text:
Oxidizing agent

An oxidizing agent (also known as an oxidant, oxidizer, electron recipient, or electron acceptor) is a substance in a redox chemical reaction that gains or "accepts"/"receives" an electron from a (called the, , or ). In other words, an oxidizer is any substance that oxidizes another substance. The oxidation state, which describes the degree of loss of electrons, of the oxidizer decreases while that of the reductant increases; this is expressed by saying that oxidizers "undergo reduction" and "are reduced" while reducers "undergo oxidation" and "are oxidized". Common oxidizing agents are oxygen, hydrogen peroxide, and the halogens.

In one sense, an oxidizing agent is a chemical species that undergoes a chemical reaction in which it gains one or more electrons. In that sense, it is one component in an oxidation–reduction (redox) reaction. In the second sense, an oxidizing agent is a chemical species that transfers electronegative atoms, usually oxygen, to a substrate. Combustion, many explosives, and organic redox reactions involve atom-transfer reactions.

Electron acceptors
Electron acceptors participate in electron-transfer reactions. In this context, the oxidizing agent is called an electron acceptor and the reducing agent is called an electron donor. A classic oxidizing agent is the ferrocenium ion, which accepts an electron to form Fe(C5H5)2. One of the strongest acceptors commercially available is "Magic blue", the radical cation derived from N(C6H4-4-Br)3.

Extensive tabulations of ranking the electron accepting properties of various reagents (redox potentials) are available, see Standard electrode potential (data page).

Atom-transfer reagents
In more common usage, an oxidizing agent transfers oxygen atoms to a substrate. In this context, the oxidizing agent can be called an oxygenation reagent or oxygen-atom transfer (OAT) agent. Examples include (permanganate),  (chromate), OsO4 (osmium tetroxide), and especially  (perchlorate). Notice that these species are all oxides.

In some cases, these oxides can also serve as electron acceptors, as illustrated by the conversion of to ,ie permanganate to manganate.

Common oxidizing agents

 * Oxygen (O2)
 * Ozone (O3)
 * Hydrogen peroxide (H2O2) and other inorganic peroxides, Fenton's reagent
 * Fluorine (F2), chlorine (Cl2), and other halogens
 * Nitric acid (HNO3) and nitrate compounds such as potassium nitrate (KNO3), the oxidizer in black powder
 * Potassium chlorate (KClO3)
 * Peroxydisulfuric acid (H2S2O8)
 * Peroxymonosulfuric acid (H2SO5)
 * Hypochlorite, chlorite, chlorate, perchlorate, and other analogous halogen oxyanions
 * Fluorides of chlorine, bromine, and iodine
 * Hexavalent chromium compounds such as chromic and dichromic acids and chromium trioxide, pyridinium chlorochromate (PCC), and chromate/dichromate compounds such as Sodium dichromate (Na2Cr2O7)
 * Permanganate compounds such as potassium permanganate (KMnO4)
 * Sodium perborate (·)
 * Nitrous oxide (N2O), Nitrogen dioxide/Dinitrogen tetroxide (NO2 / N2O4)
 * Sodium bismuthate (NaBiO3)
 * Cerium (IV) compounds such as ceric ammonium nitrate and ceric sulfate
 * Lead dioxide (PbO2)

Dangerous materials definition
The dangerous goods definition of an oxidizing agent is a substance that can cause or contribute to the combustion of other material. By this definition some materials that are classified as oxidizing agents by analytical chemists are not classified as oxidizing agents in a dangerous materials sense. An example is potassium dichromate, which does not pass the dangerous goods test of an oxidizing agent.

The U.S. Department of Transportation defines oxidizing agents specifically. There are two definitions for oxidizing agents governed under DOT regulations. These two are Class 5; Division 5.1(a)1 and Class 5; Division 5.1(a)2. Division 5.1 "means a material that may, generally by yielding oxygen, cause or enhance the combustion of other materials." Division 5.(a)1 of the DOT code applies to solid oxidizers "if, when tested in accordance with the UN Manual of Tests and Criteria (IBR, see § 171.7 of this subchapter), its mean burning time is less than or equal to the burning time of a 3:7 potassium bromate/cellulose mixture." 5.1(a)2 of the DOT code applies to liquid oxidizers "if, when tested in accordance with the UN Manual of Tests and Criteria, it spontaneously ignites or its mean time for a pressure rise from 690 kPa to 2070 kPa gauge is less than the time of a 1:1 nitric acid (65 percent)/cellulose mixture."

title: Oxidizing agent


text:
Talk:Edward Everett Horton

He may have been a sissy, but he was no dummy. He bought up property in the San Fernando Valley starting in the 1920s. He developed what he named Beleigh Acres, a 23-acre compound where he lived with his longtime partner, handsome, silver-tongued actor Gavin Gordon, and Horton’s mother, who stubbornly hung around until she was 102-years-old. Gordon and Horton met in a stage production of Coward’s Private Lives in 1934. They appeared in only one film together, the sweetly sentimental Pocketful Of Miracles (1961), alongside Bette Davis, Ann-Margret and Glenn Ford. They were together 25 years.

 * Can you prove Horton and Gordon slept together, and no one else ever joined them? Others who lived at Beleigh Acres included F. Scott Fitzgerald and Sheilah Graham.  Multiple books by her identify Horton as her landlord.  Possibly all four of them shared a bed every once in a while.  The movie Bob and Carol and Ted and Alice was based on a short story by Fitzgerald.  You also said Gordon was handsome.  He had nothing on my great-grandfather.  And Horton's mother "stubbornly hung around" until she was 102?  Oh, I get it, most mothers would have said, "When my number comes up, I'm going away."Brent Brant (talk) 23:22, 27 September 2023 (UTC)

Untitled
If you are saying that wikipedia should start generally using USA as an adjective to describe citizens of the United States, I think you should discuss it elsewhere before implementing it. I suspect most will disagree with you.

Like it or not, American is the accepted adjective used to describe citizens of the United States, just as Ni-Vanuatu is the accepted adjective used to describe citizens of Vanuatu. I understand some want to reserve "American" to refer to occupants of all of the Americas, but that isn't how it is used today.

On the other hand, if there is something specific about Edward Horton that I don't understand that would lead us to refer to him *poopy — Preceding unsigned comment added by 24.107.2.35 (talk) 19:44, 16 September 2005 (UTC) * as "USA", then I missed that entirely. — Preceding unsigned comment added by 128.193.88.82 (talk) 19:52, 18 February 2003 (UTC)

 * I certainly am aware that people from the USA commonly refer to their nationality as "American". I am not calling that "wrong". I just think that the term "USA" is clearer and more specific-- and not "wrong" either.  This is a bit tangential to Mr. Horton; perhaps the discussion should go to somewhere like Manual of Style (biographies) or WikiProject_Biography. Also, perhaps you'd like to choose a name for yourself and log in.  Cheers, -- Infrogmation 20:43 Feb 18, 2003 (UTC)

Sexual Orientation
Horton's been classified under the category "Gay actors from the United States". However, I've yet to find any proof-- despite some considerable google-fu on my part-- that Horton was gay. A few sites do say that he's gay, but they don't cite any proof, compelling or otherwise. So I'm removing it. If someone does have some proof, feel free to provide it and to revert my edit.Anonymoustom (talk) 23:38, 20 June 2008 (UTC)
 * According to director George Cukor, Horton was part of Hollywood's gay scene, although his statement probably doesn't meet WP:RS. 162.39.194.20 (talk) 22:30, 6 September 2009 (UTC)
 * Cukor died in January 1983 at age 83. People had numerous opportunities to ask him to drop names for dubious purposes. An author named Patrick McGilligan waited until Cukor had been dead for thirty years and then wrote a book claiming that a whole bunch of people who regularly visited his Los Angeles house were gay.  That sleazy type of book does not meet Wikipedia's standards as a RS. Horton's career prospered in an era when movie studio executives forbid talent from discussing their off-camera lives. Let it go.Brent Brant (talk) 23:22, 27 September 2023 (UTC)
 * My account won't meet Wikipedia standards, but as a cousin I can personally vouch that Horton was gay. He was my grandfathers generation and I knew him in the late 1950s and 1960s during the summer at Lake George, NY (in addition to him, other branches of the family including mine owned homes). While publicly and professionally he was discrete, he was fully out to his family and we dined with his friends. But I only have the oral family tradition and my personal knowledge. If there's any direct or contemporaneous proof of his orientation it could be his papers at the University of Wyoming. His orientation doesn't define him but it seems off not to recognize him for who he was (and have that obvious misdirection 1968 quote on his wikipedia page). Rogerfern (talk) 23:56, 19 June 2024 (UTC)

A kind, talented character-actor, and my friend
Mr. Horton was a frequent visitor to Rochester, NY, where he was loved and made us all giggle. I was in my teens and had the delightful chance to play, and lose, at tennis with him  in the indoor court at the Genesee Valley Club. I resent snide remarks about his sexual orientation- - -which is nobody's business, totally beside his talent and deeply in bad taste. I hope Wikipedia is able to filter slander OUT. If not, I have a slew of goodies to contribute to the sleaze faction. —Preceding unsigned comment added by Elettra (talk • contribs) 22:44, 16 August 2008 (UTC)
 * It's interesting that you want to be viewed as open minded but if you were really open minded you wouldn't think that identifying him as gay was snide, slander or "deeply in bad taste." 66.245.66.225 (talk) 23:29, 12 October 2009 (UTC)

You didn't read Elettra's comments correctly. Elettra said the remarks were "deeply in bad taste," not the sex acts. Elettra added they were "nobody's business." I agree. We don't do investigations here at Wikipedia. We only put in articles stuff that comes from legitimate sources. For example, Elettra's memories of Mr. Horton visiting his family in Rochester, New York can't go in the article because they aren't published or broadcast in a legitimate media source. —Preceding unsigned comment added by 130.166.33.92 (talk) 06:12, 27 January 2010 (UTC)
 * I have added a reference to a reliable source regarding his homosexuality. I think it would be a trivial matter to demonstrate that Mr. Horton's sexuality, while obviously a private matter for him in his lifetime, is a notable fact today, and reporting it is neither in bad taste or showing undue weight. He was in his roles an apparent homosexual stereotype, so people have naturally been curious. curious enough that their focus, and the truth of this part of his life, is notable. If we want to debate this, im ok with it, but if we decide to remove it, we MUST remove the categories related to his being gay.Mercurywoodrose (talk) 03:29, 18 May 2010 (UTC)

Your comment that Mr. Horton "was in his roles an apparent homosexual stereotype" is totally POV. It's a generalization about different film roles he played and cartoon characters he voiced over a period of several decades. You might as well say Gracie Allen was using a "girl next door" stereotype as a comic foil for her husband. In truth, neither Mr. Horton nor Ms. Allen included in his/her persona any suggestion of what he/she fantasized about. Suggesting Ms. Allen had a crush on any character besides George Burns is totally POV -- as would be calling her a girl next door or tramp.

Not that there's anything wrong with those POV interpretations of very old movies. But they belong on discussion pages, not in actual articles. This article leads readers to several of Mr. Horton's movies that can let a viewer decide for himself/herself if his character in that film was supposed to be gay. Aren't the DVD's of his films and TV series (like F Troop) substantial sources compared to a user-driven website that was created several decades after his death on which someone tries to dictate how you should interpret his performances?

With Liberace, I understand. But I've seen Edward Everett Horton's work in several Rogers / Astaire musicals and I see him as purely a comic foil -- a lot like Ed Wynn, who never seemed to have a romantic relationship onscreen, either. Rogers and Astaire had pleasant, appealing personalities, but neither could be funny no matter what lines the screenwriter gave them. (After their partnership ended, Astaire kept getting gigs for another forty years, but never in a screwball comedy or on a radio or TV sitcom. It Takes A Thief wasn't supposed to be funny. Rogers switched to drama soon after they parted, even winning an Academy Award for it.)  So the RKO Studios brought in Edward Everett Horton to elevate the non-musical scenes, which would have dragged without him. It was his indignation and "double take" that were funny -- not his refraining from making a pass at Ginger Rogers' character. —Preceding unsigned comment added by 12.101.236.250 (talk) 18:28, 4 July 2010 (UTC)

Who he is named after
Just a guess, but he's named after Edward Everett, the man who spoke before Lincoln's Gettysburg Address? Could he be a relation? I hesitate to ask, but if he is, there's always the chance his unusual voice is alike. Piano non troppo (talk) 06:49, 9 June 2009 (UTC)

Relationship to Edward Everett Hale
I removed an unsourced statement from the Early Life section indicating that Horton may be the grandson of Edward Everett Hale. Horton is named after his own father, whose last name is obviously not Hale. Edward Everett Hale's only daughter who survived to adulthood (Ellen) neither married nor had children. It's possible that he is related in some other way, but many other possibilities exist including that Edward Everett Horton Sr. was named in honor of the famous statesman and orator, Edward Everett. In any case, a reliable source is called for before addition to the article. Richigi (talk) 15:07, 29 August 2012 (UTC)

Trivia
"... Edward Everett Horton, an American comedian who appeared in several Astaire-Rogers films, collected books on tintinnabulation, and confessed to Plummer that three Dominican bellringers were constantly at work in his head." Leo Marks, Between Silk and Cyanide: A Codemaker's War 1941-1945, chap. 15. (This episode takes place at the antiquarian bookshop Marks & Co, Leo Marks being the then-owner's son.)

Who he?
Actor Gavin Gordon, 15 years Horton's junior, appears in the 1961 film Pocketful of Miracles that also features Horton, though they do not appear together in any scenes.
 * This does not seem to relate to anything before or after. Presumably it's a discreet way of hinting that Gavin Gordon was his partner. But it should indicate. Valetude (talk) 13:06, 12 October 2018 (UTC)
 * That was my thought, too. Your inference (that someone is hinting they were partners) didn't occur to me, and I was left wondering why this reference (which has since been edited to indicate that they both appeared in a second film) merited inclusion.  He had a significant part in two films with Cary Grant (having scenes and dialogs with Grant in both of them), so appearing in two films with a far-less well-known actor doesn't seem to merit mention, barring some direct statement as to why that is significant.  In a career as long as Horton's, he was bound to overlap with lots of actors more than once.  I am going to be bold and delete it. Lynn25 (talk) 01:53, 3 January 2019 (UTC)

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

title: Talk:Edward Everett Horton


text:
T'Pau (band)

T'Pau are  an English pop group formed in Shrewsbury, Shropshire, led by singer Carol Decker. They had a string of top 40 hits in the UK during the late 1980s and early 1990s, most notably "China in Your Hand", "Heart and Soul" and "Valentine", and several hits in the rest of Europe, before disbanding in 1992. Decker still performs under the name T'Pau at solo shows and 1980s nostalgia concerts, and in 2013 she reunited with original band member and co-songwriter Ronnie Rogers for a 25th-anniversary UK tour.

History
The band formed in 1986, taking their name from a Vulcan elder of the same name in the sci-fi series Star Trek. Prior to deciding on this name, they were called Talking America on early demos sent to record and publishing companies.

Their first hit was the 1987 release "Heart and Soul". Initially a flop in the UK, it first became a hit in the US Billboard Hot 100 chart, reaching No. 4 after being featured on a Pepe Jeans advertisement; it repeated the feat in the UK Singles Chart some months later. Their third single, "China in Your Hand", was their biggest UK hit; it became the 600th single to reach number 1 in the UK and spent five weeks at the top spot. It also reached the number 1 in several other European countries but made little impact in the United States. Their debut album, Bridge of Spies (simply called T'Pau in the US), also reached number 1 and went quadruple platinum in the UK. The album produced a total of five hit singles including "Valentine", "Sex Talk" (a live recording of an early flop single, "Intimate Strangers"), and "I Will Be with You".

In 1988 they released their second album Rage, which peaked at No. 4 in the UK and reached platinum status there, but failed to come close to the level of success of the band's debut album the year before. It produced the UK top 20 single "Secret Garden", though chart returns were diminishing by this point, and two following singles earned very modest success. The third album The Promise followed in 1991: it peaked at No. 10 in the UK, earning a silver disc. It included one top 20 hit, "Whenever You Need Me", but the band's commercial peak had now passed and they split up after its release. A compilation album, Heart and Soul – The Very Best of T'Pau, was released in 1993 and reached the UK top 40.

Another greatest hits release appeared in 1997, and Decker reformed the band with a new line-up in 1998, for the release of a new studio album, Red. The record was not a commercial success, but T'Pau have continued to perform live on a semi-regular basis. Two singles "With a Little Luck" and "Giving Up the Ghost" were released from the album. Decker still made guest appearances on TV shows throughout the 2000s, including Hit Me, Baby, One More Time, Just the Two of Us and The Weakest Link. In 2007, marking the 20th anniversary of T'Pau's first success, Ron Rogers and Carol Decker released a new single, "Just Dream", exclusively as an internet download. In 2008, T'Pau were part of the Here and Now 1980s nostalgia tour.

To mark the 25th anniversary of the formation of the band, Decker and Rogers embarked on a 28-date UK tour during the spring of 2013 without the original band members.

In February 2015, T'Pau released their fifth studio album, Pleasure & Pain, and embarked on a tour. The final eight shows on the Pleasure & Pain Tour had to be cancelled in early March after Decker suffered damage to her voice while suffering from bronchitis. Around the same time, it emerged that she had also grown frustrated at a lack of airplay for her band's new material, as many radio stations were only interested in playing T'Pau's classic hits. In an interview, she stated, "It's a little harder to get on the radio because all the '80s stations play the '80s stuff and they won't play your new stuff. They actually say they can't, and then the younger stations play the younger artists, the hip stations. That's the downside, and I miss hearing radio plays for the new stuff... it is a little frustrating that I can't get it out to the wider audience anymore." Despite this however the album did manage to edge into the charts at No. 98, their first chart action for more than two decades.

Band members
Current members
 * Carol Decker – vocals, songwriter (1986–1991, 1998–present)
 * Ronnie Rogers – rhythm guitars, songwriter (1986–1991, 2007, 2013–present)

Touring band members
 * Spencer Cozens – keyboards
 * James Ashby – lead guitar
 * Luke Burnet-Smith – bass
 * Dave Hattee – drums

Former members
 * Tim Burgess – drums, percussion (1986–1991)
 * Michael Chetwood – keyboards (1986–1991)
 * Paul Jackson – bass (1986–1991)
 * Taj Wyzgowski – lead guitars (1986–1987)
 * Dean Howard – lead guitars (1987–1991)
 * Jez Ashurst – lead guitars (1998)
 * Dave Hattee – drums, percussion (1998)
 * Dan McKinna – bass (1998)

Live albums

 * 2003: Greatest Hits LIVE (recorded in 1999)
 * 2006: Live in Montreux (recorded in 1998)
 * 2008: Greatest Hits Live and More!

Compilation albums

 * 1993: Heart and Soul: The Very Best of T'Pau
 * 1997: The Greatest Hits
 * 2005: Hits (reissue of The Greatest Hits)
 * 2007: The Essential
 * 2021: Essential T'Pau - 3 CD-compilation including new tracks Run (2018), Be Wonderful (2020) and Guess Who's Sorry Now (2021).

Singles

 * "Intimate Strangers" was retitled "Sex Talk" on Bridge of Spies.

title: T'Pau (band)


text:
Caroline Fox, 1st Baroness Holland

Georgiana Carolina Fox, 1st Baroness Holland, of Holland (27 March 1723 – 24 July 1774), known as Lady Caroline Lennox before 1744 and as Lady Caroline Fox from 1744 to 1762, was the eldest of the Lennox sisters.

Family background
The Lennox sisters were daughters of Charles Lennox, 2nd Duke of Richmond, and Sarah Cadogan. Charles Lennox was the grandson of Charles II of England through the King's relationship with Louise de Kérouaille, Duchess of Portsmouth. In 1744, Lady Caroline eloped with Henry Fox, a politician who was 18 years her senior. Though her parents disapproved of the marriage, it proved a happy one. The couple had four sons, including the Whig politician Charles James Fox and the general Henry Edward Fox.

Their home, Holland House, Kensington, became a social and political gathering place.

Passed over
Lady Caroline's favourite sister, Emily Lennox, married and went to live in Ireland in 1747. In 1750 and 1751, the Lennox sisters' parents died in quick succession, leaving three younger daughters, Louisa, Sarah, and Cecilia, aged eight, six, and one. The 2nd Duke of Richmond, in his will, remembered Lady Caroline's reckless elopement, passed over her and instructed that his three youngest daughters be entrusted to the care of their sister Emily and her husband, James FitzGerald, 20th Earl of Kildare. Lady Caroline resented this slight but hoped to redeem herself by presenting her younger sisters at court and helping them make good matches.

She was present at the celebrations for the Prince of Wales's birthday as there is a reference to her outfit in a letter by Mary Pendarves (became Delany) dated 29 November 1742 which notes that Lady Caroline Lenox wore the finest clothes of "gold and colours on white, embroidered by Mrs Wright".

The Kildares' allegations that the Foxes were responsible for Lady Sarah Lennox's embarrassing rejection by the young King George III, as well as her disastrous marriage to Sir Charles Bunbury soon afterwards, provoked a quarrel between the sisters that was not healed until shortly before Lady Caroline's death.

Baroness
On 3 May 1762, Caroline was created Baroness Holland of Holland in the Peerage of Great Britain. Her husband was created Baron Holland of Foxley less than a year later, on 17 April 1763.

Caroline's two spoilt sons, especially the elder one Stephen, caused her much embarrassment and distress by gambling and falling into debt.

Illness and death
Lady Caroline was already plagued by a mysterious and painful illness when her husband, who had suffered a stroke, died on 1 July 1774. She followed him 23 days later and is buried at All Saints, Farley, Wiltshire.

In media
Aristocrats, a six-part television drama based on Stella Tillyard's biography of Caroline and her sisters, first aired on BBC One in 1999.

title: Caroline Fox, 1st Baroness Holland


text:
Talk:Niels Bohr

'Quantum Physics' philosophy section and source variety
The philosophy section that discusses Bohr's perspectives on quantum physics is an obtuse amount of writing that relies on one source all throughout (other than one that was purely anecdotal.) This needs to be improved - every sentence seemingly ends with [29]. Tea-caff (talk) 06:41, 26 October 2023 (UTC)

 * What source are you recommending? Hawkeye7   (discuss)  09:46, 26 October 2023 (UTC)
 * I don't mean to be destructive. Let me explain. The source predominantly used is by Jan Faye and it is on Stanford Encyclopedia of Philosophy, which, while being a good source, it's not a good practice to use Faye's ideas and nobody else's. (Since its copying one encyclopedia over to another and using their efforts.)
 * I recommend finding the sources that he draws upon as a starting point (in the bibliography.) This will cause readers not to be redirected SEP, but to the actual source that Faye is referring to at many times. Of course, Faye is an expert on the topic, so his books, such as Niels Bohr and the Philosophy of Physics Twenty-First-Century Perspectives would be a better way to appreciate him and include the breadth of his research. Further, it would be best to use other sources that are mentioned briefly in more depth, such as "The transcendental philosophy of Niels Bohr" by J. Honner that opens this section, as upon looking at the source, whoever added it seemingly footnoted the first line of the paper and added nothing else (and the first line of that paper redirects to another footnote. That may cause a footnote 'goose chase' if you get what I mean.)
 * Academics such as L Rosenfeld (Niels Bohr's Contribution to Epistemology,) Aage Peterson (The Philosophy of Niels Bohr) who worked as Bohr's assistant from 1952-1962, and Kristian Camelliri (Bohr, Heisenberg and the divergent views of complementarity) all have good things to say on the topic which would enrich Faye's general overview. (Especially in relation to views on complementarity.) Tea-caff (talk) 01:37, 27 October 2023 (UTC)
 * I have checked out Dugald, Niels Bohr's philosophy of physics; Honner,  The description of nature : Niels Bohr and the philosophy of quantum physics; and Angeloni, Unity and continuity in Niels Bohr's philosophy of physics. I have downloaded the articles by Camelliri and Honner. Philosophy is not my field, but I will read through them and see what I come up with.  Hawkeye7   (discuss)  04:37, 27 October 2023 (UTC)

Wiki Education assignment: Cold War Science
— Assignment last updated by Yoke0003 (talk) 02:50, 2 May 2024 (UTC)

title: Talk:Niels Bohr


text:
IDEN

Integrated Digital Enhanced Network (iDEN) is a mobile telecommunications technology, developed by Motorola, which provides its users the benefits of a trunked radio and a cellular telephone. It was called the first mobile social network by many technology industry analysts. iDEN places more users in a given spectral space, compared to analog cellular and two-way radio systems, by using speech compression and time-division multiple access (TDMA).

History
The iDEN project originally began as MIRS (Motorola Integrated Radio System) in early 1991. The project was a software lab experiment focused on the utilization of discontiguous spectrum for GSM wireless. GSM systems typically require 24 contiguous voice channels, but the original MIRS software platform dynamically selected fragmented channels in the radio frequency (RF) spectrum in such a way that a GSM telecom switch could commence a phone call the same as it would in the contiguous channel scenario.

Operating frequencies
iDEN is designed and licensed to operate on individual frequencies that may not be contiguous. iDEN operates on 25 kHz channels, but only occupies 20 kHz in order to provide interference protection via guard bands. By comparison, TDMA Cellular (Digital AMPS) is licensed in blocks of 30 kHz channels, but each emission occupies 40 kHz, and is capable of serving the same number of subscribers per channel as iDEN. iDEN uses frequency-division duplexing to transmit and receive signals separately, with transmit and receive bands separated by 39 MHz, 45 MHz, or 48 MHz depending on the frequency band being used.

iDEN supports either three or six interconnect users (phone users) per channel, and six dispatch users (push-to-talk users) per channel, using time-division multiple access. The transmit and receive time slots assigned to each user are deliberately offset in time so that a single user never needs to transmit and receive at the same time. This eliminates the need for a duplexer at the mobile end, since time-division duplexing of RF section usage can be performed.

Hardware
The first commercial iDEN handset was Motorola's L3000, which was released in 1994. Lingo, which stands for Link People on the Go, was used as a logo for its earlier handsets. Most modern iDEN handsets use SIM cards, similar to, but incompatible with GSM handsets' SIM cards. Early iDEN models such as the i1000plus stored all subscriber information inside the handset itself, requiring the data to be downloaded and transferred should the subscriber want to switch handsets. Newer handsets using SIM technology make upgrading or changing handsets as easy as swapping the SIM card. Four different sized SIM cards exist, "Endeavor" SIMs are used only with the i2000 without data, "Condor" SIMs are used with the two-digit models (i95cl, for example) using a SIM with less memory than the three-digit models (i730, i860), "Falcon" SIMs are used in the three-digit phones, (i530, i710) and will read the smaller SIM for backward compatibility, but some advanced features such as extra contact information is not supported by the older SIM cards. There is also the "Falcon 128" SIM, which is the same as the original "Falcon", but doubled in memory size, which is used on new 3 digit phones (i560, i930). The interconnect-side of the iDEN network uses GSM signalling for call set-up and mobility management, with the Abis protocol stack modified to support iDEN's additional features. Motorola has named this modified stack 'Mobis'.

Each base site requires precise timing and location information to synchronize data across the network. To obtain and maintain this information each base site uses GPS satellites to receive a precise timing reference.

WiDEN
Wideband Integrated Digital Enhanced Network, or WiDEN, is a software upgrade developed by Motorola and partners for its iDEN enhanced specialized mobile radio (or ESMR) wireless telephony protocol. WiDEN allows compatible subscriber units to communicate across four 25 kHz channels combined, for up to 100 kbit/s of bandwidth. The protocol is generally considered a 2.5G wireless cellular technology.

History
iDEN, the platform which WiDEN upgrades, and the protocol on which it is based, was originally introduced by Motorola in 1993, and launched as a commercial network by Nextel in the United States in September 1996.

WiDEN was originally anticipated to be a major stepping stone for United States wireless telephone provider Nextel Communications and its affiliate, Nextel Partners. However, beginning with the December 2004 announcement of the Sprint Nextel merger, Nextel's iDEN network was abandoned in favor of Sprint's CDMA network. WiDEN was deactivated on the NEXTEL National Network in October 2005 when rebanding efforts in the 800 MHz band began in an effort to utilize those data channels as a way to handle more cellular phone call traffic on the NEXTEL iDEN network. The original Nextel iDEN network was finally decommissioned by Sprint on June 30, 2013 and the spectrum refarmed for use in the Sprint LTE network.

Subscriber Units
The first WiDEN-compatible device to be released was the Motorola iM240 PC card which allows raw data speeds up to 60 kbit/s. The first WiDEN-compatible telephones are the Motorola i850 and i760, which were released mid-summer 2005. The recent i850/i760 Software Upgrade enables WiDEN on both of these phones. The commercial launch of WiDEN came with the release of the Motorola i870 on 31 October 2005, however, most people never got to experience the WiDEN capability in their handsets. WiDEN is also offered in the i930/i920 Smartphone, however, Sprint shipped these units with WiDEN service disabled. Many in the cellular forum communities have found ways using Motorola's own RSS software to activate it. WiDEN was available in most places on Nextel's National Network. As stated above, it no longer is enabled on the Sprint-controlled towers. Since the Sprint Nextel merger the company determined that because Sprint's CDMA network was already 3G and going to EVDO (broadband speeds), and then EVDO Rev A, it would be redundant to keep upgrading the iDEN data network. WiDEN is considered a 2.5G technology.

Operators
Countries which had iDEN networks included United States of America, Canada, Mexico, Brazil, Colombia, Argentina, Peru, Chile, Puerto Rico, Costa Rica, India, Korea, Jordan, Guam, China, Philippines, Thailand, Israel, Singapore, Saudi Arabia, El Salvador, and Guatemala.

 * Sprint Nextel provided iDEN service across the United States until its iDEN network was decommissioned for additional LTE network capacity on 30 June 2013.
 * SouthernLINC Wireless provided iDEN service across the United States until its iDEN network was decommissioned for additional LTE network capacity on 1 April 2019.
 * Telus provided iDEN service under the Mike brand across most of Canada until its iDEN network was decommissioned on 29 January 2016.
 * Nextel Brazil provided iDEN service in Brazil until its iDEN network was decommissioned on 31 March 2018.
 * Nextel Argentina also provided iDEN service until decommissioning on 30 June 2019.
 * Colombian Avantel deactivated its iDEN service at the end of 2021.
 * Nextel Mexico provided iDEN service in Mexico until its iDEN network was decommissioned in 2017. The Mexican Nextel assets were purchased by AT&T in 2015 along with Iusacell the same year to form the nucleus and the revival of AT&T Mexico. AT&T gradually transitioned users from the previous CDMA and iDEN networks to AWS 3G and 4G LTE networks beginning in 2017.

Capitalization and pronunciation
Motorola originally referred to the platform as wiDEN, choosing to capitalize only the letters representing "Digital Enhanced Network," as it had with iDEN. However, subsequent promotion from Motorola and Nextel has indicated that the preferred capitalization is WiDEN.

The term has been pronounced, commonly, as a close combination to the words "why" and "den", or simply as the word "widen". The former is closer to the original pronunciation of iDEN, as "eye" and "den".

title: IDEN


text:
Reagent

In chemistry, a reagent or analytical reagent is a substance or compound added to a system to cause a chemical reaction, or test if one occurs. The terms reactant and reagent are often used interchangeably, but reactant specifies a substance consumed in the course of a chemical reaction. Solvents, though involved in the reaction mechanism, are usually not called reactants. Similarly, catalysts are not consumed by the reaction, so they are not reactants. In biochemistry, especially in connection with enzyme-catalyzed reactions, the reactants are commonly called substrates.

Organic chemistry
In organic chemistry, the term "reagent" denotes a chemical ingredient (a compound or mixture, typically of inorganic or small organic molecules) introduced to cause the desired transformation of an organic substance. Examples include the Collins reagent, Fenton's reagent, and Grignard reagents.

Analytical chemistry
In analytical chemistry, a reagent is a compound or mixture used to detect the presence or absence of another substance, e.g. by a color change, or to measure the concentration of a substance, e.g. by colorimetry. Examples include Fehling's reagent, Millon's reagent, and Tollens' reagent.

Commercial or laboratory preparations
In commercial or laboratory preparations, reagent-grade designates chemical substances meeting standards of purity that ensure the scientific precision and reliability of chemical analysis, chemical reactions or physical testing. Purity standards for reagents are set by organizations such as ASTM International or the American Chemical Society. For instance, reagent-quality water must have very low levels of impurities such as sodium and chloride ions, silica, and bacteria, as well as a very high electrical resistivity. Laboratory products which are less pure, but still useful and economical for undemanding work, may be designated as technical, practical, or crude grade to distinguish them from reagent versions.

Biology
In the field of biology, the biotechnology revolution in the 1980s grew from the development of reagents that could be used to identify and manipulate the chemical matter in and on cells. These reagents included antibodies (polyclonal and monoclonal), oligomers, all sorts of model organisms and immortalised cell lines, reagents and methods for molecular cloning and DNA replication, and many others.

Tool compounds
Tool compounds are an important class of reagent in biology. They are small molecules or biochemicals like siRNA or antibodies that are known to affect a given biomolecule—for example a drug target—but are unlikely to be useful as drugs themselves, and are often starting points in the drug discovery process.

However, many natural substances are hits in almost any assay in which they are tested, and therefore not useful as tool compounds. Medicinal chemists class them instead as pan-assay interference compounds. One example is curcumin.

title: Reagent


text:
Ramsey's theorem

In combinatorics, Ramsey's theorem, in one of its graph-theoretic forms, states that one will find monochromatic cliques in any edge labelling (with colours) of a sufficiently large complete graph. To demonstrate the theorem for two colours (say, blue and red), let $r$ and $s$ be any two positive integers. Ramsey's theorem states that there exists a least positive integer $R(r, s)$ for which every blue-red edge colouring of the complete graph on $R(r, s)$ vertices contains a blue clique on $r$ vertices or a red clique on $s$ vertices. (Here $R(r, s)$ signifies an integer that depends on both $r$ and $s$.)

Ramsey's theorem is a foundational result in combinatorics. The first version of this result was proved by Frank Ramsey. This initiated the combinatorial theory now called Ramsey theory, that seeks regularity amid disorder: general conditions for the existence of substructures with regular properties. In this application it is a question of the existence of monochromatic subsets, that is, subsets of connected edges of just one colour.

An extension of this theorem applies to any finite number of colours, rather than just two. More precisely, the theorem states that for any given number of colours, $r$, and any given integers $n1, …, nc$, there is a number, $R(n1, …, nc)$, such that if the edges of a complete graph of order $R(n1, …, nc)$ are coloured with $s$ different colours, then for some $c$ between 1 and $c$, it must contain a complete subgraph of order $i$ whose edges are all colour $c$. The special case above has $c = 2$ (and $n1 = r$ and $n2 = s$).

R(3, 3) = 6

Suppose the edges of a complete graph on 6 vertices are coloured red and blue. Pick a vertex, $ni$. There are 5 edges incident to $i$ and so (by the pigeonhole principle) at least 3 of them must be the same colour. Without loss of generality we can assume at least 3 of these edges, connecting the vertex, $v$, to vertices, $v$, $v$ and $r$, are blue. (If not, exchange red and blue in what follows.) If any of the edges, $K5$, $K3$, $(rs)$, are also blue then we have an entirely blue triangle. If not, then those three edges are all red and we have an entirely red triangle. Since this argument works for any colouring, any $(rt)$ contains a monochromatic $(st)$, and therefore $K6$. The popular version of this is called the theorem on friends and strangers.

An alternative proof works by double counting. It goes as follows: Count the number of ordered triples of vertices, $s$, $t$, $x$, such that the edge, $K3$, is red and the edge, $R(3, 3) ≤ 6$, is blue. Firstly, any given vertex will be the middle of either 0 × 5 = 0 (all edges from the vertex are the same colour), 1 × 4 = 4 (four are the same colour, one is the other colour), or 2 × 3 = 6 (three are the same colour, two are the other colour) such triples. Therefore, there are at most 6 × 6 = 36 such triples. Secondly, for any non-monochromatic triangle $(xy)$, there exist precisely two such triples. Therefore, there are at most 18 non-monochromatic triangles. Therefore, at least 2 of the 20 triangles in the $(yz)$ are monochromatic.

Conversely, it is possible to 2-colour a $(xyz)$ without creating any monochromatic $K6$, showing that $K5$. The unique colouring is shown to the right. Thus $K3$.

The task of proving that $R(3, 3) > 5$ was one of the problems of William Lowell Putnam Mathematical Competition in 1953, as well as in the Hungarian Math Olympiad in 1947.

A multicolour example: R(3, 3, 3) = 17
A multicolour Ramsey number is a Ramsey number using 3 or more colours. There are (up to symmetries) only two non-trivial multicolour Ramsey numbers for which the exact value is known, namely $R(3, 3) = 6$ and $R(3, 3) ≤ 6$.

Suppose that we have an edge colouring of a complete graph using 3 colours, red, green and blue. Suppose further that the edge colouring has no monochromatic triangles. Select a vertex $y$. Consider the set of vertices that have a red edge to the vertex $z$. This is called the red neighbourhood of $v$. The red neighbourhood of $v$ cannot contain any red edges, since otherwise there would be a red triangle consisting of the two endpoints of that red edge and the vertex $v$. Thus, the induced edge colouring on the red neighbourhood of $v$ has edges coloured with only two colours, namely green and blue. Since $K16$, the red neighbourhood of $v$ can contain at most 5 vertices. Similarly, the green and blue neighbourhoods of $v$ can contain at most 5 vertices each. Since every vertex, except for $v$ itself, is in one of the red, green or blue neighbourhoods of $v$, the entire complete graph can have at most 1 + 5 + 5 + 5 = 16 vertices. Thus, we have $K3$.

To see that $R(3, 3, 3) = 17$, it suffices to draw an edge colouring on the complete graph on 16 vertices with 3 colours that avoids monochromatic triangles. It turns out that there are exactly two such colourings on $R(3, 3, 4) = 30$, the so-called untwisted and twisted colourings. Both colourings are shown in the figures to the right, with the untwisted colouring on the left, and the twisted colouring on the right.

If we select any colour of either the untwisted or twisted colouring on $R(3, 3) = 6$, and consider the graph whose edges are precisely those edges that have the specified colour, we will get the Clebsch graph.

It is known that there are exactly two edge colourings with 3 colours on $R(3, 3, 3) ≤ 17$ that avoid monochromatic triangles, which can be constructed by deleting any vertex from the untwisted and twisted colourings on $R(3, 3, 3) = 17$, respectively.

It is also known that there are exactly 115 edge colourings with 3 colours on $K16$ that avoid monochromatic triangles, provided that we consider edge colourings that differ by a permutation of the colours as being the same.

2-colour case
The theorem for the 2-colour case can be proved by induction on $K16$. It is clear from the definition that for all $v$, $K15$. This starts the induction. We prove that $K16$ exists by finding an explicit bound for it. By the inductive hypothesis $K14$ and $r + s$ exist.

 * Lemma 1. $$R(r, s) \leq R(r-1, s) + R(r, s-1).$$

Proof. Consider a complete graph on $R(n, 2) = R(2, n) = n$ vertices whose edges are coloured with two colours. Pick a vertex $v$ from the graph, and partition the remaining vertices into two sets $n$ and $v$, such that for every vertex $M$, $N$ is in $w$ if edge $R(r, s)$ is blue, and $w$ is in $M$ if $R(r − 1, s)$ is red. Because the graph has $$R(r-1,s) + R(r,s-1) = |M| + |N| + 1$$ vertices, it follows that either $$|M| \geq R(r-1, s)$$ or $$|N| \geq R(r, s-1).$$ In the former case, if $w$ has a red $N$ then so does the original graph and we are finished. Otherwise $M$ has a blue $R(r, s − 1)$ and so $$M \cup \{ v \}$$ has a blue $Ks$ by the definition of $M$. The latter case is analogous. Thus the claim is true and we have completed the proof for 2 colours.

In this 2-colour case, if $R(r − 1, s) + R(r, s − 1)$ and $(vw)$ are both even, the induction inequality can be strengthened to:
 * $$R(r,s) \leq R(r-1,s) + R(r,s-1)-1.$$

Proof. Suppose $(vw)$ and $Kr − 1$ are both even. Let $R(r − 1, s)$ and consider a two-coloured graph of $Kr$ vertices. If $M$ is the degree of the $t$-th vertex in the blue subgraph, then by the Handshaking lemma, $$\textstyle \sum_{i=1}^t d_i $$ is even. Given that $di$ is odd, there must be an even $i$. Assume without loss of generality that $R(r, s − 1)$ is even, and that $t$ and $di$ are the vertices incident to vertex 1 in the blue and red subgraphs, respectively. Then both $$|M|= d_1$$ and $$|N|= t-1-d_1$$ are even. By the Pigeonhole principle, either $$|M| \geq p-1,$$ or $$|N| \geq q.$$ Since $$|M|$$ is even and $p = R(r − 1, s)$ is odd, the first inequality can be strengthened, so either $$|M| \geq p$$ or $$|N| \geq q.$$ Suppose $$|M| \geq p = R(r-1, s).$$ Then either the $M$ subgraph has a red $N$ and the proof is complete, or it has a blue $q = R(r, s − 1)$ which along with vertex 1 makes a blue $M$. The case $$|N| \geq q = R(r, s-1)$$ is treated similarly.

Case of more colours
Lemma 2. If $t = p + q − 1$, then $$R(n_1, \dots, n_c) \leq R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c)).$$

Proof. Consider a complete graph of $$R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c))$$ vertices and colour its edges with $Ks$ colours. Now 'go colour-blind' and pretend that $d1$ and $Kr$ are the same colour. Thus the graph is now $p – 1$-coloured. Due to the definition of $$R(n_1, \dots, n_{c-2}, R(n_{c-1}, n_c)),$$ such a graph contains either a $c$ mono-chromatically coloured with colour $c$ for some $Kr – 1$ or a $c > 2$-coloured in the 'blurred colour'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of $c − 1$ we must have either a $(c − 1)$-monochrome $1 ≤ i ≤ c − 2$ or a $Kni$-monochrome $i$. In either case the proof is complete.

Lemma 1 implies that any $KR(nc − 1, nc)$ is finite. The right hand side of the inequality in Lemma 2 expresses a Ramsey number for $c$ colours  in terms of Ramsey numbers for fewer colours. Therefore, any $R(nc − 1, nc)$ is finite for any number of colours. This proves the theorem.

Ramsey numbers
The numbers $(c − 1)$ in Ramsey's theorem (and their extensions to more than two colours) are known as Ramsey numbers. The Ramsey number $Knc − 1$ gives the solution to the party problem, which asks the minimum number of guests, $R(r,s)$, that must be invited so that at least $Knc$ will know each other or at least $c$ will not know each other. In the language of graph theory, the Ramsey number is the minimum number of vertices, $R(n1, …, nc)$, such that all undirected simple graphs of order $m$, contain a clique of order $n$, or an independent set of order $v$. Ramsey's theorem states that such a number exists for all $m$ and $n$.

By symmetry, it is true that $R(r, s)$. An upper bound for $R(m, n)$ can be extracted from the proof of the theorem, and other arguments give lower bounds. (The first exponential lower bound was obtained by Paul Erdős using the probabilistic method.) However, there is a vast gap between the tightest lower bounds and the tightest upper bounds. There are also very few numbers $m$ and $n$ for which we know the exact value of $R(m, n)$.

Computing a lower bound $r$ for $v = R(m, n)$ usually requires exhibiting a blue/red colouring of the graph $R(m, n) = R(n, m)$ with no blue $s$ subgraph and no red $L$ subgraph. Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs. Upper bounds are often considerably more difficult to establish: one either has to check all possible colourings to confirm the absence of a counterexample, or to present a mathematical argument for its absence.

Computational complexity
"Erdős asks us to imagine an alien force, vastly more powerful than us, landing on Earth and demanding the value of $R(r, s)$ or they will destroy our planet. In that case, he claims, we should marshal all our computers and all our mathematicians and attempt to find the value. But suppose, instead, that they ask for $R(r, s)$. In that case, he believes, we should attempt to destroy the aliens."

A sophisticated computer program does not need to look at all colourings individually in order to eliminate all of them; nevertheless it is a very difficult computational task that existing software can only manage on small sizes. Each complete graph $Kr$ has $R(r, s)$ edges, so there would be a total of $KL−1$ graphs to search through (for $R(5, 5)$ colours) if brute force is used. Therefore, the complexity for searching all possible graphs (via brute force) is $R(6, 6)$ for $Ks$ colourings and at most $Kn$ nodes.

The situation is unlikely to improve with the advent of quantum computers. One of the best-known searching algorithms for unstructured datasets exhibits only a quadratic speedup (c.f. Grover's algorithm) relative to classical computers, so that the computation time is still exponential in the number of nodes.

Known values
As described above, $1⁄2n(n − 1)$. It is easy to prove that $cn(n-1)/2$, and, more generally, that $c$ for all $c$: a graph on $O(cn2)$ nodes with all edges coloured red serves as a counterexample and proves that $R(3, 3) = 6$; among colourings of a graph on $n$ nodes, the colouring with all edges coloured red contains a $s$-node red subgraph, and all other colourings contain a 2-node blue subgraph (that is, a pair of nodes connected with a blue edge.)

Using induction inequalities and the handshaking lemma, it can be concluded that $R(4, 2) = 4$, and therefore $R(s, 2) = s$. There are only two $s − 1$ graphs (that is, 2-colourings of a complete graph on 16 nodes without 4-node red or blue complete subgraphs) among 6.4 × 10$s$ different 2-colourings of 16-node graphs, and only one $R(s, 2) ≥ s$ graph (the Paley graph of order 17) among 2.46 × 10$s$ colourings. (This was proven by Evans, Pulham and Sheehan in 1979.) It follows that $R(4, 3) ≤ R(4, 2) + R(3, 3) − 1 = 9$.

The fact that $R(4, 4) ≤ R(4, 3) + R(3, 4) ≤ 18$ was first established by Brendan McKay and Stanisław Radziszowski in 1995.

The exact value of $(4, 4, 16)$ is unknown, although it is known to lie between 43 (Geoffrey Exoo (1989) ) and 48 (Angeltveit and McKay (2017) ), inclusive.

In 1997, McKay, Radziszowski and Exoo employed computer-assisted graph generation methods to conjecture that $(4, 4, 17)$. They were able to construct exactly 656 $R(4, 4) = 18$ graphs, arriving at the same set of graphs through different routes. None of the 656 graphs can be extended to a $R(4, 5) = 25$ graph.

For $R(5, 5)$ with $R(5, 5)$, only weak bounds are available. Lower bounds for $R(5, 5) = 43$ and $(5, 5, 42)$ have not been improved since 1965 and 1972, respectively.

$(5, 5, 43)$ with $R(r, s)$ are shown in the table below. Where the exact value is unknown, the table lists the best known bounds. $r, s > 5$ with $R(6, 6)$ are given by $R(8, 8)$ and $R(r, s)$ for all values of $22$.

The standard survey on the development of Ramsey number research is the Dynamic Survey 1 of the Electronic Journal of Combinatorics, by Stanisław Radziszowski, which is periodically updated. Where not cited otherwise, entries in the table below are taken from the January 2021 edition. (Note there is a trivial symmetry across the diagonal since $r, s ≤ 10$.)

Asymptotics
The inequality $R(r, s)$ may be applied inductively to prove that
 * $$R(r, s) \leq \binom{r + s - 2}{r - 1}. $$

In particular, this result, due to Erdős and Szekeres, implies that when $r < 3$,
 * $$R(s, s) \leq (1 + o(1))\frac{4^{s - 1}}{\sqrt{\pi s}}.$$

An exponential lower bound,
 * $$R(s, s) \geq (1 + o(1)) \frac{s}{\sqrt{2} e} 2^{s/2},$$

was given by Erdős in 1947 and was instrumental in his introduction of the probabilistic method. There is a huge gap between these two bounds: for example, for $R(1, s) = 1$, this gives $R(2, s) = s$. Nevertheless, the exponential growth factors of either bound were not improved for a long time, and for the lower bound it still stands at $R(r, s) = R(s, r)$. There is no known explicit construction producing an exponential lower bound. The best known lower and upper bounds for diagonal Ramsey numbers are
 * $$[1 + o(1)] \frac{\sqrt{2} s}{e} 2^{\frac{s}{2}} \leq R(s, s) \leq s^{-(c \log s)/(\log \log s)} 4^s,$$

due to Spencer and Conlon, respectively; a 2023 preprint by Campos, Griffiths, Morris and Sahasrabudhe claims to have made exponential progress using an algorithmic construction relying on a graph structure called a "book", improving the upper bound to
 * $$R(s,s) \leq (4-\varepsilon)^{s}\text{ and } R(s,t) \leq e^{-\delta t+o(s)}\binom{s+t}{t}.$$

with $$\varepsilon=2^{-7}$$ and $$\delta=50^{-1}$$ where it is believed these parameters could be optimized, in particular $$\varepsilon$$.

For the off-diagonal Ramsey numbers $R(r, s)$, it is known that they are of order $R(r, s) ≤ R(r − 1, s) + R(r, s − 1)$; this may be stated equivalently as saying that the smallest possible independence number in an $26$-vertex triangle-free graph is
 * $$\Theta \left (\sqrt{n\log n} \right ).$$

The upper bound for $r = s$ is given by Ajtai, Komlós, and Szemerédi, the lower bound was obtained originally by Kim, and the implicit constant was improved independently by Fiz Pontiveros, Griffiths and Morris, and Bohman and Keevash, by analysing the triangle-free process. Furthermore, studying the more general "$s = 10$-free process" has set the best known asymptotic lower bounds for general off-diagonal Ramsey numbers, $101 ≤ R(10, 10) ≤ 48,620$
 * $$c'_s \frac{t^{\frac{s + 1}{2}}}{(\log t)^{\frac{s + 1}{2} - \frac{1}{s - 2}}} \leq R(s, t) \leq c_s \frac{t^{s - 1}}{(\log t)^{s - 2}}.$$

For $$s=4$$ the bounds become $$c'_s t^{\frac{5}{2}}(\log t)^{-2} \leq R(4, t) \leq c_s t^3(\log t)^{-2}$$, but a 2023 preprint  has improved the lower bound to  $$c'_s t^3(\log t)^{-4}$$ which settles a question of Erdős who offered 250 dollars for a proof that the lower limit has form $$c'_s t^{3}(\log t)^{-d}$$.

Induced Ramsey
There is a less well-known yet interesting analogue of Ramsey's theorem for induced subgraphs. Roughly speaking, instead of finding a monochromatic subgraph, we are now required to find a monochromatic induced subgraph. In this variant, it is no longer sufficient to restrict our focus to complete graphs, since the existence of a complete subgraph does not imply the existence of an induced subgraph. The qualitative statement of the theorem in the next section was first proven independently by Erdős, Hajnal and Pósa, Deuber and Rödl in the 1970s. Since then, there has been much research in obtaining good bounds for induced Ramsey numbers.

Statement
Let $s$ be a graph on $n$ vertices. Then, there exists a graph $H$ such that any coloring of the edges of $n$ using two colors contains a monochromatic induced copy of $G$ (i.e. an induced subgraph of $G$ such that it is isomorphic to $H$ and its edges are monochromatic). The smallest possible number of vertices of $G$ is the induced Ramsey number $√2$.

Sometimes, we also consider the asymmetric version of the problem. We define $R(3, t)$ to be the smallest possible number of vertices of a graph $H$ such that every coloring of the edges of $G$ using only red or blue contains a red induced subgraph of $G$ or blue induced subgraph of $G$.

History and bounds
Similar to Ramsey's theorem, it is unclear a priori whether induced Ramsey numbers exist for every graph $X$. In the early 1970s, Erdős, Hajnal and Pósa, Deuber and Rödl independently proved that this is the case. However, the original proofs gave terrible bounds (e.g. towers of twos) on the induced Ramsey numbers. It is interesting to ask if better bounds can be achieved. In 1974, Paul Erdős conjectured that there exists a constant $Y$ such that every graph $H$ on $c$ vertices satisfies $t2⁄log t$. If this conjecture is true, it would be optimal up to the constant $H$ because the complete graph achieves a lower bound of this form (in fact, it's the same as Ramsey numbers). However, this conjecture is still open as of now.

In 1984, Erdős and Hajnal claimed that they proved the bound
 * $$r_{\text{ind}}(H) \le 2^{2^{k^{1+o(1)}}}.$$

However, that was still far from the exponential bound conjectured by Erdős. It was not until 1998 when a major breakthrough was achieved by Kohayakawa, Prömel and Rödl, who proved the first almost-exponential bound of $R(3, t)$ for some constant $k$. Their approach was to consider a suitable random graph constructed on projective planes and show that it has the desired properties with nonzero probability. The idea of using random graphs on projective planes have also previously been used in studying Ramsey properties with respect to vertex colorings and the induced Ramsey problem on bounded degree graphs $c$.

Kohayakawa, Prömel and Rödl's bound remained the best general bound for a decade. In 2008, Fox and Sudakov provided an explicit construction for induced Ramsey numbers with the same bound. In fact, they showed that every $H$-graph $c$ with small $R(s, t)$ and suitable $H$ contains an induced monochromatic copy of any graph on $G$ vertices in any coloring of edges of $d$ in two colors. In particular, for some constant $k$, the Paley graph on $rind(H)$ vertices is such that all of its edge colorings in two colors contain an induced monochromatic copy of every $G$-vertex graph.

In 2010, Conlon, Fox and Sudakov were able to improve the bound to $rind(X,Y)$, which remains the current best upper bound for general induced Ramsey numbers. Similar to the previous work in 2008, they showed that every $rind(H) ≤ 2ck$-graph $c$ with small $rind(H) ≤ 2ck(log k)2$ and edge density $(n,d,λ)$ contains an induced monochromatic copy of every graph on $k$ vertices in any edge coloring in two colors. Currently, Erdős's conjecture that $λ$ remains open and is one of the important problems in extremal graph theory.

For lower bounds, not much is known in general except for the fact that induced Ramsey numbers must be at least the corresponding Ramsey numbers. Some lower bounds have been obtained for some special cases (see Special Cases).

Special cases
While the general bounds for the induced Ramsey numbers are exponential in the size of the graph, the behaviour is much different on special classes of graphs (in particular, sparse ones). Many of these classes have induced Ramsey numbers polynomial in the number of vertices.

If $G$ is a cycle, path or star on $k$ vertices, it is known that $n ≥ 2ck log2k$ is linear in $H$.

If $k$ is a tree on $k$ vertices, it is known that $rind(H) ≤ 2ck log k$. It is also known that $(n,d,λ)$ is superlinear (i.e. $λ$). Note that this is in contrast to the usual Ramsey numbers, where the Burr–Erdős conjecture (now proven) tells us that $1/2$ is linear (since trees are 1-degenerate).

For graphs $H$ with number of vertices $k$ and bounded degree $rind(H) ≤ 2ck$, it was conjectured that $rind(H)$, for some constant $H$ depending only on $rind(H) = O(k2 log2k)$. This result was first proven by Łuczak and Rödl in 1996, with $rind(H)$ growing as a tower of twos with height $rind(H) = ω(k)$. More reasonable bounds for $r(H)$ were obtained since then. In 2013, Conlon, Fox and Zhao showed using a counting lemma for sparse pseudorandom graphs that $Δ$, where the exponent is best possible up to constant factors.

Generalizations
Similar to Ramsey numbers, we can generalize the notion of induced Ramsey numbers to hypergraphs and multicolor settings.

More colors
We can also generalize the induced Ramsey's theorem to a multicolor setting. For graphs $rind(H) ≤ cnd(Δ)$, define $Δ$ to be the minimum number of vertices in a graph $k$ such that any coloring of the edges of $d$ into $G$ colors contain an induced subgraph isomorphic to $G$ where all edges are colored in the $r$-th color for some $d(Δ)$. Let $O(Δ2)$ ($Hi$ copies of $i$).

It is possible to derive a bound on $d(Δ)$ which is approximately a tower of two of height $rind(H) ≤ cn2Δ+8$ by iteratively applying the bound on the two-color case. The current best known bound is due to Fox and Sudakov, which achieves $H1, H2, …, Hr$, where $q$ is the number of vertices of $H$ and $k$ is a constant depending only on $H$.

Hypergraphs
We can extend the definition of induced Ramsey numbers to $c$-uniform hypergraphs by simply changing the word graph in the statement to hypergraph. Furthermore, we can define the multicolor version of induced Ramsey numbers in the same way as the previous subsection.

Let $q$ be a $d$-uniform hypergraph with $H$ vertices. Define the tower function $rind(H1, H2, …, Hr)$ by letting $1 ≤ i ≤ r$ and for $rind(H;q) := rind(H, H, …, H)$, $rind(H;q)$. Using the hypergraph container method, Conlon, Dellamonica, La Fleur, Rödl and Schacht were able to show that for $~ log q$, $rind(H;q) ≤ 2ck3$ for some constant $d$ depending on only $k$ and $c$. In particular, this result mirrors the best known bound for the usual Ramsey number when $tr(x)$.

Infinite graphs
A further result, also commonly called Ramsey's theorem, applies to infinite graphs. In a context where finite graphs are also being discussed it is often called the "Infinite Ramsey theorem". As intuition provided by the pictorial representation of a graph is diminished when moving from finite to infinite graphs, theorems in this area are usually phrased in set-theoretic terminology.

 * Theorem. Let $$X$$ be some infinite set and colour the elements of $$[X]^n$$ (the subsets of $$X$$ of size $$n$$) in $$c$$ different colours. Then there exists some infinite subset $$M$$ of $$X$$ such that the size $$n$$ subsets of $$M$$ all have the same colour.

Proof: The proof is by induction on $d$, the size of the subsets. For $t1(x) = x$, the statement is equivalent to saying that if you split an infinite set into a finite number of sets, then one of them is infinite. This is evident. Assuming the theorem is true for $i ≥ 1$, we prove it for $ti+1(x) = 2ti(x)$. Given a $q$-colouring of the $d ≥ 3, q ≥ 2$-element subsets of $n$, let $rind(H;q) ≤ td(ck)$ be an element of $c$ and let $d = 3$ We then induce a $X$-colouring of the $X$-element subsets of $c$, by just adding $n = 1$ to each $r$-element subset (to get an $n ≤ r$-element subset of $Y$). By the induction hypothesis, there exists an infinite subset $n = r + 1$ of $r$ such that every $X$-element subset of $(r + 1)$ is coloured the same colour in the induced colouring. Thus there is an element $a0$ and an infinite subset $Y = X \ {a0}.$ such that all the $a0$-element subsets of $Y$ consisting of $(r + 1)$ and $r$ elements of $Y1$ have the same colour. By the same argument, there is an element $Y1$ in $a0$ and an infinite subset $Y1$ of $(r + 1)$ with the same properties. Inductively, we obtain a sequence $a0$ such that the colour of each $Y1$-element subset $a1$ with $Y1$ depends only on the value of $Y2$. Further, there are infinitely many values of $Y1$ such that this colour will be the same. Take these ${a0, a1, a2, …}$'s to get the desired monochromatic set.

A stronger but unbalanced infinite form of Ramsey's theorem for graphs, the Erdős–Dushnik–Miller theorem, states that every infinite graph contains either a countably infinite independent set, or an infinite clique of the same cardinality as the original graph.

Infinite version implies the finite
It is possible to deduce the finite Ramsey theorem from the infinite version by a proof by contradiction. Suppose the finite Ramsey theorem is false. Then there exist integers $X$, $r$, $c$ such that for every integer $n$, there exists a $T$-colouring of $(r + 1)$ without a monochromatic set of size $k$. Let $c$ denote the $T$-colourings of $(ai(1), ai(2), …, ai(r + 1))$ without a monochromatic set of size $Ck$.

For any $c$, the restriction of a colouring in $i(1) < i(2) < … < i(r + 1)$ to $i(1)$ (by ignoring the colour of all sets containing $i(n)$) is a colouring in $T$. Define $k$ to be the colourings in $Ck$ which are restrictions of colourings in $ai(n)$. Since $[k](n)$ is not empty, neither is $C^1_k$.

Similarly, the restriction of any colouring in $Ck$ is in $C^1_k$, allowing one to define $C^1_{k+1}$ as the set of all such restrictions, a non-empty set. Continuing so, define $C^1_k$ for all integers $C^2_k$, $C^m_k$.

Now, for any integer $m$,
 * $$C_k\supseteq C^1_k\supseteq C^2_k\supseteq \cdots$$

and each set is non-empty. Furthermore, $k$ is finite as
 * $$|C_k|\le c^{\frac{k!}{n!(k-n)!}}$$

It follows that the intersection of all of these sets is non-empty, and let
 * $$D_k=C_k\cap C^1_k\cap C^2_k\cap \cdots$$

Then every colouring in $k$ is the restriction of a colouring in $[k](n)$. Therefore, by unrestricting a colouring in $Ck$ to a colouring in $Ck+1$, and continuing doing so, one constructs a colouring of $$\mathbb N^{(n)}$$ without any monochromatic set of size $Dk$. This contradicts the infinite Ramsey theorem.

If a suitable topological viewpoint is taken, this argument becomes a standard compactness argument showing that the infinite version of the theorem implies the finite version.

Hypergraphs
The theorem can also be extended to hypergraphs. An $Dk$-hypergraph is a graph whose "edges" are sets of $T$ vertices – in a normal graph an edge is a set of 2 vertices. The full statement of Ramsey's theorem for hypergraphs is that for any integers $m$ and $m$, and any integers $[k](n)$, there is an integer $k + 1$ such that if the hyperedges of a complete $m$-hypergraph of order $Ck+1$ are coloured with $c$ different colours, then for some $m$ between 1 and $c$, the hypergraph must contain a complete sub-$i$-hypergraph of order $c$ whose hyperedges are all colour $m$. This theorem is usually proved by induction on $ni$, the 'hyper-ness' of the graph. The base case for the proof is $Ck+1$, which is exactly the theorem above.

For $Dk+1$ we know the exact value of one non-trivial Ramsey number, namely $Dk+1$. This fact was established by Brendan McKay and Stanisław Radziszowski in 1991. Additionally, we have: $n1, …, nc$, $R(n1, …, nc; m)$ and $R(n1, …, nc; m)$.

Directed graphs
It is also possible to define Ramsey numbers for directed graphs; these were introduced by. Let $m = 2$ be the smallest number $i$ such that any complete graph with singly directed arcs (also called a "tournament") and with $m = 3$ nodes contains an acyclic (also called "transitive") $m$-node subtournament.

This is the directed-graph analogue of what (above) has been called $R(4, 4; 3) = 13$, the smallest number $Q$ such that any 2-colouring of the edges of a complete undirected graph with $R(4, 5; 3) ≥ 35$ nodes, contains a monochromatic complete graph on n nodes. (The directed analogue of the two possible arc colours is the two directions of the arcs, the analogue of "monochromatic" is "all arc-arrows point the same way"; i.e., "acyclic.")

We have $R(4, 6; 3) ≥ 63$, $R(5, 5; 3) ≥ 88$, $R(n)$, $≥ Q$, $R(n, n; 2)$, $≥ Z$, $R(0) = 0$, and $R(1) = 1$.

Uncountable cardinals
In terms of the partition calculus, Ramsey's theorem can be stated as $$\alef_0\rightarrow(\alef_0)^n_k$$ for all finite n and k. Wacław Sierpiński showed that the Ramsey theorem does not extend to graphs of size $$\alef_1$$ by showing that $$2^{\alef_0}\nrightarrow(\alef_1)^2_2$$. In particular, the continuum hypothesis implies that $$\alef_1\nrightarrow(\alef_1)^2_2$$. Stevo Todorčević showed that in fact in ZFC, $$\alef_1\nrightarrow[\alef_1]^2_{\alef_1}$$, a much stronger statement than $$\alef_1\nrightarrow(\alef_1)^2_2$$. Justin T. Moore has strengthened this result further. On the positive side, a Ramsey cardinal, $$\kappa$$, is a large cardinal axiomatically defined to satisfy the related formula: $$\kappa\rightarrow(\kappa)^{<\omega}_2$$. The existence of Ramsey cardinals cannot be proved in ZFC.

Relationship to the axiom of choice
In reverse mathematics, there is a significant difference in proof strength between the version of Ramsey's theorem for infinite graphs (the case n = 2) and for infinite multigraphs (the case n ≥ 3). The multigraph version of the theorem is equivalent in strength to the arithmetical comprehension axiom, making it part of the subsystem ACA0 of second-order arithmetic, one of the big five subsystems in reverse mathematics. In contrast, by a theorem of David Seetapun, the graph version of the theorem is weaker than ACA0, and (combining Seetapun's result with others) it does not fall into one of the big five subsystems. Over ZF, however, the graph version implies the classical Kőnig's lemma, whereas the converse implication does not hold, since Kőnig's lemma is equivalent to countable choice from finite sets in this setting.

title: Ramsey's theorem


text:
Talk:Jesus and John the Baptist

Ed, you asked about this claim:
 * It has long been a source of consternation to UC members that many Christians impute to the UC the view that Jesus failed, although the UC clearly teaches it was rather John who failed.

I think the reason 'mainstream' Christians impute to the UC the view that Jesus failed, is that the UC teaches that Jesus did not do everything that needed doing, leaving a messiah-type role for Rev. Moon to fill. Historic Christian teaching is that Jesus accomplished everything needed for our salvation; if you say Christ's work is incomplete and we need Rev. Moon as well, that's going to be interpreted as saying that Jesus failed. Maybe you're saying Jesus did everything he could given John the Baptist's failure to support him; even so, it sounds like you're still saying Jesus failed to do everything necessary, just that it wasn't his fault.

Personally, my biggest gripe with the UC on this and most other points of difference is that there was a pretty consistent way of interpreting the life and role of John the Baptist (and Jeus Christ, etc. etc.) for 1,900 years of church history, and then Rev. Moon comes along and says all that is backwards. Putting them side by side, I'm going to stick with all the Christians from the first 1,900 years, rather than massive theological innovations dating from the 20th century (or from the 16th century or the 11th or the 6th, in the case of other innovations). But that's just me. Me and Vincent of Lerins. :-) Wesley 20:27 Feb 18, 2003 (UTC)

Why is this entire article from the Unification Church point of view? I can see a general article on the relationship, with inputs from many different denominations. If we leave it as it is, let's rename it to Unification Church viewpoint on Jesus and John the Baptist. -- Zoe

 * Zoe, just roll it in with Unification Church.

 * Either that, or this needs to be a paragraph or two in the longer article on John the Baptist. (Note, the paragraph above this was not written by me.) Vicki Rosenzweig

AFD
This article was nominated for deletion. The result was merge with Unification Church. Anyone may do so at their leisure. See Articles for deletion/Jesus and John the Baptist · Katefan0(scribble) 04:22, 23 November 2005 (UTC)

CRC
Roles of John the Baptist — Preceding unsigned comment added by 197.231.239.115 (talk) 09:30, 27 April 2022 (UTC)

title: Talk:Jesus and John the Baptist


text:
Talk:Latin/Archive 1

Should this be at "Latin"? "Latin" as a word on its own means "Latin language", as far as I'm aware. -- Oliver P. 20:33 Feb 18, 2003 (UTC)

We seem to be differentiating all languages -- Spanish language vs. Spanish, for example. -- Zoe

 * I think it's best to leave it as it is. "Latin" can mean other things, as in "Latin American". Deb 20:39 Feb 18, 2003 (UTC)

 * But Latin redirects to it anyway... Perhaps that should be a disambiguation page, but it would mean correcting hundreds of pages that link to Latin in the sense of "Latin language", which would be quite a horrible task... How would people feel if we had it at "Latin", but put a disambiguation note at the top, clarifying that the term is also used to mean "Latin American" and whatever else, with links to the relevant other articles? -- Oliver P. 20:46 Feb 18, 2003 (UTC)

 * I've made an attempt at turning "Latin" into a disambiguation page (probably not very complete) along the lines of what's already been done for English, French, etc. -- for consistency's sake. Deb 19:43 Feb 19, 2003 (UTC)

 * No, no, we'll have none of that consistency malarky round here! ;) If we did that for all languages, we'd end up with articles at Esperanto language and Volapük language, too, and I hope no-one is suggesting that we do that... We went through the argument with Sanskrit and with programming languages quite recently, and it was decided that if the name on its own made it clear what was being talked about, the disambiguator "language" or "programming language" wasn't needed. In this case, I admit, it is true that the word "Latin" is sometimes used to mean other things, but Wikipedia is not a dictionary, and I don't think we need to cover every way in which a word may be used in the article for that word. As a topic to be discussed, "Latin" means "Latin language", and nothing else, as far as I'm aware. Of course, "Latin America" is also a topic to be discussed, but I honestly can't imagine anyone looking for that anywhere other than at Latin America, so I don't see the need for a disambiguation page. I hope I'm not causing you offence here, Deb, as your disambiguation page is very nice and all that, but it does just look like a dictionary entry. -- Oliver P. 21:20 Feb 19, 2003 (UTC)

 * Well, yes, but like I said, it's not complete - I think there may be other connotations that have passed me by. After all, the Romans called it "lingua latina", not "latina". Deb 22:25 Feb 19, 2003 (UTC)

 * Erm, but this is the English-language Wikipedia... :) We always called it "Latin" at school, not "Latin language"... -- Oliver P. 22:33 Feb 19, 2003 (UTC)

 * Didn't you do any Latin literature? Deb 22:40 Feb 19, 2003 (UTC)

 * Yup. Latin literature is literature written in Latin. Note that it's not called "Latin language literature", although that's what it means. It's not analogous to "French literature", because French literature (in the usual usage of the term) is literature from France, rather than literature written in the French language. The term "Latin literature" refers to the language of origin, not the country of origin, because "Latin" (in this sense) refers to a language, not a country. -- Oliver P. 22:49 Feb 19, 2003 (UTC)

 * Ah, but by the same token, the language was called Latin because it was spoken by the Latin people. Admittedly the use of the term Latin to refer to the culture is not nearly so common as that of English, Spanish, and the like.  Still, isn't consistency something to aim for?  Note that languages like Esperanto do not have comparable groups to refer to, so aren't the same thing.

 * Hey, where were you during the great English literature debate? Deb 22:54 Feb 19, 2003 (UTC)

 * Nobody was at the great English literature debate. :) Atorpen 22:57 Feb 19, 2003 (UTC)

This is effectively the same argument (reversed) as for British. That's not a disambiguation page because, unlike many other nationalities, there is no confusion with a language. I don't think we need consistency at all costs, different words need different treatment. Personally I would prefer Oliver's suggestion of having "Latin language" as the redirect and a short disambiguation note at the top of "Latin" to cover the use of the word in other phrases. But I think either as the redirect would work. Just to explain my interest - I've been working at removing links to language disambiguation pages and changing all the pages linking to "Latin" would be a lot of work (but I'm prepared to do it if that's the way the consensus ends up :) - sannse 17:19 Feb 20, 2003 (UTC)

 * My point is that the situation isn't analagous, because unlike Esperanto or the like, there is a Latin people that one could be referring to. They don't exist any more, having been absorbed into the Roman Republic fairly early on, and to be fair aren't as often mentioned as the language - we don't have an article on them yet - but decisions of importance are not so straightforward as ones of existance, and I think that ultimately the people should get the priority.  I suppose, though, there's no need to worry about that now.  Maybe what we should do is move the page for the time being, and then consider whether or not to move it back in the event someone adds more discussion about the Latins themselves?

 * OK, I see your point. I always thought of "Latin people" as another way of saying "of Italian descent", I didn't realise it was (or had been) the name of a people in itself.  I think your solution would work well for now, but maybe a better solution would be for someone to write the necessary article so dumbbells like me would know these things ;)  -- sannse 15:18 Feb 21, 2003 (UTC)

It looks as though no one dislikes the current arrangement enough to change it (yet at least), so I thought I'd make a start on changing the links to the Latin disambiguation page tomorrow (there are only 500 ;). It won't affect anything if we do make the change above later of course, as Latin language would be a redirect, but please let me know of any objections to this.

Also, I've left a more general disambiguation notice in the place of the disambiguation of Latinus and Latins to Latin, I wasn't sure how to word it, so please check and change if there is a better way. Thanks -- sannse 21:26 Feb 24, 2003 (UTC)

 * Sorry, I think I missed something by not keeping an eye on "Recent Changes". I noticed that there was an entry for "Latins", which appeared to break the rules for article naming, so I just plunged in and moved the content to the disambiguation page (singular).  But I see it's already been improved, so I assume I haven't offended anyone. (Hope not, anyway.) Deb 21:36 Feb 24, 2003 (UTC)

 * Not me, and I'm sure no one else. You just beat me to finishing the job :) -- sannse 21:42 Feb 24, 2003 (UTC)

Surely the statement 'Latin grammar is not reflected in English' a bit of a contentious one? Olivia Curtis 21:49 Feb 24, 2003 (UTC)

 * Could you elaborate on why you think it is contentious? soulpatch

 * Sorry if I intrude, but there are some obvious parallels, for example the existence, in both languages, of present and past participles (and participle constructions in general). Of course you could argue that this does not affect the surface structure of the English language and, accordingly, should not be mentioned here. --KF 22:08 Feb 24, 2003 (UTC)

 * There are similarities, of course, largely since both are Indo-European languages. But the rest of the paragraph makes it clear that it's just saying that English grammar isn't just a corrupted offspring of Latin grammar, this being a strange idea which some attempts to "improve" English grammar seem to have been based on. -- Oliver P. 22:22 Feb 24, 2003 (UTC)

 * But they failed to really improve the language, didn't they? KF 22:33 Feb 24, 2003 (UTC)

I still object to the page on Latin being at Latin language. What is the purpose of disambiguation pages? As far as I'm aware, their purpose is to catch different people who are linking to a term but with different meanings, to draw their attention to the conflict of meanings, and to allow them to correct (disambiguate) their links as appropriate. The point of disambiguation pages is not to act as dictionary definition pages just listing every conceivable way in which a term is used.

So a disambiguation page is needed at Latin if and only if there are significant numbers of people linking to the term but using it in different senses. So the question is this: Is there a significant number of people linking to Latin, but in a sense other than the language? As far as I can see, there is not. In fact, are there any pages linking to Latin that don't mean the language? I can't find any.

Anyone new to the Wikipedia who didn't know about our strange naming conventions would link to the language by writing " Latin ". I honestly couldn't dream of anyone ever linking any other meaning of "Latin" by writing " Latin ". If they were talking about Latin America, they'd link to Latin America, obviously. And the argument about the Latin people doesn't count for much, because as the above comments make clear, (a) they don't exist as a distinct people any more, and haven't done for several millennia, and (b) most people aren't even aware that there ever was a distinct people of that name. So who's going to accidentally write " Latin ", meaning the people rather than the language?

There are about 500 pages that link to Latin, and only about 300 that link to Latin language. Why not just move Latin language to Latin? It's a lot less work! And it will save everyone work in the long run. Why should we have to type " Latin " instead of " Latin ", every time we want to link to the language? -- Oliver P. 22:05 Feb 24, 2003 (UTC)

P.S. - In any case, disambiguators are better in parentheses (e.g. " Gloop (language) ", for consistency with other disambiguators, and to allow people to use the pipe trick. -- Oliver P. 22:06 Feb 24, 2003 (UTC)

 * I must protest -- I just amended the links to "Latins" (meaning the people), which I had changed to the singular, "Latin". There were about fifteen of them. Deb 22:13 Feb 24, 2003 (UTC)

 * Well done for doing that, Deb! But is fifteen a significant fraction of five hundred? I'm not entirely sure that it is, although I suppose the word "significant" is not very precisely defined... -- Oliver P. 22:22 Feb 24, 2003 (UTC)

OK, so where are we at the moment? The proposals seem to be:

1. Move the text currently at "Latin language" to "Latin" and have a short disambiguation notice at the top, perhaps something like:

As an adjective, Latin can be used to refer to those of extraction from the Romance-speaking peoples of the Mediterranean, such as in terms such as "Latin America" and "Latin lover." For the use of the word Latins as jargon in linguistics see Romance languages. For the ancient people known as the Latins see Latin (people).

"Latin language" would become a redirect. We would have a new page at "Latin (people)", "Latins" would be a redirect to there.

2. Leave "Latin" as a disambiguation page and leave the current text at "Latin language" where it is. Links to "Latin" will have to be disambiguated (but I volunteer to do this if necessary).

3. Leave "Latin" as a disambiguation page and move the text at "Latin language" to "Latin (language)". Again links to "Latin" will have to be disambiguated.

My personal preference is for the first option. I think Oliver's arguments make sense, "Latin" on its own almost invariably means the language and the disambiguation notice would cover all other meanings without interfering with the article on the language. Opinions?

 * I've moved the article to Latin with a disambiguation block to make the links work. I don't have any major preference as to whether it should be at "Latin" or "Latin language", so I'm open to having the main article there if people prefer it. Enchanter

Latin Scholar Needed!
See: Talk:Greek numerical prefixes Leonard G. 14:47, 20 May 2004 (UTC)

title: Talk:Latin/Archive 1