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26 large mural paintings have been executed either in Gambier Parry's medium or in the paraffin-copal medium. Some of these works have been painted directly on plastered walls, some on canvases which have been afterwards affixed by marouflage to the surfaces prepared to receive them. To the latter category belong nearly all the paintings in the Ambulatory of the Royal Exchange, London. Each of the compartments has been very carefully arranged with a view to secure dryness and freedom from soluble saline matter. In front of the wall itself has been fixed a slate slab slightly inclined forwards at the top and having a ventilated airspace behind it. Upon the slate the finished picture has been attached (or maroufle) by means of a thick paste of white-lead, oil, and copal-varnish, spread not only upon the slate, but simultaneously upon the back of the canvas. It may be affirmed that paintings so secured are free from all risk of injury from the back. In an atmosphere like that of London the surface of the painting must either be protected by glass or be periodically cleansed from deposits of dust, soot, tarry matters, and the other impurities which are described in Chapter XXV. of this handbook. Several fresh materials have been recently employed as painting-grounds. They are either patent or secret preparations, dependent in general for their solidification upon reactions between insoluble earthy and alkaline earthy matters, such as china-clay, asbestos, and compounds of lime and magnesia, with solutions of such salts as magnesium chloride, aluminium sulphate, and alum. There is sometimes a lack of tenacity, and always a lack of toughness in these mixtures, but some artists find them to possess precisely the texture and absorptive character they desire in grounds not only for tempera, but also for oil


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Rh painting, and they may be spread on canvas as well as on more rigid supports. There is some danger of want of adhesion between the paint and the ground. It is also necessary to make sure that the materials of the ground do not affect sensitive pigments such as ultramarine. The hardening or petrifying liquids which in most cases are used in association with solid preparations to make the grounds in question, are invariably acid to test-paper, unlike the alkaline silicates described in Chapter IX.

Slate may be used as the ground for spirit-fresco and oil-painting; but its freedom from crystals of iron-pyrites, which present a brass-yellow colour, must be first ascertained. The firm adhesion of any priming, or other layers of oil-paint which may be applied subsequently, to slate may be secured in the following manner. The slate is slowly warmed in a water-oven, and thus becomes quite dry. While still warm, it receives a very thin coat of oil-copal varnish, largely diluted with turpentine or with toluol, and applied warm. When this film is hard, the painting may be carried out as in the ordinary way of using oil-colours; a priming of flake-white ground in oil and mixed with a little copal-varnish and turpentine, may be first applied, if desired. Terra-cotta and stone may be treated in the same way, but, being more absorbent than slate, the process recommended on p. 31 is preferable.

Owing to the presence of sulphuric acid in urban air painting-grounds containing calcium carbonate are liable to an injurious change, the carbonate being turned in part into the hydrous sulphate (gypsum) with a considerable increase of bulk. Then, through such expansion, the surface-pigment becomes fissured and even detached. It will be readily understood that grounds consisting chiefly of sulphate of lime are not susceptible of such

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'''Leviticus 26:28 If you follow My statutes and carefully keep My commandments, I will give you rains in their season, and the land will yield its produce, and the trees of the field will bear their fruit. Your threshing will continue until the grape harvest, and the grape harvest will continue until sowing time; you will have your fill of food to eat and will dwell securely in your land.

And I will give peace to the land, and you will lie down with nothing to fear. I will rid the land of dangerous animals, and no sword will pass through your land. You will pursue your enemies, and they will fall by the sword before you. Five of you will pursue a hundred, and a hundred of you will pursue ten thousand, and your enemies will fall by the sword before you.

I will turn toward you and make you fruitful and multiply you, and I will establish My covenant with you. You will still be eating the old supply of grain when you need to clear it out to make room for the new.

And I will make My dwelling place among you, and My soul will not despise you. I will walk among you and be your God, and you will be My people. I am the LORD your God, who brought you out of the land of Egypt so that you would no longer be slaves to the Egyptians. I broke the bars of your yoke and enabled you to walk in uprightness.

Punishments for Disobedience (Leviticus 26:14–39; Deuteronomy 28:15–68)

If, however, you fail to obey Me and to carry out all these commandments, and if you reject My statutes, despise My ordinances, and neglect to carry out all My commandments, and so break My covenant, then this is what I will do to you: I will bring upon you sudden terror, wasting disease, and fever that will destroy your sight and drain your life. You will sow your seed in vain, because your enemies will eat it. And I will set My face against you, so that you will be defeated by your enemies. Those who hate you will rule over you, and you will flee when no one pursues you.

And if after all this you will not obey Me, I will proceed to punish you sevenfold for your sins. I will break down your stubborn pride and make your sky like iron and your land like bronze, and your strength will be spent in vain. For your land will not yield its produce, and the trees of the land will not bear their fruit.

If you walk in hostility toward Me and refuse to obey Me, I will multiply your plagues seven times, according to your sins. I will send wild animals against you to rob you of your children, destroy your livestock, and reduce your numbers, until your roads lie desolate.

And if in spite of these things you do not accept My discipline, but continue to walk in hostility toward Me, then I will act with hostility toward you, and I will strike you sevenfold for your sins. And I will bring a sword against you to execute the vengeance of the covenant. Though you withdraw into your cities, I will send a plague among you, and you will be delivered into the hand of the enemy. When I cut off your supply of bread, ten women will bake your bread in a single oven and dole out your bread by weight, so that you will eat but not be satisfied.

But if in spite of all this you do not obey Me, but continue to walk in hostility toward Me, then I will walk in fury against you, and I, even I, will punish you sevenfold for your sins.


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 Every 584 days Venus appears in the west as an evening star, first appearing about 20 minutes after sunset, near the horizon. After being visible but a short time, it disappears from view not far from the point where the sun went down.

Every succeeding day this star appears at a higher elevation and glows brighter and brighter; this continues for several months until it appears like a great lamp, outshining every other object in the heavens, except the moon. It then begins to appear at a lower elevation each evening and to set a little earlier, until in the course of two or three weeks it disappears in the early twilight. About two weeks after her departure from the evening sky, Venus appears in the east shining brightly in the cool, gray hours just before Aurora opens the gates for the Sun-god.

Day by day, the Morning Star increases in brilliance just as did the Evening Star until it appears at the highest point above the horizon; then its light gradually declines as each day its stay becomes shorter and shorter, until finally it pales in the morning light and is immersed in the rays of the sun.

Venus passes through its period of greatest brilliance every eighth year. This last occurred in 1921 and will next occur in 1929. At such a time the planet is so bright that it may be seen in daylight, if one knows where to look for it.

The discovery through a telescope that Venus had phases firmly established the theories of Copernicus. When this Polish astronomer, in 1543, announced the astounding theory that the sun was the center of our solar system, he said that if the planets between the earth and the sun could be

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28 change. So, where damp can be excluded, they may be used for mural paintings, ground flints or fine sand being admixed with the burnt gypsum employed. The paintings of the buried cities of Chinese Turkestan explored by Sir Aurel Stein were executed on grounds of this kind—grounds, that is, of nearly pure plaster of Paris.


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'''156  You will eat the flesh of your own sons and daughters. I will destroy your high places, cut down your incense altars, and heap your lifeless bodies on the lifeless remains of your idols; and My soul will despise you.

I will reduce your cities to rubble and lay waste your sanctuaries, and I will refuse to smell the pleasing aroma of your sacrifices. And I will lay waste the land, so that your enemies who dwell in it will be appalled. But I will scatter you among the nations and will draw out a sword after you as your land becomes desolate and your cities are laid waste.

Then the land shall enjoy its Sabbaths all the days it lies desolate, while you are in the land of your enemies. At that time the land will rest and enjoy its Sabbaths. As long as it lies desolate, the land will have the rest it did not receive during the Sabbaths when you lived in it.

As for those of you who survive, I will send a faintness into their hearts in the lands of their enemies, so that even the sound of a windblown leaf will put them to flight. And they will flee as one flees the sword, and fall when no one pursues them. They will stumble over one another as before the sword, though no one is behind them. So you will not be able to stand against your enemies.

You will perish among the nations, and the land of your enemies will consume you. Those of you who survive in the lands of your enemies will waste away in their iniquity and will decay in the sins of their fathers.

God Remembers Those Who Repent

But if they will confess their iniquity and that of their fathers in the unfaithfulness that they practiced against Me, by which they have also walked in hostility toward Me— and I acted with hostility toward them and brought them into the land of their enemies—and if their uncircumcised hearts will be humbled and they will make amends for their iniquity, then I will remember My covenant with Jacob and My covenant with Isaac and My covenant with Abraham, and I will remember the land.

For the land will be abandoned by them, and it will enjoy its Sabbaths by lying desolate without them. And they will pay the penalty for their iniquity, because they rejected My ordinances and abhorred My statutes.

Yet in spite of this, when they are in the land of their enemies, I will not reject or despise them so as to destroy them and break My covenant with them; for I am the LORD their God. But for their sake I will remember the covenant with their fathers, whom I brought out of the land of Egypt in the sight of the nations, that I might be their God. I am the LORD.”

These are the statutes, ordinances, and laws that the LORD established between Himself and the Israelites through Moses on Mount Sinai. Rules about Valuations

Then the LORD said to Moses, “Speak to the Israelites and say to them, ‘When someone makes a special vow to the LORD involving the value of persons, if the valuation concerns a male from twenty to sixty years of age, then your valuation shall be fifty shekels of silver, according to the sanctuary shekel. Or if it is a female, then your valuation shall be thirty shekels. And if the


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 clearly seen, they would show phases like the moon. When in 1610, Galileo turned his telescope on Venus and beheld the beautiful little crescent, the followers of the ancient system of Ptolemy—who claimed that the earth was the central body around which the sun and planets revolved—were, for the first time, silenced, for they saw that the orbit of the earth must enclose the orbit of Venus, and that they must both revolve around the sun. Even a small telescope will show the phases of Venus, the crescent being far more beautiful than one would imagine.

When first seen through a telescope, in the evening twilight, Venus is gibbous, or nearly round. It has then just emerged on the farther side of the sun from us. After first becoming visible as a disk, it gradually assumes the figure of a half moon (this is at the point of its greatest elongation when it is at its highest in the sky); then as it again approaches the sun, it gradually becomes more and more like a slender crescent of light. When Venus is showing nearly a full face, it is the whole diameter of its orbit farther away than when it shows a slender crescent. In this position it is 160 millions of miles distant from the earth; when nearest the earth it is only 22 millions of miles distant. This is the reason why it appears much smaller when its full hemisphere is turned toward us than when it is seen as a small crescent. The proportion of this disk to the crescent is as 10 to 65. After becoming a "silver bow," Venus disappears from the evening sky, later to appear on the other side of the sun in the

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'''Leviticus 27:27 person is from five to twenty years of age, then your valuation for the male shall be twenty shekels, and for the female ten shekels.

Now if the person is from one month to five years of age, then your valuation for the male shall be five shekels of silver, and for the female three shekels of silver. And if the person is sixty years of age or older, then your valuation shall be fifteen shekels for the male and ten shekels for the female. But if the one making the vow is too poor to pay the valuation, he is to present the person before the priest, who shall set the value according to what the one making the vow can afford.

If he vows an animal that may be brought as an offering to the LORD, any such animal given to the LORD shall be holy. He must not replace it or exchange it, either good for bad or bad for good. But if he does substitute one animal for another, both that animal and its substitute will be holy.

But if the vow involves any of the unclean animals that may not be brought as an offering to the LORD, the animal must be presented before the priest. The priest shall set its value, whether high or low; as the priest values it, the price will be set. If, however, the owner decides to redeem the animal, he must add a fifth to its value.

Now if a man consecrates his house as holy to the LORD, then the priest shall value it either as good or bad. The price will stand just as the priest values it. But if he who consecrated his house redeems it, he must add a fifth to the assessed value, and it will belong to him.

If a man consecrates to the LORD a parcel of his land, then your valuation shall be proportional to the seed required for it—fifty shekels of silver for every homer of barley seed. If he consecrates his field during the Year of Jubilee, the price will stand according to your valuation.

But if he consecrates his field after the Jubilee, the priest is to calculate the price in proportion to the years left until the next Year of Jubilee, so that your valuation will be reduced. And if the one who consecrated the field decides to redeem it, he must add a fifth to the assessed value, and it shall belong to him.

If, however, he does not redeem the field, or if he has sold it to another man, it may no longer be redeemed. When the field is released in the Jubilee, it will become holy, like a field devoted to the LORD; it becomes the property of the priests.

Now if a man consecrates to the LORD a field he has purchased, which is not a part of his own property, then the priest shall calculate for him the value up to the Year of Jubilee, and the man shall pay the assessed value on that day as a sacred offering to the LORD. In the Year of Jubilee the field shall return to the one from whom it was bought—the original owner of the land. Every valuation will be according to the sanctuary shekel, twenty gerahs to the shekel.

But no one may consecrate a firstborn of the livestock, because a firstborn belongs to the LORD. Whether it is an ox or a sheep, it is the LORD’s. But if it is among the unclean animals, then he may redeem it according to your valuation and add a fifth of its value. If it


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'''158 is not redeemed, then it shall be sold according to your valuation.

Nothing that a man sets apart to the LORD from all he owns—whether a man, an animal, or his inherited land—can be sold or redeemed; everything so devoted is most holy to the LORD.

No person set apart for destruction may be ransomed; he must surely be put to death.

Instruction on Tithes (Deuteronomy 14:22–29; Deuteronomy 26:1–15; Nehemiah 13:10–14)

Thus any tithe from the land, whether from the seed of the land or the fruit of the trees, belongs to the LORD; it is holy to the LORD. If a man wishes to redeem part of his tithe, he must add a fifth to its value.

Every tenth animal from the herd or flock that passes under the shepherd’s rod will be holy to the LORD. He must not inspect whether it is good or bad, and he shall not make any substitution. But if he does make a substitution, both the animal and its substitute shall become holy; they cannot be redeemed.’”

These are the commandments that the LORD gave to Moses for the Israelites on Mount Sinai.


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 90 synonymous with 'official'; and I have heard my Hindu clerks, of Kayath caste, described as Turks merely because they were in Government employ. On the Biloch frontier, also, the word Turk is commonly used as synonymous with Mughal."

But though Oriental writers make use of the tribal name of Turk to denote a nomadic people, similar inconsistencies are not wanting in European languages. The way in which the French apply the word Bohémien to the gipsies is a parallel instance. The gipsies, though in no way belonging to the same race as the natives of Bohemia, acquired their name in France, on account of certain social habits and customs which they were believed to have brought with them from Bohemia, and because they were known to wander into France from that country. An almost similar instance, though not precisely parallel, was the use in English of the word Indian, during the seventeenth and eighteenth centuries, to define the aborigines of North America, the Caribbean, and many other islands. In this case it was not the social condition and proclivities of the inhabitants that caused the misapplication of name, but their colour, the climate and products of their countries, and other circumstances, which reminded those who came into contact with them, of the India of the East. The process and result, however, are much the same. But if we leave out of consideration the fact that Turk happened also to be a race-name, its employment to designate the pastoral tribes of unsettled abodes becomes no more anomalous than such appellations as Kohistano, Baduin, etc., in Asia, or the familiar Mountaineer, Islander, etc., in Europe.

Misapplication, or change in the application, of race-names is a practice so commonly met with, that it is almost superfluous to mention it here. It may, however, be briefly pointed out, in regard to the names we are dealing with, that the term Tájik has been made, in one instance, to take exactly the opposite meaning to that which it usually bears. Mr. A. G. Ellis, of the British Museum, informs me that while early Armenian writers applied it to the Arabs, modern Armenians


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Translation:Demonstratio nova altera theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse

1.
Although the proof of the theorem on the resolution of algebraic integral functions into factors, which I presented in a paper sixteen years ago, seems to leave nothing to be desired in terms of both rigor and simplicity, I hope that geometers will not be ungrateful if I return to the same most serious question again, and attempt to construct another demonstration from entirely different principles, which will be no less rigorous. Indeed, that previous proof relies at least in part on geometric considerations: on the contrary, the one which I undertake to explain here will be based solely on analytic principles. I have reviewed the most notable analytic methods through which other geometers had attempted to prove our theorem up to that time, and I have extensively exposed the defects they suffered from. The most serious and truly fundamental defect common to all those efforts, as well as to more recent ones that have come to my attention, I declared however not to be inevitable in an analytic demonstration. It is now up to the experts to judge whether the trust once given has been fully justified through these new efforts.

2.
Certain preliminary considerations will precede the main discussion, both to ensure that nothing appears to be lacking, and because the treatment itself may shed new light on even those matters which have already been considered by others. First of all, we will deal with the highest common divisor of two integral algebraic functions of one indeterminate. It should be noted here that we are only concerned with integral functions: from two such functions, if their product is formed, each of them is called a divisor of this product. The order of the divisor is determined by the exponent of the indeterminate in the highest power it contains, without any consideration of numerical coefficients. That which applies to the common divisors of functions can be concluded more briefly, since it is entirely analogous to that which applies to the common divisors of numbers.

Let two functions $Y,$ $Y^{\prime}$  of the indeterminate $x$  be given, where the former is of higher order or at least not of lower order than the latter. We then form the following equations,

$$\begin{array}{c} Y = q Y^{\prime} + Y^{\prime \prime} \\ Y^{\prime} = q^{\prime} Y^{\prime \prime} + Y^{\prime \prime \prime} \\ Y^{\prime \prime} = q^{\prime \prime} Y^{\prime \prime \prime} + Y^{\prime \prime \prime \prime} \\ \text{etc.} \text{ up to } \\ Y^{(\mu-1)} = q^{(\mu-1)} Y^{(\mu)} \end{array}$$

namely by the rule that first $Y$ is divided in the usual way by $Y^{\prime};$  then $Y^{\prime}$  by the remainder of the first division $Y^{\prime \prime},$  which will be of lower order than $Y^{\prime};$  then again the first remainder by the second $Y^{\prime \prime \prime},$  and so on, until we arrive at a division without remainder, which must necessarily happen, as the order of the functions $Y^{\prime},$  $Y^{\prime \prime},$  $Y^{\prime \prime \prime},$  etc., continually decreases. It is scarcely necessary to remark that these functions, as well as the quotients $q,$ $q^{\prime},$  $q^{\prime \prime},$  etc., are integral functions of $x.$  With these preliminaries, it is clear that

I. by going back from the last of these equations to the first, that the function $Y^{(\mu)}$ is a divisor of each of the preceding ones, and consequently a certain common divisor of the given functions $Y,$  ${Y}^{\prime}.$

II. by proceeding from the first equation to the last, any common divisor of the functions ${Y},$ ${Y}^{\prime}$  also measures each of the following ones, and therefore also the last $Y^{(\mu)}.$  Therefore, the functions $Y,$  $Y^{\prime}$  cannot have any common divisor of higher order than $Y^{(\mu)},$  and every common divisor of the same order as $Y^{(\mu)},$  such as ${Y}^{(\mu)},$  will be in the same ratio to this as one number to another, wherefore it must itself be considered as the highest common divisor.

III. If $Y^{(\mu)}$ is of order 0, i.e., a number, then no function of the indeterminate $x$  properly so called can measure the functions $Y,$  ${Y}^{\prime}{:}$  in this case, therefore, it must be said that these functions do not have a common divisor.

IV. If we extract the penultimate equation from our equations, and from this we eliminate $Y^{(\mu-1)}$ by means of the equation before the penultimate one; then again we eliminate ${Y}^{(\mu-2)}$  by means of the preceding equation, and so on, then we will have

$$\begin{aligned} Y^{(\mu)} & =+k Y^{(\mu-2)}-k^{\prime} Y^{(\mu-1)} \\ & =-k^{\prime} Y^{(\mu-3)}+k^{\prime \prime} Y^{(\mu-2)} \\ & =+k^{\prime \prime} Y^{(\mu-4)}-k^{\prime \prime \prime} Y^{(\mu-3)} \\ & =-k^{\prime \prime \prime} Y^{(\mu-5)}+k^{\prime \prime \prime \prime} Y^{(\mu-4)} \\ & \text{ etc., } \end{aligned}$$

provided we suppose the functions $k,$ $k^{\prime},$  $k^{\prime \prime},$  etc., to be formed according to the following rule:

$$\begin{aligned} & k=1 \\ & k^{\prime}=q^{(\mu-2)} \\ & k^{\prime \prime}=q^{(\mu-3)} k^{\prime}+k \\ & k^{\prime \prime \prime}=q^{(\mu-4)} k^{\prime \prime}+k^{\prime} \\ & k^{\prime \prime \prime \prime}=q^{(\mu-5)} k^{\prime \prime \prime}+k^{\prime \prime} \\ & \text{ etc. } \end{aligned}$$

Therefore, we will have

$$\pm k^{(\mu-2)} Y \mp k^{(\mu-1)} Y^{\prime}=Y^{(\mu)}$$

with the upper signs holding when $\mu$ is even and the lower signs holding when it is odd. In the case where $Y$ and $Y^{\prime}$  do not have a common divisor, we can find in this way two functions $Z,$  $Z^{\prime}$  of the indeterminate $x$  such that

$$Z Y+Z^{\prime} Y^{\prime}=1.$$

V. This proposition manifestly also holds in reverse, namely, if the equation

$$Z Y+Z^{\prime} Y^{\prime}=1$$

can be satisfied, such that $Z,$ $Z^{\prime}$  are integral functions of the indeterminate $x,$  then the functions $Y$  and $Y^{\prime}$  cannot have a common divisor.

3.
Another preliminary discussion will concern the transformation of symmetric functions. Let $a,$ $b,$  $c$  etc. be indeterminate quantities, their multitude being $m,$  and let us denote by $\lambda^{\prime}$  the sum of these quantities, by $\lambda^{\prime \prime}$  the sum of their products taken two at a time, by $\lambda^{\prime \prime \prime}$  the sum of their products taken three at a time, etc., so that from the expansion of the product

$$(x-a)(x-b)(x-c) \ldots$$

arises

$$x^{m}-\lambda^{\prime} x^{m-1}+\lambda^{\prime \prime} x^{m-2}-\lambda^{\prime \prime \prime} x^{m-3}+\text{etc. }$$

It follows that the quantities $\lambda^{\prime},$ $ \lambda^{\prime \prime},$  $ \lambda^{\prime \prime \prime},$  etc., are symmetric functions of the indeterminates $a,$  $b,$  $c,$  etc., i.e., such that these indeterminates occur in them in the same manner, or more clearly, such that they are not changed by any permutation of the indeterminates. It is clear that, in general, any integral function of the quantities $\lambda^{\prime},$ $ \lambda^{\prime \prime},$  $ \lambda^{\prime \prime \prime},$  etc., (whether it involves only these indeterminates or still others independent of $a,$  $b,$  $c,$  etc.) will be a symmetric integral function of the indeterminates $a,$  $b,$  $c,$  etc.

4.
The inverse theorem is somewhat less obvious. Let $\rho$ be a symmetric function of the indeterminates $a,$  $b,$  $c,$  etc., which will thus be composed of a certain number of terms of the form

$$M a^{\alpha} b^{\beta} c^{\gamma} \ldots$$

where $\alpha,$ $\beta,$  $\gamma,$  etc. are non-negative integers, and $M$  is a coefficient that is either determinate or at least does not depend on $a,$  $b,$  $c,$  etc. (if it happens that indeterminates besides $a,$  $b,$  $c,$  etc., are involved in the function $\rho$ ). Let us first of all establish a certain order among these terms, to which end we will initially arrange the indeterminates $a,$ $b,$  $c,$  etc., in a definite order, entirely arbitrary in itself, e.g. such that $a$  occupies the first place, $b$  the second, $c$  the third, etc. Then, out of two terms

$M a^{\alpha} b^{\beta} c^{\gamma} \ldots$ and $M a^{\alpha^{\prime}} b^{\beta^{\prime}} c^{\gamma^{\prime}} \ldots$ |undefined

we will assign a higher order to the former than to the latter if either

$\alpha>\alpha^{\prime},$ or $\alpha=\alpha^{\prime}$ and $\beta>\beta^{\prime},$  or $\alpha=\alpha^{\prime},$  $\beta=\beta^{\prime}$  and $\gamma>\gamma^{\prime},$  or etc.,

i.e. if among the differences $\alpha-\alpha^{\prime},$ $\beta-\beta^{\prime},$  $\gamma-\gamma^{\prime},$  etc. the first non-vanishing one turns out to be positive. Therefore, since terms of the same order differ only with respect to the coefficient $M,$ and thus can be merged into one term, we assume that each term of the function $\rho$  belongs to different orders.

Next, we observe that if $M a^{\alpha} b^{\beta} c^{\gamma} \ldots$ has the highest order among all the terms of the function $\rho,$  then $\alpha$  must be greater than, or at least not less than, $\beta.$  For if $\beta>\alpha,$  then the term $M a^{\beta} b^{\alpha} c^{\gamma} \ldots,$  which the function $\rho,$  being symmetric, also involves, would be of higher order than $M a^{\alpha} b^{\beta} c^{\gamma} \ldots,$  contrary to the hypothesis. Similarly, $b$ will be greater than, or at least not less than, $\gamma;$  furthermore, $\gamma$  will be not less than the subsequent exponent $\delta,$  etc.; thus, each of the differences $\alpha-\beta,$  $\beta-\gamma,$  $\gamma-\delta,$  etc., will be non-negative integers.

Secondly, let us consider that if a product is formed from any number of indeterminate integral functions of $a,$ $b,$  $c$  etc., then the highest term of this product must necessarily be the product of the highest terms of those factors. It is equally clear that the highest terms of the functions $\lambda^{\prime},$ $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., respectively, are $a,$  $a b,$  $a b c$  etc. Hence it follows that the highest term produced by the product

$$p=M \lambda^{\prime \alpha-\beta} \lambda^{\prime \prime \beta-\gamma} \lambda^{\prime \prime \prime \gamma-\delta} \ldots$$

will be $M a^{\alpha} b^{\beta} c^{\gamma} \ldots;$ hence by setting $\rho-p=\rho^{\prime},$  the highest term of the function $\rho^{\prime}$  will certainly be of lower order than the highest term of the function $\rho.$  Moreover, it is clear that $p,$  and therefore $\rho^{\prime},$  will become an integral symmetric function of $a,$  $b,$  $c$  etc. Therefore, treating $\rho^{\prime}$  just as $\rho$  was treated before, it will be split into $p^{\prime}+\rho^{\prime \prime},$  so that $p^{\prime}$  is a product of powers of $\lambda^{\prime},$  $ \lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., with coefficients that are either determinate or at least independent of $a,$  $b,$  $c$  etc., and $\rho^{\prime \prime}$  is an integral symmetric function of $a,$  $b,$  $c$  etc. whose highest term belongs to a lower order than the highest term of the function $p^{\prime}.$  Continuing in the same way, it is clear that $\rho$  will eventually  be reduced to the form $p+p^{\prime}+p^{\prime \prime}+p^{\prime \prime \prime}$  etc., i.e. it will be transformed into an integral function of $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc.

5.
The theorem demonstrated in the preceding article can also be stated as follows: Given any symmetric integral function of the indeterminates $a,$ $b,$  $c$  etc., $\rho,$  another integral function of an equal number of other indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., can be assigned, such that through substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., it passes into $\rho.$  Moreover, it can be easily shown that this can only be done in one way. For suppose that there are two different functions $r$ and $r^{\prime}$  of the indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., such that upon substituting $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc. these functions yield the same function of $a,$  $b,$  $c$  etc. Then $r-r^{\prime}$  will be a function of $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which does not vanish by itself, but which is identically destroyed after those substitutions. However, it is easy to see that this is absurd, by considering that $r-r^{\prime}$ must necessarily be composed of a certain number of parts of the form

$$M l^{\prime \alpha} l^{\prime \prime \beta} l^{\prime \prime \prime \gamma} \ldots$$

whose coefficients $M$ do not vanish, and which are different with respect to their exponents, so that the highest order terms coming from each of these parts are of the form

$$M a^{\alpha+\beta+\gamma+\text{etc.}} b^{\beta+\gamma+\text{etc.}} c^{\gamma+\text{etc.}} \ldots$$

and therefore must have different orders, so that the highest order term cannot be destroyed in any way.

Moreover, the calculation of such transformations can be significantly abbreviated by several methods, which we do not dwell on here, since for our purpose only the possibility of the transformation is sufficient.

6.
Consider the product of $m(m-1)$ factors

$$\begin{aligned} & (a-b)(a-c)(a-d) \ldots \\ \times & (b-a)(b-c)(b-d) \ldots \\ \times & (c-a)(c-b)(c-d) \ldots \\ \times & (d-a)(d-b)(d-c) \ldots \\ & \text{ etc. } \end{aligned}$$

which we will denote by $\pi.$ Since it involves the indeterminates $a,$  $b,$  $c$  etc. symmetrically, let us assume that it has been reduced to the form of a function of $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc. Let this function transform into $p,$  if we substitute $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc. in place of $\lambda^{\prime}\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc.,  Having done this, we will call $p$  the determinant of the function

$$y=x^{m}-l^{\prime} x^{m-1}+l^{\prime \prime} x^{m-2}-l^{\prime \prime \prime} x^{m-3}+\text{etc.}$$

So, for $m=2,$ we have

$$p=-l^{\prime 2}+4 l^{\prime \prime}$$

Similarly for $m=3,$ we find

$$p=-l^{\prime 2} l^{\prime \prime 2}+4 l^{\prime 3} l^{\prime \prime \prime}+4 l^{\prime \prime 3}-18 l^{\prime} l^{\prime \prime} l^{\prime \prime \prime}+27 l^{\prime \prime \prime 2}$$

The determinant of the function $y$ is therefore a function of coefficients $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which through the substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., is transformed into the product of all differences between pairs of the quantities $a,$  $b,$  $c$  etc. In the case where $m=1,$  i.e. where only one indeterminate $a$  is present, and thus there are no differences at all, it will be convenient to adopt the number $1$  as the determinant of the function $y.$

In fixing the notion of determinant, the coefficients of the function $y$ should be regarded as indeterminate quantities. The determinant of a function with determined coefficients

$$Y=x^{m}-L^{\prime} x^{m-1}+L^{\prime \prime} x^{m-2}-L^{\prime \prime \prime} x^{m-3}+\text{etc. }$$

will be a determined number $P,$ namely the value of the function $p$  when $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc. Therefore, if we suppose that $Y$  can be resolved into simple factors, as

$$Y=(x-A)(x-B)(x-C) \ldots$$

or that $Y$ arises from

$$\upsilon=(x-a)(x-b)(x-c)$$

by setting $a=A,$ $b=B,$  $c=C$  etc., and thus by the same substitutions by which $\lambda^{\prime}, \lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., become $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime}$  etc., it is clear that $P$  will be equal to the product of factors

$$\begin{aligned} & (A-B)(A-C)(A-D) \ldots \\ \times & (B-A)(B-C)(B-D) \cdots \\ \times & (C-A)(C-B)(C-D) \cdots \\ \times & (D-A)(D-B)(D-C) \cdots \\ &\text{etc.} \end{aligned}$$

Therefore, it is clear that if $P=0,$ at least two of the quantities $A,$  $B,$  $C$  etc., must be equal; on the contrary, if $P \neq 0,$  then $A,$  $B,$  $C$  etc. must all be different. Now we observe that if we set $\frac{\operatorname{d} Y}{\operatorname{d} x}=Y^{\prime},$ or

$$Y^{\prime}=m x^{m-1}-(m-1) L^{\prime} x^{m-2}+(m-2) L^{\prime \prime} x^{m-3}-(m-3) L^{\prime \prime \prime} x^{m-4}+\text{etc. }$$

then we have

$$\begin{aligned} Y^{\prime}=\phantom{+}& (x-B)(x-C)(x-D) \ldots \\ +&(x-A)(x-C)(x-D) \ldots \\ +&(x-A)(x-B)(x-D) \ldots \\ +&(x-A)(x-B)(x-C) \ldots \\ +& \text{ etc. } \end{aligned}$$

Therefore, if two of the quantities $A,$ $B,$  $C$  etc., are equal, e.g. $A=B,$  then $Y^{\prime}$  will be divisible by $x-A,$  meaning that $Y$  and $Y^{\prime}$  will have a common divisor $x-A.$  Vice versa, if it is supposed that $Y^{\prime}$  and $Y$  have a common divisor, then necessarily $Y^{\prime}$  must involve some simple factor from among $x-A,$  $x-B,$  $x-C$  etc., e.g. the first $x-A,$  which clearly cannot happen unless $A$  is equal to one of the other $B,$  $C,$  $D$  etc. From all this, we obtain two theorems:

I. If the determinant of the function $Y$ becomes $=0,$  then certainly $Y$  and $Y^{\prime}$  have a common divisor, and therefore, if $Y$  and $Y^{\prime}$  do not have a common divisor, the determinant of the function $Y$  cannot be $=0.$

II. If the determinant of the function $Y$ is not $=0,$  then certainly $Y$  and $Y^{\prime}$  cannot have a common divisor; or, if $Y$  and $Y^{\prime}$  have a common divisor, the determinant of the function $Y$  must necessarily be $=0.$

7.
It should be noted that the entire force of this very simple demonstration relies on the assumption that the function $Y$ can be resolved into simple factors: which assumption, in this place, where the general demonstration of this resolvability is discussed, would be nothing but begging the question. And yet not all have avoided fallacies entirely similar to this, who have attempted analytic demonstrations of the principal theorem, the specious illusion of whose origin we have already observed, as all have sought only the form of the roots of equations, whereas it should have been necessary to demonstrate their existence. However, enough has been said already, in the previously cited commentary, about such a method of proceeding, which deviates too much from rigor and clarity. Therefore, we will now place the theorems in the preceding article, of which we cannot afford to be without at least one for our purpose, on a more solid foundation. We will start from the second, this being the easier one.

8.
Let us denote by $\rho$ the function

$$\begin{aligned} & \frac{\pi(x-b)(x-c)(x-d) \dots \dots}{(a-b)^{2}(a-c)^{2}(a-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-c)(x-d) \dots \dots}{(b-a)^{2}(b-c)^{2}(b-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-b)(x-d) \dots \dots}{(c-a)^{2}(c-b)^{2}(c-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-b)(x-c) \dots \dots}{(d-a)^{2}(d-b)^{2}(d-c)^{2} \dots \dots} \\ + & \text{ etc. } \end{aligned}$$

which, since $\pi$ is divisible by each of the denominators, becomes an integral function of the variables $x,$  $a,$  $b,$  $c$  etc. Let us further set $\frac{\operatorname{d} \upsilon}{\operatorname{d} x}=\upsilon^{\prime},$  so that we have

$$\begin{aligned} \upsilon^{\prime}=\phantom{+} &(x-b)(x-c)(x-d) \dots \dots \\ +&(x-a)(x-c)(x-d) \dots \dots\\ +&(x-a)(x-b)(x-d) \dots \dots\\ +&(x-a)(x-b)(x-c) \dots \dots\\ +&\text{ etc. } \end{aligned}$$

Clearly, for $x=a,$ we have $\rho \upsilon^{\prime}=\pi,$  hence we conclude that the function $\pi$  - $\rho \upsilon^{\prime}$  is indefinitely divisible by $x-a,$  and similarly by $x-b,,$  $x-c,$  etc., as well as by the product $v.$  Therefore, if we set

$$\frac{\pi-\rho\upsilon^{\prime}}{\upsilon}=\sigma$$

then $\sigma$ will be an integral function of the variables $x,$  $a,$  $b,$  $c$  etc., and indeed, like $\rho,$  it will symmetric with respect to the variables $a,$  $b,$  $c,$  etc. Consequently, two integral functions $r,$  $s,$  of the variables $x,$  $l^{\prime}, l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., can be derived, which upon substituting $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime},$  etc. are transformed into $\rho,$  $\sigma,$  respectively. Therefore, following the analogy, if the function

$$m x^{m-1}-(m-1) l^{\prime} x^{m-2}+(m-2) l^{\prime \prime} x^{m-3}-(m-3) l^{\prime \prime \prime} x^{m-4}+\text{etc.}$$

i.e. the differential quotient $\frac{\operatorname{d} y}{\operatorname{d} x},$ is denoted by $y^{\prime},$  then since $y^{\prime}$  is transformed into $v^{\prime}$  by the same substitutions, it is evident that $p-s y-r y^{\prime}$  is transformed into $\pi- \sigma \upsilon-\rho \upsilon^{\prime},$  i.e. into $0,$  and therefore must vanish identically (art. 5): thus, we have the identity

$$p=s y+r y^{\prime}$$

and hence, if we suppose that substituting $l^{\prime}=L^{\prime},$ $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc. produces $r=R,$  $s=S,$  then we have the identity

$$P=S Y+R Y^{\prime}$$

where since $S,$ $R$  are integral functions of $x,$  and $P$  is a determined quantity or number, it is evident that $Y$  and $Y^{\prime}$  cannot have a common divisor unless $P=0.$  This is precisely the second theorem of art. 6.

9.
We conclude the demonstration of the previous theorem by showing that, in the case where ${Y}$ and ${Y}^{\prime}$  do not have a common divisor, it cannot be that $P=0.$  To this end, we first, following the instructions of article 2, find two integral functions $f x$  and $\varphi x$  of the variable $x,$  such that we have the identity

$$f x. Y+\varphi x. Y^{\prime}=1$$

which we write as

$$f x. \upsilon+\varphi x. \upsilon^{\prime}=1+f x. (\upsilon-Y)+\varphi x. \frac{\operatorname{d}(\upsilon-Y)}{\operatorname{d} x}$$

or equivalently, since we have

$$\begin{aligned} \upsilon^{\prime} =&(x-b)(x-c)(x-d) \ldots \\ &+(x-a). \frac{\operatorname{d}[(x-b)(x-c)(x-d) \ldots]}{\operatorname{d} x} \end{aligned}$$

in the form

$$\begin{aligned} &\phantom{+.} \varphi x. (x-b)(x-c)(x-d) \dots \\ &+ \varphi x. (x-a). \frac{\operatorname{d}[(x-b)(x-c)(x-d) \dots ]}{\operatorname{d} x} \\ &+ f x. (x-a)(x-b)(x-c)(x-d) \dots=1+f x. (\upsilon-Y)+\varphi x. \frac{\operatorname{d}(\upsilon-Y)}{\operatorname{d} x} \end{aligned}$$

For brevity, let us express

$$f x. (y-Y)+\varphi x. \frac{\operatorname{d}(y-Y)}{\operatorname{d} x}$$

which is a function of the variables $x,$ $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., by

$$ F (x, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})$$

Then we will have the identities [1]

$$\begin{aligned} \varphi & a. (a-b)(a-c)(a-d) \ldots=1+F (a, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.}) \\ \varphi & b. (b-a)(b-c)(b-d) \ldots=1+F (b, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.}) \\ \varphi & c. (c-a)(c-b)(c-d) \ldots=1+F (c, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})\\ & \text{etc.} \end{aligned}$$

Supposing, therefore, that the product of all

$$\begin{aligned} 1 & + F(a, l', l, l', \text{etc.}) \\ 1 & + F(b, l', l, l', \text{etc.}) \\ 1 & + F(c, l', l, l', \text{etc.}) \\ & \text{etc.} \\ \end{aligned}$$

which will be a complete function of the indeterminates $a, b, c,$ etc., $l', l, l',$  etc., and indeed a symmetric function with respect to the same $a, b, c,$  etc., can be represented by

$$\psi(\lambda', \lambda, \lambda, \text{etc.}, l', l, l, \text{etc.})$$

and by multiplying all the equations [1], the result is a new identity [2]

$$\pi \varphi a \cdot \varphi b \cdot \varphi c \ldots = \psi(\lambda', \lambda, \lambda, \text{etc.}, \lambda', \lambda, \lambda, \text{etc.})$$

Moreover it is clear, since the product $\varphi a. \varphi b. \varphi c \ldots$ involves the indeterminates $a,$  $b,$  $c$  etc. symmetrically, that it is possible to find an integral function of the indeterminates $l',$  $l,$  $l'$  etc., which is transformed into $\varphi a. \varphi b. \varphi c \ldots$ by the substitutions $l' = \lambda',$  $l = \lambda,$  $l = \lambda$  etc.  Letting $t$  be that function, we will have the identity [3]

$$p t = \psi(l', l, l, \text{etc.}, l', l, l, \text{etc.})$$

since the substitutions $l' = \lambda',$ $l = \lambda,$  $l = \lambda,$  etc. transform this equation into the identity [2].

Now, from the very definition of the function $F,$ it follows that

$$F(x, L', L, L', \text{etc.}) = 0$$

identically. Hence we also have identities

$$\begin{aligned} 1 & + F(a, L', L, L', \text{etc.}) = 1 \\ 1 & + F(b, L', L, L', \text{etc.}) = 1 \\ 1 & + F(c, L', L, L', \text{etc.}) = 1 \\ & \text{etc.} \\ \end{aligned}$$

and consequently

$$\psi(\lambda', \lambda, \lambda, \text{etc.}, L', L, L, \text{etc.}) = 1$$

and therefore also [4]

$$\psi(l', l, l, \text{etc.}, L', L, L, \text{etc.}) = 1$$

Wherefore, by the combination of equations [3] and [4], and by substituting $l' = L',$ $l = L,$  $l = L$  etc., we will have [5]

$$P T = 1$$

if we denote by $T$ the value of the function $t$  corresponding to those substitutions. Since this value necessarily becomes a finite quantity, $P$ cannot certainly be $=0.$  Q.E.D.

10.
From the foregoing, it is now clear that any integral function $Y$ of a single indeterminate $x,$  whose determinant is $=0,$  can be decomposed into factors, none of which has determinant $0.$  Indeed, once we have found the highest common divisor of the functions $Y$  and $\frac{\operatorname{d} Y}{\operatorname{d} x},$  it will already be resolved into two factors. If one of these factors again has a determinant of $0,$ it can be resolved into two factors in the same way, and thus we shall continue until $Y$  is finally resolved into factors of such a nature that none of them has a determinant of $0.$

Moreover, it is easy to see that among the factors into which $Y$ is resolved, there must be at least one such that among the prime factors of its order, two occurs no more often than among the factors of the order $m$  of the function $Y{:}$  more precisely, if we assume that $m = k \cdot 2^{\mu},$  where $k$  denotes an odd number, then among the factors of the function $Y$  there will be at least one whose order is $k' \cdot 2^{\nu},$  where $k'$  is also odd, and either $\nu=\mu,$  or $\nu<\mu.$  The truth of this assertion follows automatically from the fact that $m$  is the sum  of all of the orders of all of the factors of ${Y}.$

11.
Before we proceed further, we will exhibit a certain expression, the introduction of which is of great utility in all investigations concerning symmetric functions, and which will also be very convenient for us. Let $M$ be a function of some of the indeterminates $a,$  $b,$  $c$  etc., and let $\mu$  be the number of these which enter into the expression of $M,$  without any consideration of the other indeterminates that $M$  may happen to involve. By permuting these $\mu$ indeterminates in all possible ways, both among themselves and with the remaining $m-\mu$  from $a,$  $b,$  $c,$  etc., other expressions similar to $M$  will arise from $M,$  and altogether there will be

$$m(m-1)(m-2)(m-3) \ldots (m-\mu+1)$$

expressions, including $M$ itself, which we shall simply call the complex of all $M$ . Hence it is clear what is meant by the sum of all $M,$ the product of all $M,$  etc. Thus, for example, $\pi$  will denote the product of all $a-b,$  $\upsilon$  the product of all $x-a,$  $\upsilon'$  the aggregate of all $\frac{\upsilon}{x-a},$  etc.

If it happens that $M$ is a symmetric function with respect to some of the $\mu$  indeterminates it contains, then those permutations do not alter the function $M.$  Consequently, in the complex of all $M,$  each term will be repeated, at least $1.2.3 \dots \nu$  times, where $\nu$  is the number of indeterminates with respect to which $M$  is symmetric. However, if $M$ is not only symmetric with respect to $\nu$  indeterminates, but also with respect to $\nu'$  others, and also with respect to $\nu$  others, etc., then $M$  will not be altered, whether the pairs from the first $\nu$  indeterminates are permuted among themselves, or the pairs from the second $\nu',$  or the pairs from the third $\nu,$  etc., so that

$$1 . 2 . 3 \ldots \nu, \quad 1. 2 . 3 \ldots \nu', \quad 1. 2 . 3 \ldots \nu'' \text{ etc.}$$

permutations always give rise to identical terms. Therefore, if we always retain only one of these identical terms, we will have altogether

$$\frac{m(m-1)(m-2)(m-3) \ldots (m-\mu+1)}{1. 2 . 3 \ldots \nu. 1 . 2 . 3 \ldots \nu'. 1 . 2 . 3 \ldots \nu'' \text{etc.}}$$

terms, which we shall call the complex of all $M$ excluding repetitions, to distinguish it from the complex of all $M$  including repetitions. Whenever nothing is expressly stated to the contrary, repetitions will always be understood to be admitted.

Of course, it is easy to see that the sum of all $M,$ or the product of all $M,$  or generally any symmetric function of all $M$  always becomes a symmetric function of the indeterminates $a,$  $b,$  $c$  etc., whether repetitions are allowed or excluded.

12.
We now let $u,$ $x$  be  indeterminates, and consider the product of all $u-(a+b) x+a b,$  excluding repetitions, which we will denote by $\zeta.$  Thus, $\zeta$  will be the product of $\frac{1}{2} m(m-1)$  factors

$$ \begin{array}{c} u-(a+b) x+a b \\ u-(a+c) x+a c \\ u-(a+d) x+a d \\ \text{ etc. } \\ u-(b+c) x+b c \\ u-(b+d) x+b d \\ \text{ etc. } \\ u-(c+d) x+c d \\ \text{ etc. etc. etc. } \end{array} $$

This function, since it symmetrically implicates the indeterminates $a,$ $b,$  $c,$  etc., can be assigned an entire function of the indeterminates $u,$  $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime},$  etc., denoted by $z,$  which will pass into $\zeta$  if the indeterminates l^{\prime}, $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., are replaced by $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc. Finally, let us denote by $Z$  the function of the indeterminates $u,$  $x$  alone, which is transformed into $z$  if we assign determinate values $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime \prime},$  etc., to the indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc.

These three functions $\zeta,$ $z,$  $Z$  can be considered as integral functions of order $\frac{1}{2} m(m-1)$  of the indeterminate $u,$  with indeterminate coefficients, which

for $\zeta,$ will be functions of the indeterminates $x,$ $a,$ $b,$  $c$  etc. for $z,$  will be functions of the indeterminates $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc. for $Z,$  will be functions of the indeterminate $x$  alone.

However, the individual coefficients of $z$ will be transformed into the coefficients of $\zeta$  by substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime},$  etc., and into the coefficients of $Z$  by substitutions $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime},$  etc. The same considerations regarding the coefficients will also hold for the determinants of the functions $\zeta,$  $z,$  $Z.$  And we will now inquire more closely into these, with the aim of demonstrating the following theorem:

Whenever $P \neq 0,$ the determinant of the function $Z$  cannot be identically $=0.$

13.
The proof of this theorem would indeed be very easy if we were allowed to assume that $Y$ can be resolved into simple factors

$$(x-A)(x-B)(x-C)(x-D) \ldots$$

Then it would also be certain that $Z$ is the product of all $u-(A+B) x+A B,$  and that the determinant of the function $Z$  is the product of the differences between pairs of the quantities

$$\begin{array}{c} (A+B) x-A B \\ (A+C) x-A C \\ (A+D) x-A D \\ \text{etc.} \\ (B+C) x-B C \\ (B+D) x-B D \\ \text{etc.} \\ (C+D) x-C D \\ \qquad \text{etc. etc.}\\ \end{array}$$

However, this product cannot identically vanish unless one of the factors becomes identically $=0.$ It would then follow that two of the quantities $A,$  $B,$  $C$  etc. would be equal, and consequently, the determinant $P$  of the function $Y$  would be $=0,$  contrary to the hypothesis.

Setting aside such reasoning, which is clearly begging the question (just as in art. 6), we now proceed to a rigorous demonstration of the theorem of art. 12.

14.
The determinant of the function $\zeta$ will be the product of all differences between pairs $(a+b) x-a b,$  the number of which is:

$$\frac{1}{2} m(m-1) \left(\frac{1}{2} m(m-1)-1\right)=\frac{1}{4}(m+1) m(m-1)(m-2)$$

This number represents the order of the determinant of the function $\zeta$ with respect to the indeterminate $x.$  The determinant of the function $z$  will indeed be of the same order. However, the determinant of the function $Z$ may belong to a lower order in the case where the coefficients of some of the highest powers of $x$  vanish. It is now our task to demonstrate that all coefficients in the determinant of the function $Z$ cannot vanish.

Upon closer examination of these differences, whose product is the determinant of the function $\zeta,$ we will find that some of them (namely, the differences between pairs of $(a+b) x-ab$  which have a common element) will provide

the product of all $(a-b)(x-c)$

while others (namely, the differences between pairs $(a+b) x-a b,$ whose elements are different) will result in:

the product of all $(a+b-c-d) x-a b+c d$ excluding repetitions.

Each factor in the former product will appear $m-2$ times, and each factor $x-c$  will appear $(m-1)(m-2)$  times. Therefore, we conclude that this product becomes:

$$=\pi^{m-2} \upsilon^{(m-1)(m-2)}$$

If we denote this latter product by $\rho,$ the determinant of the function $\zeta$  will be:

$$=\pi^{m-2} \upsilon^{(m-1)(m-2)} \rho$$

Furthermore, if we let $r$ denote the function of the indeterminates $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which is transformed into $\rho$  by the substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., and by ${R}$  the function of only $x,$  into which $r$  is transformed by the substitutions $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc., it is clear that the determinant of the function $z$  becomes

$$=p^{m-2} y^{(m-1)(m-2)} r$$

and the determinant of the function $Z$ becomes

$$=P^{m-2} Y^{(m-1)(m-2)} R$$

Therefore, since $P$ is not $=0$  by hypothesis, it remains to be demonstrated that $R$  cannot vanish identically.

15.
To this end, we introduce another indeterminate $w,$ and consider the product of all

$$(a+b-c-d) w+(a-c)(a-d)$$

excluding repetitions. Since this product involves $a,$ $b,$  $c$  etc. symmetrically, it can be represented as an integral function of the indeterminates $w,$  $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc. We will denote this function as $f (w, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}).$  The number of these factors $(a+b-c-d) w+(a-c)(a-d)$  is

$$\frac{1}{2} m(m-1)(m-2)(m-3)$$

from which we easily deduce that

$$f (0, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})=\pi^{(m-2)(m-3)}$$

and therefore also

$$f (0, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.})=p^{(m-2)(m-3)}$$

and finally

$$f (0, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})=P^{(m-2)(m-3)}$$

Generally speaking, the function $f (w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.})$ will be of order

$$\frac{1}{2} m(m-1)(m-2)(m-3)$$

However, in special cases, it may belong to a lower order, if it happens that some coefficients vanish from the highest power of $w.$ Nevertheless, it is impossible for that function to be identically $=0,$  since the equation just found demonstrates that at least the final term of the function does not vanish. Let us take the highest term of the function $f (w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$ which indeed has a non-vanishing coefficient, to be $N w^{\nu}.$  If we substitute $w=x-a,$  it is clear that $f (x-a, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.})$  is an integral function of the indeterminates $x,$  $a,$  or in other words, it is a function of $x$  with coefficients depending on the indeterminate $a,$  in such a way that the highest term is $N x^{\nu},$  and thus it has a coefficient determined by $a$  that is not $=0.$  Likewise, $f (x-b, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$  $f (x-c, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$  etc., will be integral functions of $x,$  whose highest terms are $N x^{\nu},$  while the coefficients of the subsequent terms depend on the indeterminates $a,$  $b,$  $c,$  etc.

Let us now consider the product of the following $m$ factors:

$$\begin{array}{c} f (x-a, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ \text{etc.} \end{array}$$

Since this product is a function of indeterminates $x,$ $a,$  $b,$  $c,$  etc., $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime},$  etc., and it is symmetric with respect to $a,$  $b,$  $c,$  etc., it can be represented as a function of indeterminates $x,$  $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc., $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which we denote by

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.})$$

Therefore,

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})$$

will be the product of the factors

$$\begin{array}{c} f (x-a, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ \text{etc.} \end{array}$$

and therefore it will be indefinitely divisible by $\rho,$ as it is easy to see that any factor of $\rho$  is involved in some of these factors. Let us therefore set

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})=\rho \psi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})$$

where the character $\psi$ represents an integral function. From this it is easily deduced that

$$\varphi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})=R \psi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})$$

But as we have shown above, the product of factors

$$\begin{array}{c} f (x-a, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ \text{ etc. } \end{array}$$

which will be $=\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})$ will have $N^{m} x^{m v}$  as its highest term; therefore, the same highest term will be present in the function $\varphi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}),$  and hence it cannot be identically $=0.$  Therefore, $R$  cannot be identically $=0,$  and neither can the determinant of the function $Z.$  Q.E.D.

16.
Let $\varphi(u, x)$ denote the product of any number of factors, in which the indeterminates $u, x$ enter linearly only, or equivalently, which are of the form

$$\begin{array}{c} \alpha+\beta u+\gamma x \\ \alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x \\ \alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x \\ \text{ etc. } \end{array}$$

''and let $w$ be another indeterminate. Then the function''

$$\varphi (u+w . \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}, x-w . \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u})=\Omega$$

will be indefinitely divisible by $\varphi(u, x).$

Proof. Setting

$$\begin{aligned} \varphi(u, x)= & (\alpha+\beta u+\gamma x) Q \\ = & (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x) Q^{\prime} \\ = & (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x) Q^{\prime \prime} \\ & \text{ etc. } \end{aligned}$$

where $Q, Q^{\prime}, Q^{\prime \prime},$ etc., are integral functions of the indeterminates $u, x, \alpha, \beta, \gamma, \alpha^{\prime}, \beta^{\prime}, \gamma^{\prime},$  $\alpha^{\prime \prime}, \beta^{\prime \prime}, \gamma^{\prime \prime},$  etc., we have

$$\begin{aligned} \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x} & =\gamma Q+(\alpha+\beta u+\gamma x). \frac{\operatorname{d} Q}{\operatorname{d} x} \\ & =\gamma^{\prime} Q^{\prime}+ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x). \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x} \\ & =\gamma^{\prime \prime} Q^{\prime \prime}+ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x). \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x} \\ & \text{ etc. } \\ \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} & =\beta Q+(\alpha+\beta u+\gamma x). \frac{\operatorname{d} Q}{\operatorname{d} u} \\ & =\beta^{\prime} Q^{\prime}+ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x). \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u} \\ & =\beta^{\prime \prime} Q^{\prime \prime}+ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x). \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u} \\ & \text{ etc.} \end{aligned}$$

Substituting these values into the factors from which the product $Q$ is formed, namely

$\begin{array}{c} \alpha+\beta u+\gamma x+\beta w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x+\beta^{\prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma^{\prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x+\beta^{\prime \prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma^{\prime \prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \end{array}$ etc. resp.

yields the following values:

$\begin{array}{c} (\alpha+\beta u+\gamma x) \left(1+\beta w . \frac{\operatorname{d} Q}{\operatorname{d} x}-\gamma w . \frac{\operatorname{d} Q}{\operatorname{d} u}\right) \\ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x) \left(1+\beta^{\prime} w . \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x}-\gamma^{\prime} w . \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u}\right) \\ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x) \left(1+\beta^{\prime \prime} w . \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x}-\gamma^{\prime \prime} w . \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u}\right) \\ \end{array}$ etc.|undefined

Therefore, $\Omega$ will be the product of $\varphi(u, x)$  with the factors

$$\begin{aligned} & 1+\beta w. \frac{\operatorname{d} Q}{\operatorname{d} x}-\gamma w. \frac{\operatorname{d} Q}{\operatorname{d} u} \\ & 1+\beta^{\prime} w. \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x}-\gamma^{\prime} w. \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u} \\ & 1+\beta^{\prime \prime} w. \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x}-\gamma^{\prime \prime} w. \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u} \end{aligned}$$

etc., i.e. of $\varphi(u, x)$ with an integral function of the indeterminates $u,$  $x,$  $w,$  $\alpha,$  $b,$  $\gamma,$  $\alpha^{\prime},$  $b^{\prime},$  $\gamma^{\prime},$  $\alpha^{\prime \prime},$  $b^{\prime \prime},$  $\gamma^{\prime \prime}$  etc. Q.E.D.

17.
The theorem of the previous article clearly applies to the function $\zeta,$ which we suppose to be given by

$$f (u, x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

so that

$$f (u+w . \frac{\operatorname{d} \zeta}{\operatorname{d} x}, x-w . \frac{\operatorname{d} \zeta}{\operatorname{d} u}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

will be indefinitely divisible by $\zeta.$ We will represent the quotient, which is a function of the indeterminates $u,$  $x,$  $w,$  $a,$  $b,$  $c$  etc., and is symmetric with respect to the same $a,$  $b,$  $c$  etc., by

$$\psi (u, x, w, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

From this, we deduce the identities

$$f (u+w . \frac{\operatorname{d} z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} z}{\operatorname{d} u}, l^{\prime}, l^{\prime \prime}, l^{\prime \prime} \text{ etc.})=z \psi (u, x, w, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})$$

and

$$f (u+w . \frac{\operatorname{d} Z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} Z}{\operatorname{d} u}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime} \text{ etc.})=Z \psi (u, x, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

Therefore, if we simply represent the function $Z$ as ${F}({u}, x),$  so that we have

$$f (u, x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})=F(u, x)$$

we obtain the identity

$$F (u+w . \frac{\operatorname{d} Z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} Z}{\operatorname{d} u})=Z \psi (u, x, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

18.
Thus, if for determinate values $u=U,$ $x=X$  of $u,$  $x,$  we obtain

$$\frac{\operatorname{d} Z}{\operatorname{d} x}=X^{\prime}, \quad \frac{\operatorname{d} Z}{\operatorname{d} u}=U^{\prime}$$

we will have an identity

$$F (U+w {X}^{\prime}, {X}-w U^{\prime})={F}(U, {X}). \psi (U, X, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

Whenever $U^{\prime}$ does not vanish, it will be permissible to establish

$$w=\frac{X-x}{U^{\prime}}$$

and from this we obtain

$$F (U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}, x)=F(U, X). \psi (U, X, \frac{X-x}{U^{\prime}}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

which can also be stated as follows:

 * If we set $u=U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}$ in the function $Z,$  it is transformed into

$$F(U, X). \psi (U, X, \frac{X-x}{U^{\prime}}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

19.
Therefore, since in the case where $P=0$ does not hold, the determinant of the function $Z$  is a function of the indeterminate $x$  which does not vanishing by itself, it is clear that the multitude of determinate values of $x$  through which this determinant can attain the value $0$  will be a finite number, so that infinitely many determinate values of $x$  can be assigned that make this determinant different from $0.$  Let $X$  be such a value of $x$  (which can also be assumed to be real). Then the determinant of the function ${F}(u, X)$ will not be $=0,$  and hence it follows, by theorem II. of art. 6, that the functions

$F(u, X)$ and $ \frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$

cannot have any common divisor. Furthermore, let us suppose that there exists some determinate value of $u,$ say $U$  (whether it be real or imaginary, i.e. of the form $g+h \sqrt{-1}\,$ ), which makes $F(u, X)=0,$  i.e. ${F}(U, X)=0.$  Therefore, $u-U$  will be an indefinite factor of the function ${F}(u, {X}),$  and consequently the function $\frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$  will not be divisible by $u-U.$  Therefore, assuming that this function $\frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$  attains the value $U^{\prime}$  when $u=U,$  it cannot be true that $U^{\prime}=0.$  However, $U^{\prime}$  will clearly be the value of the partial differential quotient $\frac{\operatorname{d} Z}{\operatorname{d} u}$  for $u=U,$  $x={X}.$  Therefore, if we denote the value of the partial differential quotient $\frac{\operatorname{d} Z}{\operatorname{d} x},$  for the same values of $u,$  $x,$  by $X^{\prime},$  then it is evident from what has been demonstrated in the previous article that the function $Z$  will vanish identically upon substituting

$$u=U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}$$

and therefore it will be indefinitely divisible by the factor

$$u+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)$$

Consequently, by setting $u=x x,$ it is evident that ${F}(x x, x)$  will be divisible by

$$x x+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)$$

and therefore it will obtain the value $0$ if $x$  is taken to be the root of the equation

$$x x+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)=0$$

i.e. if we set

$$x=\frac{-X^{\prime} \pm \sqrt{(4 U U^{\prime} U^{\prime}+4 X X^{\prime} U^{\prime}+X^{\prime} X^{\prime})}}{2 U^{\prime}}$$

whose values are known to be either real or of the form $g+h \sqrt{-1}.$

It is easy to show that, for the same values of $x,$ the function $Y$  must also vanish. For clearly $f (x x, x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}\text{ etc.})$ is the product of all $(x-a)(x-b)$  without repetitions, and thus it is $=v^{m-1}.$  Hence it follows automatically that

$$\begin{aligned} & f (x x, x, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})=y^{m-1} \\ & f (x x, x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})=Y^{m-1} \end{aligned}$$

or ${F}(x x, x)=Y^{m-1},$ whose determined value therefore cannot vanish unless the value of $Y$  vanishes simultaneously.

20.
By the aid of the preceding investigations, the solution of the equation $Y=0,$ i.e., the finding of a determined value of $x,$  either real or in the form $g+h \sqrt{-1},$  which satisfies it, has been reduced to the solution of the equation ${F}(u, X)=0,$  if indeed the determinant of the function $Y$  is not $=0.$  It is worth noting that if all coefficients in $Y,$  i.e., the numbers $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime \prime},$  etc., are all real quantities, then all coefficients in $F(u, X)$  will also be real, provided that, as is permissible, a real value is taken for $X.$  The order of the secondary equation $F(u, X)=0$  is $\frac{1}{2} m(m-1){:}$  hence, whenever $m$  is a even number of the form $2^{\mu} k,$  where $k$  is an indefinite odd number, the order of the secondary equation will be of the form $2^{\mu-1} k.$

In the case where the determinant of the function $Y$ is $=0,$  we can find (by art. 10) another function $\mathfrak{Y}$  which divides it, whose determinant is not $=0,$  and whose order is of the form $2^{\nu} k,$  with either $\nu<\mu$  or $\nu=\mu.$  Any solution of the equation $\mathfrak{Y}=0$  will also satisfy the equation $Y=0{:}$  the solution of the equation $\mathfrak{Y}=0$  can then be reduced to the solution of another equation, whose order is of the form $2^{\nu-1} k.$

From these considerations, we infer that the general solution of any equation, whose order is an even number of the form $2^{\mu} k,$ can be reduced to the solution of another equation, whose order is of the form $2^{\mu^{\prime}} k,$  such that $\mu^{\prime}<\mu.$  If this number is even, i.e. if $\mu^{\prime}$  is not $=0,$  the same method can be applied again, and so we can continue until we reach an equation whose order is expressed by an odd number; and the coefficients of this equation will all be real, since all coefficients of the original equation were real. Such an equation of odd order is known to be solvable, and indeed has a real root, whence each preceding equation will also be soluble, either by real roots or by roots of the form $g+h \sqrt{-1}.$

It is therefore evident that any function $Y$ of the form $x^{m}-L^{\prime} x^{m-1}+L^{\prime \prime} x^{m-2}-$  etc., where $L^{\prime},$  $L^{\prime \prime}$  etc., are determined real quantities, involves an indefinite factor $x-A,$  where $A$  is a quantity either real or contained in the form $g+h \sqrt{-1}.$  In the latter case, it is easily seen that $Y$  also obtains the value $0$  upon substituting $x=g-h \sqrt{-1},$  thus it is divisible by $x-(\dot{g}-h \sqrt{-1}),$  and therefore also by the product $x x-2 g x+g g+h h.$  Therefore, any function $Y$  certainly involves an indefinite real factor of the first or second order, and since the same applies again to the quotient, it is evident that $Y$  can be resolved into real factors of the first or second order. This demonstrates the proposition of this commentary.


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Page:Holy Bible Berean Standard Bible.pdf/167

 The First Census of Israel (Numbers 26:1–4)

On the first day of the second month of the second year after the Israelites had come out of the land of Egypt, the LORD spoke to Moses in the Tent of Meeting in the Wilderness of Sinai. He said: “Take a census of the whole congregation of Israel by their clans and families, listing every man by name, one by one.

You and Aaron are to number those who are twenty years of age or older by their divisions—everyone who can serve in Israel’s army. And one man from each tribe, the head of each family, must be there with you.

The Leaders of the Tribes

These are the names of the men who are to assist you:

These men were appointed from the congregation; they were the leaders of the tribes of their fathers, the heads of the clans of Israel.

The Number of Every Tribe

So Moses and Aaron took these men who had been designated by name, and on the first day of the second month they assembled the whole congregation and recorded their ancestry by clans and families, counting one by one the names of those twenty years of age or older, just as the LORD had commanded Moses.

So Moses numbered them in the Wilderness of Sinai:


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Page:The Music of the Spheres.djvu/264

 same guise, and gradually, as the days go by, changing to a half moon and on again through all its phases.

The orbit of Venus is tipped slightly from the plane of the earth's orbit; thus only on very rare occasions does it pass directly in front of the sun. Such a passage is called a transit. These transits occur in pairs—two transits being 8 years apart and the successive pairs being separated by 122 years. The last transit occurred on December 9th, 1874, and December 6th, 1882. They will not occur again until June 8th, 2004, and June 6th, 2012. There is only a slight possibility that some young reader may see this.

When passing before the face of the sun, Venus appears as a round, black spot surrounded by an illuminated ring of atmosphere. The spectroscope has shown that this atmosphere is about the same as the earth's, possibly a little more dense, and, strange to say, there is now strong evidence that water vapor and oxygen are conspicuous by their absence.

The dense atmosphere on Venus causes it to reflect light most beautifully but it also obscures its surface so that it is very difficult to see any markings. From time to time, when Venus passes between the sun and ourselves, it approaches to within 25 million miles of the earth, and even less—which is about 10 million miles closer than Mars comes at its nearest approach—yet we can clearly see the surface of Mars for its atmosphere is very tenuous, while the surface of Venus is almost completely hidden under thick white clouds. At such a time, Venus is also a slender crescent while Mars shows a full round face. The spectroscope combined with observations made upon such surface markings as may be seen in full daylight has, however, enabled us to determine that Venus rotates so slowly upon her axis that one face is always turned toward the sun. Thus the day on Venus must last forever and its night-side has only one long night. It has been conjectured that this condition may cause storms of frightful intensity upon

title: Page:The Music of the Spheres.djvu/264


text:
Page:Bostock v. Clayton County (2020).pdf/20

16 in enacting Title VII or certain expectations about its operation. They warn, too, about consequences that might follow a ruling for the employees. But none of these contentions about what the employers think the law was meant to do, or should do, allow us to ignore the law as it is.

Maybe most intuitively, the employers assert that discrimination on the basis of homosexuality and transgender status aren’t referred to as sex discrimination in ordinary conversation. If asked by a friend (rather than a judge) why they were fired, even today’s plaintiffs would likely respond that it was because they were gay or transgender, not because of sex. According to the employers, that conversational answer, not the statute’s strict terms, should guide our thinking and suffice to defeat any suggestion that the employees now before us were fired because of sex. Cf. (, dissenting); (, dissenting).

But this submission rests on a mistaken understanding of what kind of cause the law is looking for in a Title VII case. In conversation, a speaker is likely to focus on what seems most relevant or informative to the listener. So an employee who has just been fired is likely to identify the primary or most direct cause rather than list literally every but-for cause. To do otherwise would be tiring at best. But these conversational conventions do not control Title VII’s legal analysis, which asks simply whether sex was a but-for cause. In Phillips, for example, a woman who was not hired under the employer’s policy might have told her friends that her application was rejected because she was a mother, or because she had young children. Given that many women could be hired under the policy, it’s unlikely she would say she was not hired because she was a woman. But the Court did not hesitate to recognize that the employer in Phillips discriminated against the plaintiff because of her sex. Sex


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Page:Cicero (Collins 1871).djvu/113

Rh There is anecdote and pleasantry enough in this particular oration; but the scandals of Roman society of that day, into which the defence of Cælius was obliged to enter, are not the most edifying subject for any readers. Cælius was a young man of "equestrian" rank, who had been a kind of ward of Cicero's, and must have given him a good deal of trouble by his profligate habits, if the guardianship was anything more than nominal. But in this particular case the accusation brought against him—of trying to murder an ambassador from Egypt by means of hired assassins, and then to poison the lady who had lent him the money to bribe them with—was probably untrue. Clodia, the lady in question, was the worthy sister of the notorious Clodius, and bore as evil a reputation as it was possible for a woman to bear in the corrupt society of Rome—which is saying a great deal. She is the real mover in the case, though another enemy of Cælius, the son of a man whom he had himself brought to trial for bribery, was the ostensible prosecutor. Cicero, therefore, throughout the whole of his speech, aims the bitter shafts of his wit and eloquence at Clodia. His brilliant invectives against this lady, who was, as he pointedly said, "not only noble but notorious," are not desirable to quote. But the opening of the speech is in the advocate's best style. The trial, it seems, took place on a public holiday, when it was not usual to take any cause unless it were of pressing importance.

"If any spectator be here present, gentlemen, who knows nothing of our laws, our courts of justice, or

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text:
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 350 $29$ Lyndon Baines Johnson, The Vantage Point (New York: Holt, Rinchard, and Winston, 1971), pp. 385–415 passim; Msg, ACSFOR, DA, to JCS, and other addresses, 859314, 11 Apr 68, Mobilization Order, Msg, ACSFOR, DA, to JCS, and other addresses, 860342, 18 Apr 68, sub: Mobilization Station Change, 18 Apr 68, and ORC-OPT, Fact Sheet, sub: 12 Apr 68, Selection of Reserve Components for Mobilization Vietnam, Reserve Component file, DAMH-HSO.

$30$ Tolson, Airmobility, pp. 195–98; Msg, ACSFOR, DA, to CinC, USARPAC, 866236, 28 May 68, sub: Second Airmobile Division, 101st Abn Div file, DAMH-HSO.

$31$ Tolson, Airmobility, p. 197; Msg, ACSFOR, DA, to CinC, USARPAC, 907931, 6 May 69, sub: Activation/Inactivation/ Reorganization of Units in RVN, and GO 577, USARPAC, 1969, both 1st Cav Div file, DAMH-HSO,

$32$ Draft SS, ACSFOR for CofS, undated, sub: Redesignation of 101st Air Cavalry Division, FOR PP FP, and Msg, ACSFOR, DA, to CinC, USARPAC, 877510, 26 Aug 68, sub: Redesignation of Airmobile Division, 101st Abn Diy file, DAMH-HSO.

$33$ Cover Sheet (OPS 28) ACSFOR, 15 Nov 68, sub: Reorganization of the 23d Infantry Division (Americal), FOR OT OR PO, and GO USARV 113, 69, both 23d Inf Div file, DAMH-HSO.

$34$ Richard P. Weinert, The Role of USCONARC in the Army Buildup, FY 1968 (Fort Monroe, Va. U.S. Continental Army Command, 1969), p. 71; Ltr, TAG to CG, CONARC, and other addresses, 7 Mar 67, sub: Activation of STARF Units, 7 Mar 67, AGSD-C (M) (14 Mar 67) ACSFOR, Ltr, TAG to CG, CONARC, 20 Oct 67, sub: Activation of STRAF Units, AGSD-C (M) (16 Oct 67) ACSFOR, and Ltr, TAG to CinC, USARPAC, 2 Oct 67, sub: Activation of Units, AGSD-C (M) (22 Sep 67) ACSFOR, all AG Reference files, DAMH-HSO. In addition to the infantry battalions assigned to divisions and brigades, one nondivisional infantry unit served in Vietnam in a military police role.

$35$ Mahon and Danysh, Infantry, pp. 115–16; Westmoreland, A Soldier Reports, p. 283.

$36$ Mahon and Danysh, Infantry, p. 115; George E. Dexter, "Search and Destroy in Vietnam," Infantry 56 (Jul–Aug 1966): 36–42.

$37$
Starry, Mounted Combat, pp. 142, 234–37; John H. Hay, Jr., Tactical and Materiel Innovations (Washington, D.C.; Government Printing Office, 1974), pp. 111–16; Army Concept Team in Vietnam, Final Report; Optimum Mix of Armored Vehicles for Use in Stability Operations, p. H–20, CGSC Library, Fort Leavenworth, Kans, Although nine mechanized infantry battalions served in Vietnam, only eight served there at any given time. The assets of the 5th Battalion, 60th Infantry, an element of the 9th Infantry Division, were used to organize the 1st Battalion, 16th Infantry, an element of the 1st Infantry Division, in 1968.

$38$
David Ewing Ott, Field Artillery 1964–1973 (Washington, D.C; Government Printing Office, 1972), pp. 168–73.

$39$
USCONARC/ARSTRIKE (Army Strike) "Annual Historical Summary, FY 1969," p. 87.

$40$
Andrew P. O'Meara, "Drawdown for Vietnam: How USAREUR Makes Do." Army 16 (Oct 1966): 73–75+; James H. Polk, "A Changing US Army Europe: Building Combat Capability for Tomorrow," Army 17 (Oct 1967): 65–66+.

$41$
Aviation Requirements for Combat Structures of the Army, Annex II, 1965; TOE 17G, Armored Division, 1966; TOE 37G, Infantry Division (Mechanized), 1966; TOE 1–87G, Division Aviation Support Detachment, 1966.

$42$
Msg, ACSFOR to CinC, USAREUR, DA 737567, 22 Oct 65, sub: Reorganization of USAREUR Division, Division General file, Ltr, TAG to CinC, USAREUR, 12 Apr 66, sub: Change in Status of Units, AGSD-C (M) (12 Apr 66) ACSFOR, both (26 Sep 66), Ltr, TAG to CinC, USAREUR, 3 Oct 66, sub: Activation of US Army Europe Units, AGSD-C-C (M)(26 Sep 66) ACSFOR, AG Reference files, Msg. ACSFOR to CinC, USAREUR, and CG, CONARC, 112158Z Feb 66, sub: Transfer of the 5th Bn 32d Armor, DA 750634, 5th Bn, 32d Armor, file, all DAMH-HSO; TOE 17G, Armored Division, 1966, TOE 37G, Mechanized Infantry Division, 31 Mar 1966; TOE 7–45G, Mechanized Infantry Battalion, 1966; TOE 17–35G, Tank Battalion, 31 Mar 1966; John S. Yakshe, "REDEYE Ground to Air," Infantry 56 (Jul–Aug 1966): 22–23.

$43$
"US, UK, and Germany Conclude Trilateral Talks," Department of State Bulletin 56 (22 Nov 67): 788–89; Richard P. Weinert, USCONARC and Redeployment of Forces from Europe (Fort Monroe, Va. U.S, Continental Army Command, 1969).


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102 our national customs, he will not fail to wonder what can be the atrocious nature of this case, that on a day of national festival and public holiday like this, when all other business in the Forum is suspended, this single trial should be going on; and he will entertain no doubt but that the accused is charged with a crime of such enormity, that if it were not at once taken cognisance of, the constitution itself would be in peril. And if he heard that there was a law which enjoined that in the case of seditious and disloyal citizens who should take up arms to attack the Senate-house, or use violence against the magistrates, or levy war against the commonwealth, inquisition into the matter should be made at once, on the very day;—he would not find fault with such a law: he would only ask the nature of the charge. But when he heard that it was no such atrocious crime, no treasonable attempt, no violent outrage, which formed the subject of this trial, but that a young man of brilliant abilities, hard-working in public life, and of popular character, was here accused by the son of a man whom he had himself once prosecuted, and was still prosecuting, and that all a bad woman's wealth and influence was being used against him,—he might take no exception to the filial zeal of Atratinus; but he would surely say that woman's infamous revenge should be baffled and punished I can excuse Atratinus; as to the other parties, they deserve neither excuse nor forbearance."

It was a strange story, the case for the prosecution, especially as regarded the alleged attempt to poison Clodia. The poison was given to a friend of Cælius,

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Rh he was to give it to some slaves of Clodia whom he was to meet at certain baths frequented by her, and they were in some way to administer it. But the slaves betrayed the secret; and the lady employed certain gay and profligate young men, who were hangers-on of her own, to conceal themselves somewhere in the baths, and pounce upon Cælius's emissary with the poison in his possession. But this scheme was said to have failed. Clodia's detectives had rushed from their place of concealment too soon, and the bearer of the poison escaped. The counsel for the prisoner makes a great point of this.

"Why, 'tis the catastrophe of a stage-play—nay, of a burlesque; when no more artistic solution of the plot can be invented, the hero escapes, the bell rings, and—the curtain falls! For I ask why, when Licinius was there trembling, hesitating, retreating, trying to escape—why that lady's body-guard let him go out of their hands? Were they afraid lest, so many against one, such stout champions against a single helpless man, frightened as he was and fierce as they were, they could not master him? I should like exceedingly to see them, those curled and scented youths, the bosom-friends of this rich and noble lady; those stout men-at-arms who were posted by their she-captain in this ambuscade in the baths. And I should like to ask them how they hid themselves, and where? A bath?—why, it must rather have been a Trojan horse, which bore within its womb this band of invincible heroes who went to war for a woman! I would make them answer this question,—why they, being so many

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18. Calibration Data on General Electric Recording Spectrophotometer. , General Electric Company.

Due to the general interest shown by members of the Optical Society of America in spectrophotometry, it has occurred to the authors that it would be interesting to present calibration data obtained from various General Electric spectrophotometers. Therefore, during the past two years we have recorded rather complete data on the performance of these instruments, which will be presented.

Bibliography:, J. Opt. Soc. Am. and Rev. Sci. Inst. 18, 96 (1929)., J. Opt. Soc. Am. 28, 365 (1938)., J. Opt. Soc. Am. 30, 276 (1940).

19. The Importance of Optically Clean Absorption Cells in the Determination of the Concentration of Dye Solutions. , Massachusetts Institute of Technology.

Spectrophotometric transmission data are often used in determining the strength of dye solutions. Customarily, effects due to the presence of the absorption cell and to the solvent are compensated by placing a duplicate cell filled with undyed solvent in the comparison beam of the photometer. It is the purpose of this paper to investigate the errors in dye strength determinations which may result when this compensation is imperfect. Such imperfect compensation may occur during a series of measurements due to contamination of the outer cell surfaces by fingerprints, spilled dye solution, etc. By the application of differential calculus to Beer’s law, it can be shown that the greatest change in transmission for a given percentage change in dye concentration occurs when the transmission is 37 percent. It can further be shown that the errors arising from imperfect cell compensation can be slightly reduced by using a lower value of transmission. However, it is never possible to reduce the error to negligible proportions by this device. For example, with a pair of cells which, when filled with solvent, yield an apparent transmission factor of 99 percent, the resulting error in ∆C/C is greater than 50 percent even when the sample transmits no more than 15 percent. This leads to the conclusion that the use of concentrated solutions does not excuse the operator from checking the condition of his cells before each measurement. The treatment is extended to show the effect of imperfect cell compensation on data taken with the “five times,” cam available for the Hardy recording spectrophotometer.

20. The Viewing Angle of Reflectometry. , Lamp Development Laboratory, General Electric Company, Nela Park.

Many instruments for the rapid measurement of the reflectance of matte surfaces provide for illumination, or viewing, at an angle of 45°, although it has been known for some time that the angle of equivalence with diffuse illumination is, for most matte surfaces, appreciably greater than 45°. Thus and C. H. Sharp, in discussing the Taylor absolute reflectometer, speak of viewing at 50°, while McNicholas reports; “Thus for matt samples of the kind herein studied (of not extremely low reflectance), one would be quite safe in choosing an angle of observation of 55° and assuming an accuracy of 1 or 2 percent.” It appears not to have been pointed out that there is a priori reason for illuminating at an angle of approximately this value: If diffuse illumination is to be replaced by illumination from a finite number of points, the nearest equivalence will be secured by locating the sources at the corners of a regular polyhedron, i.e., of a regular tetrahedron, octahedron or cube, the number of sources being, respectively, 4, 6 and 8. For an opaque plane surface, the sources behind the plane will not contribute to the illumination, hence may be omitted. If the remaining sources are so located as to make equal angles with the normal to the surface, and if the reflectance of the surface does not vary with azimuth, the normal brightness will, by symmetry, be unchanged if the 2, 3 or 4 sources are replaced by a single source of the total luminous intensity, at the location of one of them. The angle from the normal will be, in each case, that between the threefold and fourfold axes in the cubic crystallographic system. This angle is tan$-1$$$\sqrt{2}=$$54°44′8+″. To the degree of accuracy of the reciprocity principle, the apparent reflectance will be the same for normal illumination and viewing at an angle of 54'44’8",

21. An Extreme Case of the Performance of the Eye versus that of the Spectrophotometer. , E. I. duPont de Nemours and Company.

The addition of small amounts of Crocein Scarlet to Tartrazine was supposed by Draves to be a case where the eye can detect smaller additions than a spectrophotometer; but it was shown by Nutting, using the Hardy spectrophotometer, that the reverse is true if the measurements are not confined to a single wave-length, as was done by Draves. This case is one in which the two spectral absorption curves involved are very different in shape. The opposite extreme, where the curves have similar shapes, includes cases where, at least under industrial working conditions, the spectrophotometer cannot detect as small additions as the eye. This extreme is illustrated by mixtures of Pontacyl Carmines 2B and 6B Conc., which are not chemically identical but have very similar spectral absorption curves. For equal weights of standardized dyes, the long wave portions of the curves for the solutions can be practically superimposed, the short wave portions being parallel; somewhat different relations hold for the reflection curves of dyeings. Two very experienced dye testers find that 2.5 percent of the latter dye mixed with the former can be seen to change the color of the skeins, but the mixture would usually be passed as not “off-shade,” while a 5 percent admixture would cause positive rejection. Using the curves of the buffered solutions of the two dyes normally employed by us for spectrophotometric standardization, the hue change due to a 25 percent admixture of the latter dye probably can be detected by inspection of the curves; but the curve due to a 10 percent admixture (with 90


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 330 the 196th had to undergo additional training for combat in Southeast Asia. Also, support units normally attached to separate brigades in Vietnam had to be organized, including signal and military police platoons and chemical, military intelligence, Army Security Agency, military history, and public information detachments, The training process began in June, and by August 1966 the 196th had deployed to Vietnam.

After July 1966 no further increase took place in the number of divisions and brigades until the spring of 1967, In March of that year, responding to Westmoreland's request for additional forces, the Army Staff considered organizing either an infantry or a mechanized infantry brigade for service along the demilitarized zone between North Vietnam and South Vietnam. The original request called for a mechanized infantry brigade with personnel drawn mostly from the 1st and 2d Armored Divisions at Fort Hood. Before the unit could be activated, Westmoreland decided that he needed a standard separate infantry brigade. On 10 May the 198th Infantry Brigade was activated using personnel from the 1st and 2d Armored Divisions. Two days later, at the insistence of General Ralph E. Haines, Vice Chief of Staff and former commander of the 1st Armored Division, the brigade's three infantry battalions and artillery battalion were inactivated and replaced with units taken from regiments assigned to the 1st and 2d Armored Divisions, In turn, new battalions from those regiments were activated to replace the units taken from the two divisions.

Westmoreland's plan to use the 198th along the demilitarized zone between the two Vietnams went astray, Unable to wait for the brigade to arrive, he established a blocking force in April 1967 with units already in the theater. Designated "Task Force Oregon," it included the 196th Infantry Brigade; the 3d Brigade, 25th Infantry Division; and the 1st Brigade, 101st Airborne Division.

In August 1967 a complex organizational exchange took place in Vietnam due in large part to the awkward location of units in relation to their parent divisions. Both the 4th and the 25th Infantry Divisions had "orphan" brigades that operated outside their parent division's areas. To correct the problem, Headquarters and Headquarters Company, 3d Brigade, 4th Infantry Division, and the brigade's non-color-bearing elements were transferred (less personnel and equipment) from Tay Ninh to Task Force Oregon at Chu Lai; Headquarters and Headquarters Company, 3d Brigade, 25th Infantry Division, and its non-color-bearing units (less personnel and equipment) concurrently joined the 25th Infantry Division at Tay Ninh. The color-bearing units (infantry and artillery battalions) attached to the brigades were relieved from the divisions in place and reassigned. These administrative actions gave the commander of the 25th Infantry Division operational control of his 3d Brigade for the first time in Vietnam. The 3d Brigade, 4th Infantry Division, however, remained under the operational control of Task Force Oregon.