormones), and also plays an important role in carbohydrate metabolism, detoxification of the cell, and calcium storage.

Proteins that are transported by the ER and from there throughout the cell are marked with an address tag that are called a signal sequence. Günter Blobel was awarded the 1999 Nobel Prize in Physiology or Medicine for his discovery of these signal sequences in 1975. The N-terminus (one end) of a polypeptide chain (e.g., a protein) contains a few amino acids that work as an address tag, which are removed when the polypeptide reaches its destination. Proteins that are destined for places outside the ER are packed into transport vesicles and moved along the cytoskeleton towards their destination.

Endoplasmic Reticulum | Mitochondria and Chloroplasts >>

Cell Biology | ../Parts of the cell/ | ../Organelles/

title: Cell Biology/Endoplasmic Reticulum


text:
Latin/Lesson 4-Ablative

The Ablative Case
The ablative case in Latin has 9 main uses:

 * 1) With certain prepositions, eg. in, cum, sub, ab
 * 2) Instrumental ablative, expressing the equivalent of English "by", "with" or "using"
 * 3) Ablative of manner, expressing how an action is done, only when an adjective is used alongside it. Example : Magnā cūrā id scrīpsit: he wrote it with great care.
 * 4) Ablative of time when or within which. note: not to be mistaken with the accusative of time which indicates for how long a period of time an action occurs.
 * 5) Locative Ablative, using the ablative by itself to mean "in", locating an action in space or time
 * 6) Ablative of separation or origin, expressing the equivalent of English "from"
 * 7) Ablative of comparison; when the first element to be compared was in the nominative or accusative case, quam was often omitted and the second element followed in the ablative case.
 * 8) Ablative with special deponents, which is technically the instrumental ablative but is used idiomatically with a few deponent verbs, such as ūtor, fruor, fungor, potior, vēscor
 * 9) Ablative Absolute, which is a type of participial phrase generally consisting of a noun (or pronoun) and a modifying participle in the ablative case; usually set off by commas, the phrase describes some general circumstances under which the action of the sentence occurs.

The different uses of the ablative will be dealt progressively. For a summary of all forms of the ablative, please consult the Appendix.

Grammar Part 5: The Power of the Ablative Case
Ablative generally indicates position in time and/or space (i.e. when and where). It can also indicate the idea of ways of getting to a location, abstractly or concretely.

Exercise
How would you translate "I made the toga by hand"?

 * Hint: You would not (and should not) use the genitive. The case you are studying right now can be used by itself for this goal.
 * Hint: Remember that you won't need to use the pronoun "I," since Latin is based not on word order, but on the endings!
 * Glossary:
 * "to make" - Facio ("I make"), facere ("to make"), feci ("I made"), factus ("made")
 * "toga" - Toga, togae feminine
 * "hand" - Manus, manus feminine (This is fourth declension)

Answer
Answer: Togam manu feci.

In this case, the word "manu" is in the ablative (see fourth declension list) and thus means "by hand."

Exercise
I have my wisdom by means of my teacher.

 * 'Glossary:
 * "wisdom" - Sapientia, sapientiae feminine
 * "to have" - Habeo ("I have"), habere ("to have"), habui ("I had"), habitus ("had")
 * "teacher" - Magister, magistri masculine (This is a second declension word, despite the 'r' at the end, like puer.)

Answer
Answer: Habeo sapientiam magistro.

Ablative of Time
How would you say: I will arrive at the 5th hour.

'at the 5th hour' is indicating position of time. Thus, it can be put into the ablative case, giving:

adveniam quinta hora

In general, therefore, in order to say "In the morning", "At nine O'clock," or "In the tenth year," use ablative. It is generally used to refer to a specific time in which something has, does, or will occur.

Example: I will leave in the night.

Hint: Future tense can be looked up in the appendices of this Wikibook!

Hint: to leave- discedo, discedere; night- nox, noctis(This is a third declension word!)

Answer
Answer: Discedam nocte.

Note the simplicity in which Latin translates the six words into simply two. The ending based language completely negates the need for the words "I," "will," "in," and "the."

Ablative of Place
Naves navigabant mari. The ships were sailing on the sea.

The ablative is also useful for showing the location of things, in general where you would use the words on, in, or at. There is an exception for the slightly more archaic locative, which is used with the words domi (from domus, domus, f., home), ruri (from rus, ruris, n., country [as opposed to city]), and Romae (from Roma, Romae, f., Rome), as well as with the names of towns, cities and small islands.

Latin has its own way of handling prepositions depending on the nouns and their cases in the sentence, including the versatile in, which can take many different meanings depending upon the case of the object.

Ablative with prepositions
Here are a few prepositions that can take the ablative (for a fuller list, see the lesson on adverbs and prepositions in the previous chapter):

As a general rule, when motion is implied, use the accusative instead.

Example 3
Servus ex agris venit.
 * "The slave came from the fields."

Note: Ager (ager, agri, m., field) must take an ablative suffix to match the preceding preposition, in this case e/ex.

Incidentally, both ager and campus mean "field," but ager, like its English derivative "agriculture", connotes a farming field, while campus (think "camping" or "college campus") means "open field." The Campus Martius was a large field in Rome used for military training.

The Vocative Case
While you will rarely need to ask Lupus where the bathroom is in Latin, you may find yourself reading either quotes or letters in which a person is being directly addressed. The case it will be in is the vocative.

For example, "Hail, Augustus" will appear in Latin as Ave Auguste, and not Ave Augustus.

Each declension has its own form of the vocative singular and plural. They are listed in the table below.

Furthermore, in all but the second declension, the nominative and vocative are exactly the same!
 * In the second declension singular, there are three separate possibilities for the vocative, depending on its nominative ending. Hence, if it is a us word, it will become an e and so forth.

Examples for different declensions in the second declension

 * -us:
 * Lupus ->Lupe (given name, wolf)
 * -ius:
 * Filius -> Fili (son)
 * Horatius-> Horati (given name)
 * -r:
 * Puer-> Puer (boy)

 * In all cases, the plural vocative is exactly the same as the plural nominative. This extends to those words which are neuter, which always have an 'a' for the nominative and vocative.

Examples

 * 1) Hello, Sextus.(Hello= Salve)*
 * Salve, Sexte.
 * 1) Speak, girl! (Speak= dico, dicere, dixi)*
 * Dic, puella.
 * 1) Knee, run!*(Knee= genu; run= curro, currere, cucurri)*
 * Genu, curre!
 * Oh, heart, why do you lead me? (Oh-o; heart- cor, cordis-f.; lead-duco, ducere;
 * O, cor! Cur ducis tu me?

 * Note that the first three also require use of the imperative. The imperative is used when ordering or telling someone what to do, e.g.- "Stop," or "Get away from me."

The basic form of the imperative is created by dropping the "re" off of the infinitive form of the verb, as in: Amare, which becomes Ama; at least in the singular active form, which is all that these exercises require. More can be found about this subject in the chapter on verbs.

title: Latin/Lesson 4-Ablative


text:
Latin/Appendix A

The pronunciation of Church Latin might be easier, for an average European, than Classical Latin. The way introduced here reflect the medieval way.

C is pronounced as [ts] or [tS] before e, i, y, ae, or oe

G is alway hard, except before e, i, or y, then soft (like j in jut)

H is not pronounced.

QU might become [k] rather than [kv] or [kw].

title: Latin/Appendix A


text:
Botany/Kinds of plants

<< Contents Page | Chapter 6 | Chapter 7 | Chapter 8 >>

Chapter 7. Plant Systematics

The Kinds of Plants
The total number of species of plants is tremendous. Any attempt at representing their great diversity requires a system of ordering or arranging the many plant types—hopefully a system that itself contributes to our knowledge and understanding. We could take the straightforward approach of listing all plants alphabetically by their common name, or perhaps by their species name. This index approach would be handy, but would not tell us much about the plants themselves. We would have to know the plants whose names are included for the list itself to have any meaning to us. However, botanists developed a hierarchical system based on the fact—considering a variety of characteristics—that some plants are clearly more similar to each other than to other plants. Arranging all of the world's plants (and animals and minerals) by similarities to each other was an idea first promoted by Carolus Linnaeus.

 * Reread Plants ( Links need not be perused )
 * Read Linnaean taxonomy ( follow links )

Categorizing is an important process by which we humans gain understanding of the world around us, and something we all do to some degree as part of our observation of things and events that we encounter. In biology, as the concept of evolution was formulated, it became obvious that this concept could be the basis for categorization. If plants that are similar in form are indeed closely related—at least more closely related than plants that are dissimilar in form—then a system of classification could be devised that reflected these relationships. This approach has important implications. Related plants have common properties, a fact that can be exploited in agriculture and other practical botanical fields.

Initially, botanists had but one approach: physical examination. The careful examination (and detailed description) of plant structures allowed for arranging each species within a system that placed all more or less similar plants (in certain "important" features) together. This approach is not as easy as it sounds, but played off of and contributed to the expansion of descriptive botany in the 18th and 19th centuries. One problem that became evident is that as species evolved, unrelated plants could come to resemble each other in many respects. After all, form and function are closely related. Within similar habitats (say deserts), species of very distantly related plants might well evolve towards a similar form. Species do not have (or certainly did not have over geological time) unrestricted access to all places on the planet and species distribution is then an important part of interpreting the evolutionary process. As species evolved, they did so within the constraints to dispersion that existed at the time. This fact provides an important clue: unrelated but similar plants are likely to be distributed far from each other on the Earth's surface; and the corollary: plants that have similar structures but have widely separated distributions, may not be so closely related in an evolutionary sense.

At this point it is worthwhile to consider some examples. There are many succulent plants, as this form (typically thick, fleshy stems and/or leaves; often reduction or complete loss of leaves) incorporates adaptations necessary for a plant to survive very dry conditions. Non-botanists are tempted to classify all such plants as types of cacti. In fact, cacti evolved in the New World (the Americas), yet there are many succulents (and many plants that resemble cacti) that are not native to the New World, and evolved independently on the African continent. A large group of such plants are known as the euphorbs.

title: Botany/Kinds of plants


text:
Internet Technologies/History of the Web

The Web grew out of a project at CERN, beginning around 1989, where Tim Berners Lee and Robert Cailliau built the prototype system that became the core of what is now the World Wide Web. The original intent of the system was to make it easier to share research papers among colleagues. The original name of the first prototype was Enquire Within Upon Everything, after a famous 19th century reference work of how-tos. Berners-Lee released files describing his idea for the "World Wide Web" onto the Internet on August 6, 1991.


text:
Internet Technologies/The Web

The World Wide Web (the "Web" or "WWW" for short) is a hypertext system that operates over the Internet. To view the information, you use a software program called a web browser to retrieve pieces of information (called "documents" or "web pages") from web servers (or "web sites") and view them on your screen. You can then follow hyperlinks on the page to other documents or even send information back to the server to interact with it. The act of following hyperlinks is often called "surfing" the web.

Looking further at web browsers, a web browser is an application program that accesses the World Wide Web, which then searches for wanted information on the Internet. The first web browser named Mosaic was developed in the early 1990s. The ease of information access provided by web browsers greatly added to the popularity of the Internet. Companies and individual users alike can use a browser to access untold amounts of information, and its as easy to find as clicking a mouse. The four most popular web browsers are Internet Explorer, Chrome, Firefox, and Netscape. Tight competition has caused for continual improvement in the programs and associated technologies. Web browsers are loaded with ease-of-use features and are customizable to an individual user’s preference.

URLs, HTTP and HTML
The core functionality of the Web is based on three standards: the Uniform Resource Locator (URL), which specifies how each page of information is given a unique "address" at which it can be found; Hyper Text Transfer Protocol (HTTP), which specifies how the browser and server send the information to each other; and Hyper Text Markup Language (HTML), a method of encoding the information so it can be displayed on a variety of devices.

Tim Berners-Lee now heads the World Wide Web Consortium, which develops and maintains these standards and others that enable computers on the Web to effectively store and communicate all kinds of information.

Beyond text
The initial "www" program at CERN only displayed text, but later browsers such as Pei Wei's Viola (1992) added the ability to display graphics as well. Marc Andreessen of NCSA released a browser called "Mosaic for X" in 1993 that sparked a tremendous rise in the popularity of the Web among novice users. Andreesen went on to found Mosaic Communications Corporation (now Netscape Communications, a unit of AOL Time Warner). Additional features such as dynamic content, music and animation can be found in modern browsers.

Frequently, the technical capability of browsers and servers advances much faster than the standards bodies can keep up with, so it is not uncommon for these newer features to not work properly on all computers, and the web as seen by Netscape is not at all the same as the web seen by Internet Explorer. The ever-improving technical capability of the WWW has enabled the development of real-time web-based services such as webcasts, web radio and live web cams.

Java and Javascript
Another significant advance in the technology was Sun Microsystems' Java programming language, which enabled web servers to embed small programs (called applets) directly into the information being served that would run on the user's computer, allowing faster and richer user interaction.

The similarly named, but actually quite different, JavaScript is a scripting language developed for Web pages. In conjunction with the Document Object Model (DOM), JavaScript has become a much more powerful language than its creators originally envisaged.

Sociological Implications
The exponential growth of the Internet was primarily attributed to the emergence of the web browser Mosaic, followed by another, Netscape Navigator during the mid-1990s.

It brought unprecedented attention to the Internet from media, industries, policy makers, and the general public.

Eventually, it led to several visions of how our society might change, although some point out that those visions are not unique to the Internet, but repeated with many new technologies (especially information and communications technologies) of various era.

Because the web is global in scale, some suggested that it will nurture mutual understanding on a global scale.

Publishing web pages
The web is available to individuals outside the mass media. To "publish" a web page, one does not have to go through a publisher or other media institution, and the potential reader is around the globe, some thought. This to some is an opportunity to enhance democracy by giving voices to alternative and minority views. Some others took it as a path to anarchy and unrestrained freedom of expression. Yet others took it as a sign that hierarchically organized society, mass media being a symptomatic part of it, will be replaced by the so-called network society.

Also, the hyper-text seemed to promote a non-hierarchical and non-linear way of expression and thinking. Unlike books and documents, hypertext does not have a linear order from the beginning to the end. It is not broken down into the hierarchy of chapters, sections, subsections, etc. This reminded some of the ideas of Marshall McLuhan that new media change people's perception of the world, mentality, and way of thinking. While not unique issue to the web, hypertext in this sense is closely related to the notion of "death of author" and intertextuality in structuralist literary theory.

These bold visions are at least not fully realized yet. We can find both supporting and countering aspects of web usage.

First, regarding the increased global unity, indeed, many different kinds of information are now available on the web, and for those who wish to know other societies, their cultures, and people, it became easier. When one travels to a foreign country or a remote town, s/he might be able to find some information about the place on the web, especially if the place is in one of the developed countries. Local newspapers, government publications, and other materials are easier to access, and therefore the variety of information obtainable with the same effort may be said to have increased, for the users of the Internet.

At the same time, there are some obvious limitations. The web is so far a very text-centered medium, and those who are illiterate cannot make much use of it. Even among the literate, usage of a computer may or may not be easy enough. It has been known during the late 1990s, though with ample exceptions, that web users are dominantly young males in college or with a college degree. Now the trend has been changing and female and elderly are also using the web, level of education and income are related to the web use, some think (See also the Wikipedia article Digital divide). Another significant obstacle is language. Currently, only a limited number of languages are useable on the web, due to software and standard issues, and none would understand all the available languages. These factors would challenge the notion that the World Wide Web will bring unity to the world.

Second, the increased opportunity to individuals is certainly observable in the countless personal pages, as well as other groups such as families, small shops, which are not among those who publish materials. The emergence of free web hosting services is perhaps an important factor in bringing this possibility into reality. The activities of alternative media expanded into the web as well.

Yet not a small part of those pages seem to be either prematurely abandoned or one-time practice. Very few of those pages, even when they are well-developed, are popular. When it comes to the expression of ideas and provision of information, it seems that the major media organizations and those companies who became major organizations through their online operations are still favored by the dominant majority. Besides, the Web is not necessarily a tool for political self-education and deliberation. The most popular uses of the Web include searching and downloading pornography, which perhaps has a very limited effect in improving democracy. The most intensively accessed web pages include the document detailing the former U.S. president Bill Clinton's sexual misconduct with Monica Lewinsky, as well as the lingerie fashion show by Victoria's Secret. In sum, both in terms of writers and readers, the Web is not popularly used for democracy. While this is not enough to categorically reject the possibility of the Web as a tool for democracy, the effect so far seems to be smaller than some of the expectations for a quite simple reason, lack of interest and popularity. Anarchistic freedom of expression may be enjoyed by some, but many web hosting companies have developed their acceptable use policy over time, sometimes prohibiting some sensitive and potentially illegal expressions. And again, those expressions may not reach great many. The web is still largely a hierarchical place, some may argue.

Third, regarding the non-linear and non-hierarchical structure of the Web, the effect of those on people's perception and psychology are still largely unknown. Some argue that our culture is changing to that of postmodernity, which is closely related to a non-linear and non-hierarchical way of thinking, being, and even social organization. Yet the counter-evidence is available as well. Among the most notable would be the existence of web directories and search engines. Those sites often provide navigations to the most popular sites to visitors. Besides, it is quite obvious that many web sites are organized according to a simple hierarchy, having the "home page" at the top. At least the present state of the Web and web users seem to suggest the change has not been as great as envisioned by some.

title: Internet Technologies/The Web


text:
Spanish/Exercises/Possessive Adjectives

^Lesson 5^

Rellena los espacios en blanco Fill in the blank
1. Manuel tiene una bicicleta. _____ bicicleta es azul. ''Manuel no has a bike. His bike is blue''.

2. _____ clases son a las nueve de la mañana. My classes are at 9:00 am.

3. Nosotros visitamos a _____ abuelos. We visit our grandparents.

4. Los mexicanos tienen _____ día de independencia el 16 de septiembre. The Mexicans have their independence day on September 16

5. ¿Pueden mostrarme _____ casa? Can you (plural) show me your house?

6. La clase de Señor Ford es fantástica. _____ clase es fantástica. Mr. Ford's (his) class is fantastic.

Soluciones a los ejercicios

^Lesson 5^

title: Spanish/Exercises/Possessive Adjectives


text:
GCSE Science/Electromagnetism

Electric currents produce magnetic fields and moving magnetic fields produce electric currents. This module looks that this property of electric and magnetic fields in detail. You will learn about the magnetic fields generated by a wire and a solenoid, then go on to look at motors generators and transformers.

Review of Magnetism
From work in lower school you should already be familiar with the basic properties of magnets. They are:
 * Magnets strongly attract magnetic materials such as iron, steel, nickel, cobalt, and some alloys and ceramics.
 * Magnets have two poles: a north and a south.
 * Like poles repel, unlike poles attract.

Notice the similarity between magnetism and electricity. Electricity attracts neutral materials (recall a rubbed balloon picking up small bits of paper). Electricity comes in two forms:positive and negative. Like charges repel, unlike charges attract. Physicists long ago realised that these similarities were important. It is now realised that electricity and magnetism are closely related forces. They are really two different aspects of the same force! We call that force electromagnetism. This next section of the electricity module is probably the hardest. But it is also the most interesting. Work the problems in the text as you go along and you will be able to cope.

 * GCSE Science/Magnetic effects of a current
 * GCSE Science/Uses of electromagnets
 * GCSE Science/The motor effect
 * GCSE Science/Induction
 * GCSE Science/Generating electricity
 * GCSE Science/Transformers

title: GCSE Science/Electromagnetism


text:
Organic Chemistry

Front Matter

 * 1) Welcome
 * 2) Authors
 * 3) Licensing
 * 4) /Foreword/

Chapters

 * 1) Foundations of Organic Chemistry
 * 2) /Alkanes/
 * 3) /Stereochemistry/
 * 4) /Chirality/
 * 5) /Cycloalkanes/
 * 6) /Alkenes/
 * 7) /Alkynes/
 * 8) /Overview of Functional Groups/
 * 9) /Haloalkanes/
 * 10) /Alcohols/
 * 11) /Ethers/
 * 12) /Amines/
 * 13) /Dienes/
 * 14) /Aromatics/
 * 15) /Aromatic reactions/
 * 16) /Ketones and aldehydes/
 * 17) /Carboxylic acids/
 * 18) /Carboxylic acid derivatives/
 * 19) /Analytical techniques/
 * 20) /Questions and Discussions/

Supplemental material

 * Appendix A: Introduction to reactions
 * Appendix B: Index of reactions
 * Appendix C: Introduction to functional groups
 * Glossary
 * Periodic table
 * Short periodic table
 * Other online resources

Administrative

 * To-Do List
 * Help organize the book structure.
 * Page element templates

title: Organic Chemistry


text:
Wikibooks:Natural sciences bookshelf

For Wikibooks relating to medicine and health, see Health science bookshelf.

Introductory textbooks

 * General Chemistry
 * General Biology
 * The Scientific Method
 * Biology, Answering the Big Questions of Life

High school and children's courses

 * Middle School Science
 * GCSE Science for secondary school students
 * NZ Level 6: Experimental sciences
 * Wikijunior Big Book of Fun Science Experiments
 * Introduction to Physical Science
 * Regents Earth Science (High School)
 * Biology
 * A-level Applied Science
 * 8th Grade Science

Astronomy

 * Astronomy
 * Introduction to Astrophysics
 * Solar System

Chemistry

 * A-level Chemistry
 * AP Chemistry
 * Biochemistry
 * Chemical Engineering
 * Computational chemistry
 * Crystallography
 * General Chemistry
 * IB Chemistry
 * Inorganic Chemistry
 * Mathematics for Chemistry
 * Nanowiki - The opensource handbook of nanoscience
 * Organic Chemistry
 * Physical Chemistry
 * Problems In Highschool Chemistry
 * Proteomics
 * Solved Question Papers - IIT JEE
 * VCE Chemistry

Earth sciences

 * A Chronological Survey of Sedimentary Landforms in the Continental United States
 * Geodesy
 * Geology
 * Oceanography
 * Soil mechanics
 * Volcanoes
 * The Geology of Indonesia

Ecology

 * See also: Biology_bookshelf

 * Ecology
 * Applied Ecology

Environmental sciences

 * Climate change
 * Environmental Science
 * Environmental Science in New York City
 * Meteorology
 * Oceanography - An introduction
 * Regents Earth Science (High School)
 * Soil Science

Library and information science

 * Applications of ICT in Libraries
 * Introduction to Library and Information Science by Richard E. Rubin
 * Library and Information Science

Physical sciences
he:מדף:מדעי הטבע

title: Wikibooks:Natural sciences bookshelf


text:
Wikibooks:Languages bookshelf

Languages of Africa

 * Afrikaans
 * Bambara
 * Kinyarwanda
 * Swahili
 * Zulu

Languages of the Americas
(For English..., French..., Portuguese..., Spanish... see Languages of Europe.)

 * Arapaho
 * Cherokee
 * Inuktitut, Inuinnaqtun (Inuit, Eskimo)
 * Jumieka
 * Mapudungun
 * Miskito
 * Miwok (Central Sierra)
 * Mohawk
 * Nahuatl
 * Navajo
 * Quechua

Languages of Asia
(For Hebrew see Languages of the Middle East.)

 * Azerbaijani
 * Bengali
 * Bikol

 * Gujarati

 * Indonesian
 * Japanese
 * Kannada
 * Kapampangan
 * Korean
 * Marathi
 * Manchu
 * Mongolian
 * Nepal Bhasa
 * Persian
 * Punjabi
 * Sanskrit
 * Tagalog (Filipino)
 * Tajik
 * Tamil
 * Telugu
 * Thai
 * Turkmen
 * Urdu
 * Uzbek
 * Vietnamese

Languages of Europe
(For Afrikaans see Languages of Africa. For Azerbaijani see Languages of Asia. For Hebrew see Languages of the Middle East.)

 * Albanian
 * Armenian
 * Austrian
 * Belarusian
 * Bulgarian
 * Catalan
 * Chechen
 * Croatian
 * Danish
 * Dutch

 * Gothic

 * Hungarian

 * Irish Gaelic
 * Italian

 * Latvian
 * Lithuanian
 * Lower Sorbian grammar
 * Lowland Scots
 * Macedonian
 * Manx
 * Neapolitan

 * Polish

 * Provençal
 * Romanian
 * Russian
 * Scottish Gaelic
 * Serbian
 * Sicilian
 * Silesian
 * Slovenian

 * Swedish
 * Turkish
 * Ukrainian
 * Welsh

Languages of the Middle East
(For Yiddish... see Languages of Europe.)

 * Arabic
 * Modern Hebrew
 * Sumerian

Languages of the Pacific

 * Fijian
 * Hawaiian
 * Tetum
 * Tok Pisin

Sign Languages

 * American Sign Language

Constructed Languages

 * D'ni
 * Esperanto
 * Folksprak
 * Ido
 * Easy Ido
 * Interlingua
 * Klingon
 * Lingua Franca Nova
 * Lojban
 * Neo
 * Novial
 * Fasili Novial
 * Quenya (Grammar)
 * Rohingya
 * Slovio
 * Tengwar (The Elven writing systems invented by J.R.R. Tolkien)
 * Toki Pona
 * Unilingua
 * Volapük
 * Al Bakiyye

Other Language Books

 * Authoring Foreign Language Textbooks
 * Becoming a linguistic mastermind
 * East Asian Calligraphy
 * False Friends of the Slavist
 * How to Learn a Language
 * How to Teach a Language
 * Understanding Scientific Terminology — Greek and Latin roots of Technical terms
 * Visual Language Interpreting

title: Wikibooks:Languages bookshelf