text:
Talk:Yom Kippur War

“Israeli victory”
In the results sections, we see the outcome of two fronts that played out very differently get simplified to “Israeli victory”, implying the Arabs completely failed to achieve every objective (to recapture the eastern bank of the suez and the golan). The outcome on the Syrian front was undeniably an Israeli victory, nobody is challenging that, but reducing the outcome on the Egyptian front to “Israeli victory” is wildly inaccurate. Not only was Egypt able to achieve their goal in the Yom Kippur war, which was to establish a foothold in the eastern bank of the suez, but they were also able to halt the Israeli counterattack and prevent the encirclement of their army at the battles of ismailia and suez. At most it’s a stalemate, not an Israeli victory when most of the occupied suez bank was now under Egyptian control The Great Mule of Eupatoria (talk) 05:37, 23 November 2023 (UTC)

 * Do you have reliable sources to back this up? Also, it is not the job of Wikipedia articles to help you with implications you may be concluding incorrectly. —OuroborosCobra (talk) 05:40, 23 November 2023 (UTC)
 * For Egypt’s claims
 * https://time.com/6322802/yom-kippur-war-israel-history/
 * “ In 1973, Egypt’s goal in crossing the Suez Canal was to force Israel to the negotiation table to make a peace deal and get back control of the Sinai peninsula. According to Avi Shilon, a historian who teaches at Tel-Hai College in Israel, “The Egyptian and the Syrians didn't plan to conquer Israel. They planned to hit Israel and to force Israel to go into negotiations. For them, it was enough to hit Israel to show that they can beat Israel in the first days, and they preferred to stop, so it was easier for Israel to launch a retaliation attack.””
 * This outlines Egypts goal of the war, which was to cross the suez and not conquer Sinai or Israel proper.
 * The war ended with an Israeli counterattack, however it did not restore control over the suez bank and was defeated in the closing battle of Sinai, failing to defeat the Egyptian army in the city
 * Archived citations:
 * As for the implications, it’s not that they’re incorrect but just heavily simplified as it doesn’t even specify what kind of victory Israel attained (It used to say “Israeli military victory”) as it was in no way shape or form a political victory for Israel, and neither could it be a military victory against Egypt, though it is was an undeniable victory in Syria The Great Mule of Eupatoria (talk) 05:57, 23 November 2023 (UTC)
 * I Agree, if we talk Land gains the northern front was an Egyptian victory and the southern was israeli victory so it was a military stalemate Kelcoz (talk) 21:24, 26 March 2024 (UTC)
 * I have been trying for ages but the editors are adamant on labelling this an “Israeli victory” The Great Mule of Eupatoria (talk) 06:36, 20 April 2024 (UTC)
 * So it's Unanimous, in the Suez front the north is Egyptian victory and the south is a partial israeli victory so logically it's a Stalemate, thats excluding the Egyptian political victory Kelcoz (talk) 00:05, 27 May 2024 (UTC)
 * Exactly. Despite all my explanations and several sources including admissions from Israeli military staff, the editors seem adamant to simplify the entire outcome of this war as “Israeli victory” The Great Mule of Eupatoria (talk) 06:20, 29 May 2024 (UTC)
 * So it's Unanimous, in the Suez front the north is Egyptian victory and the south is a partial israeli victory so logically it's a Stalemate, thats excluding the Egyptian political victory Kelcoz (talk) 00:05, 27 May 2024 (UTC)
 * Exactly. Despite all my explanations and several sources including admissions from Israeli military staff, the editors seem adamant to simplify the entire outcome of this war as “Israeli victory” The Great Mule of Eupatoria (talk) 06:20, 29 May 2024 (UTC)

 * The citations seem to refer to tactical/operational results, not the definitive result of the whole war. What page of the source is the "stunning defeat" quote on? (Hohum @ ) 11:48, 24 November 2023 (UTC)

 * Yes, I’m talking about the Egyptian front specifically and their closing victories at suez and Ismailia, not the entire war which included a northeastern front. Im having a hard time specifying the page for the first citation as much of it seems to have been lost in the archive The Great Mule of Eupatoria (talk) 18:47, 24 November 2023 (UTC)
 * Don't be ridiculous. Egypt's military resigned in disgrace after the war. It was a disaster for Egypt. More importantly, your bold change lacks consensus or reliable sources. Andre🚐 22:06, 29 November 2023 (UTC)
 * I think you’re talking about the six day war. I’ve brought several sources to prove that Egypt managed to achieve its goal in the war, and the suggestion for an “inconclusive” result has also been brought up with sources by another user in the talk topic above The Great Mule of Eupatoria (talk) 02:01, 30 November 2023 (UTC)
 * Those sources are WP:PRIMARY and outdated, not high quality reference sources, and don't say what you are claiming. And it would be WP:UNDUE to blow away the whole academic field of Arab-Israeli studies for some random cherry-picked military studies you happen to like because they say the ahistorical thing you want it to say. As any student knows, in 1973, Israel was taken by surprise and fought to the brink of trouble before the US bailed them out and defeated Egypt. Therefore, an Egyptian defeat; your argument that they achiveved their strategic aims is not supported by any historical or academic material. Andre🚐 02:11, 30 November 2023 (UTC)
 * Defeated Egypt how exactly? Egypt still controlled most of the suez by the ceasefire and prevented the Israelis from capturing suez city in the closing stages of the war The Great Mule of Eupatoria (talk) 03:57, 30 November 2023 (UTC)
 * Egypt and Syria did not win the war militarily, but they got what they needed out of it in terms of domestic and regional legitimation and cracking Israel’s veneer of invincibility. If fighting continued, the Arab states likely faced the prospect of another defeat. For Israel, even though they controlled more territories than at the start of the war, it was nevertheless a major political defeat, so again, a military defeat of Egypt on the battlefield for Egypt even though Israel had political problems as a result of the war. They still controlled more territory as a result of the war. Andre🚐 04:36, 30 November 2023 (UTC)
 * This is also including Syria, which absolutely did not win in this war and had their capital Damascus threatened by the Israeli advance. I have given my citations to explain how this was not a full victory for Israel on this specific front, and I will also quote your citation
 * ”For Israel, even though they controlled more territories than at the start of the war, it was nevertheless a major political defeat, and the country was reeling from the loss of life on a scale it had not experienced since 1948.” The “controlled more territories” here more likely refers to the advance on the Syrian front as aside from a crossing and occupation southwest of Sinai Israel had lost control of most of the Suez Canal The Great Mule of Eupatoria (talk) 05:04, 30 November 2023 (UTC)
 * The Suez was Egyptian territory occupied since 67. The Egyptian had made gains and then Israelis were surrounding Suez when the ceasefire was agreed to - but that doesn't mean Egypt got more territory. That was Egyptian territory. Again, as it says, it was a military defeat for Egypt, but even though it was a military win for Israel, it was a political defeat. But, the infobox refers to military defeat in the war. Israel’s victory came at the cost of heavy casualties, and Israelis criticized the government’s lack of preparedness. In April 1974, the nation’s prime minister, Golda Meir (1898-1978), stepped down. Although Egypt had again suffered military defeat at the hands of its Jewish neighbor, the initial Egyptian successes greatly enhanced Sadat’s prestige in the Middle East and gave him an opportunity to seek peace It's History.com which isn't RS but it's still better than the weird old CIA studies or whatever. Andre🚐 05:14, 30 November 2023 (UTC)
 * 1967-1973 Egypt only controlled the western bank of the suez, the eastern bank of the suez was occupied by Israel. by the end of the war Egypt controlled most of the eastern bank and after the ceasefire was signed Egypt retained control of both banks. In the war Israel failed to retake the eastern bank of the suez which negates an “Israeli victory” as this article simplifies it, while Egypt achieved its goal. The Israeli advance in the south however threatened Egypts encircled army however they inflicted two defeats on israel (battle of Ismailia and battle of suez)
 * The combined Egyptian paratrooper-commando force managed to achieve a tactical and strategic victory at a time when Egypt's general situation on the battlefield was deteriorating, and GHQ was in a state of confusion. Sharon's advance toward Ismailia had been halted, and Second Army's logistical lines remained secure.
 * not to mention the admission of David Elazar, 9th chief of staff during the war (Not sure if this specific one is RS but there are many sources mentioning the quote)
 * https://yom-kippur-1973.info/eng/west.htm#:~:text=As%20for%20the%20third%20army,or%20conquered%20them)%20David%20Elazar.
 * “As for the third army, in spite of our encircling them they resisted and advanced to occupy in fact a wider area of land at the east. Thus, we can not say that we defeated or conquered them”
 * Your initial article admits that Israel suffered a major political failure, so Egypt managing to achieve its goals in the war reasserts my view that simplifying the entire conflict to a “isralei victory” is inaccurate, and far from Egypt being “severely defeated” in this war The Great Mule of Eupatoria (talk) 05:41, 30 November 2023 (UTC)
 * Certainly not, you've proven no such thing. Sounds like WP:IDHT Andre🚐 05:43, 30 November 2023 (UTC)
 * @Andrevan I thank you for sharing this very recent source.
 * well since it mentions "For Israel, even though they controlled more territories than at the start of the war, it was nevertheless a major political defeat" then I'll have to agree with the @The Great Mule of Eupatoria on this one, you actually gave support to his point , the result was edited back from "israeli military victory" to "israeli victory" by this guy's logic @Modrenebe :-
 * Done Liu1126 (talk) 09:09, 18 October 2023 (UTC)[reply]
 * Thanks, but could you also remove the word "military" from "Israeli military victory"? It wasn't there before. Its use can give the impression that Israel only won militarily, while the political or other aspects were not or instead were won by the other parties. Modrenebe (talk) 13:58, 18 October 2023 (UTC)[reply]
 * Done. Sorry, missed that in the last edit. Liu1126 (talk) 14:43, 18 October 2023 (UTC)
 * His argument was that political or other aspects were not included in "military victory", the other aspects he hadn't mentioned . Osmarion (talk) 15:36, 30 November 2023 (UTC)
 * Well, you might have an argument that Israel had a political defeat, but that doesn't mean it belongs in the infobox, since the infobox is about a war. Andre🚐 20:07, 30 November 2023 (UTC)
 * war is only means to achieve political objectives and if there is a political defeat then the war has failed. Osmarion (talk) 23:36, 17 February 2024 (UTC)
 * I agree, I think there should be an RFC owing to the proven Egyptian inconclusive front as well as a major political victory for Egypt The Great Mule of Eupatoria (talk) 14:15, 28 February 2024 (UTC)

Time Styles
There are inconsistencies in the page on time between 12h style and 24h style. I suggest edits to make consistent across and specifically recommend 24h being superior for the purposes of both the narratives and the thematic content Hans K Pauley (talk) 14:37, 24 April 2024 (UTC)

Pakistan
Why shouldn't it be in infobox as it's pilots participated in the war and there's a whole article on a pilot who took part in this war Sattar Alvi and north Korea is mentioned which didn't even score a kill. Waleed (talk) 11:17, 1 May 2024 (UTC)

 * @Hohum, @Skitash Waleed  (talk) 11:18, 1 May 2024 (UTC)
 * Expeditionary forces, and Sattar Alvi, have multiple sources detailing Pakistani involvement @Skitash and @Hohum Waleed  (talk) 15:53, 1 May 2024 (UTC)

 * The source would need to state that the country in question was a belligerent, not that there were just members of that nationality taking part. The source would also need to be WP:RELIABLE, ideally in a historical work on the war itself. Wikipedia can not be a source per WP:CIRCULAR. North Korea probably shouldn't be there either, I think it has been removed several times. (Hohum @ ) 18:05, 1 May 2024 (UTC)
 * Well the source states that they were
 * sent by Pakistan and they were active duty members of Pakistan armed for who joined this war and allegedly shot down an Israeli plane ces
 * were Waleed  (talk) 00:01, 2 May 2024 (UTC)
 * There are some news articles which I cited but you've removed, moreover the ones I mentioned are cited, and used those citations including three already in this article about Pakistani involvement, books I mean and you've removed, so i gave others but again removed Waleed  (talk) 00:04, 2 May 2024 (UTC)
 * @Hohum Waleed  (talk) 11:50, 4 May 2024 (UTC)
 * I believe they were self published sources, and the uninterpreted recollections of a primary source and/or didn't name Pakistan as a belligerent. (Hohum @ ) 11:53, 4 May 2024 (UTC)
 * Can I give news articles stating so, would that be fine Waleed  (talk) 22:31, 4 May 2024 (UTC)

Extended-confirmed-protected edit request on 7 May 2024
the result should be changed to:

Both sides claim victory

 * Military stalemate
 * Egyptian political victory
 * Egypt ultimately regains full control of the Sinai peninsula
 * Military stalemate on Syrian front
 * 1978 Camp David accords 2603:6010:1C00:325:E9E2:1FBA:BE32:1F0 (talk) 02:33, 7 May 2024 (UTC)
 * Red information icon with gradient background.svg Not done for now: please establish a consensus for this alteration before using the template. M.Bitton (talk) 00:29, 9 May 2024 (UTC)
 * i think a consensus has been reached in February 2024, as the argument has been settled since Kelcoz (talk) 20:02, 7 June 2024 (UTC)
 * There was no consensus reach. I have shown several sources and no rfc or no discussion has taken place. “Israeli victory” is incorrect and a distortion, that not even the Hebrew Wikipedia article of this war states The Great Mule of Eupatoria (talk) 07:48, 8 June 2024 (UTC)
 * I have been trying this and discussed this with several sources in a previous discussion. I’m not sure why but Wikipedia editors seem to have an allergy to admitting Israel lost anything. Even the Hebrew Wikipedia admits that Egypt’s front was inconclusive The Great Mule of Eupatoria (talk) 04:05, 9 May 2024 (UTC)

Extended-confirmed-protected edit request on 9 May 2024
To fix grammar in the Disengagement section of the article, I recommend adding a comma to change "After the failed conference Henry Kissinger started conducting shuttle diplomacy, meeting with Israel and the Arab states directly." to "After the failed conference, Henry Kissinger started conducting shuttle diplomacy, meeting with Israel and the Arab states directly." Anonymous Libertarian (talk) 20:00, 9 May 2024 (UTC)
 * ✅ Charliehdb (talk) 10:22, 10 May 2024 (UTC)

Article length
The article is about 22,000 words readable prose. This is far beyond the upper limit of WP:SIZERULE. Any article should be readable in a single reasonable length sitting per WP:CANYOUREADTHIS. Subsections is not a cure for this; an encyclopedia caters to people who know nothing or little about a subject, and need an overview. However complex an many faceted an article subject is, is not an excuse for excessively long articles; they can always be made more concise, with additional articles made for detailed aspects. (Hohum @ ) 14:25, 18 May 2024 (UTC)

 * Yes, it's horrendous. It's also rather hard to edit down while retaining necessary detail and balance. Iskandar323 (talk) 15:38, 18 May 2024 (UTC)

 * A potential first step: Any section that has an article dedicated to it should probably be reduced to a single overview paragraph. (Hohum @ ) 09:34, 19 May 2024 (UTC)

Extended-confirmed-protected edit request on 5 June 2024
I believe some parts of this article should be rewritten so they are not just copied from "The Yom Kippur War" by Abraham Rabinovich. Most notably this part is word-for-word from the book and quite awkward, in my opinion: On the night of 25 September, Hussein secretly flew to Tel Aviv to warn Meir of an impending Syrian attack. "Are they going to war without the Egyptians, asked Mrs. Meir. The king said he didn't think so. 'I think they [Egypt] would cooperate.' Shlokie (talk) 11:20, 5 June 2024 (UTC)
 * Pictogram voting comment.svg Note: Given that it's a quote, you expect it to be copied word for word. M.Bitton (talk) 15:04, 6 June 2024 (UTC)
 * Red information icon with gradient background.svg Not done for now: Please address the concerns presented. Geardona (talk to me?) 01:03, 3 July 2024 (UTC)

The result is inconclusive, not “Israeli victory”
It is getting tiring with how useless bringing up this topic is and how it falls on deaf ears, but the second the result of this war gets changed from “Israeli victory” it gets reverted immediately. This is a bias that is not even present on the Hebrew Wikipedia, and makes Wikipedia’s policy of neutrality falter

Here are why the result of this war are “inconclusive”

https://time.com/6322802/yom-kippur-war-israel-history/

“ In 1973, Egypt’s goal in crossing the Suez Canal was to force Israel to the negotiation table to make a peace deal and get back control of the Sinai peninsula. According to Avi Shilon, a historian who teaches at Tel-Hai College in Israel, “The Egyptian and the Syrians didn't plan to conquer Israel. They planned to hit Israel and to force Israel to go into negotiations. For them, it was enough to hit Israel to show that they can beat Israel in the first days, and they preferred to stop, so it was easier for Israel to launch a retaliation attack.”

This outlines Egypts goal of the war, which was to cross the suez and not conquer Sinai or Israel proper. Israel counterattacked, but they failed to repulse the Egyptian army occupying most of the suez

The Israeli military failure to capture two small towns in their supposed legendary encirclement that Wikipedia uses to construct the basis of the delusion of “Israeli victory”

A declassified CIA document

https://www.cia.gov/readingroom/docs/1975-09-01A.pdf

The CIA asserts is as a fact that the war was inconclusive.

Not only that, but mentioning Israel was “100 kn from Cairo” is another perpetuation of Wikipedia delusion. If I am 50 metres from a house, and I step outside and take two steps it is not worth mentioning I am 48 metres from it. Israel’s starting point was the Suez Canal, “100 km from Cairo” is another form of coping, to legitimise a victory that doesn’t exist The Great Mule of Eupatoria (talk) 08:01, 24 June 2024 (UTC)
 * The first conclusion is your own, not a specifically stated result in the source. The second, while it does at least say "the fact is that the war ended on a militarily inconclusive note." on page 24. However, this is a 49 year old primary source report made 2 years after the events - it lacks any information gathered since. It was released in 2012, so it has been available for historians to use in secondary works for over a decade.
 * You will get a lot more traction using WP:secondary sources from reliable military historians. The article already uses these to support a different conclusion though. (Hohum @ ) 15:13, 24 June 2024 (UTC)
 * Several sources do describe it as a stalemate or inclusive, far from the “Israeli victory” oversimplification, which isn’t even correct neither militarily nor politically
 * Many sources regard the war as a stalemate, even on the Syrian front
 * https://www.theatlantic.com/magazine/archive/1974/03/who-lost-the-yom-kippur-war-a-military-inventory-of-the-middle-east/670833/
 * Here Henry Kissinger states “it would be a nightmare if either side won”, implying neither side got a conclusive victory
 * https://nsarchive.gwu.edu/briefing-book/henry-kissinger/2019-08-09/kissinger-told-soviet-envoy-during-1973-arab-israeli-war-my-nightmare-victory-either-side-soviet
 * This one is hidden by a paywall
 * https://www.lemonde.fr/en/history/article/2023/10/02/50th-anniversary-of-yom-kippur-war-the-enduring-ambiguity-of-pax-americana_6143516_157.html
 * The new Arab, I am not sure if it’s RS, also states that it is a stalemate
 * https://www.newarab.com/analysis/october-war-nothing-bloody-stalemate?amp
 * “Which had nearly resulted in Israel’s defeat, but ended in a stalemate”
 * https://www.historytoday.com/archive/no-victor-no-vanquished-yom-kippur-war
 * This is ignoring Israel’s defeat at two small towns during the final battles of the war, as well as an air battle, which saw an Egyptian victory. “Israeli victory” also COMPLETELY ignores the result of the war, seeing Israel actually come to negotiate Sinai which is had adamantly refused in the Rogers Plan and Jarrings Plan, both before the war The Great Mule of Eupatoria (talk) 17:35, 24 June 2024 (UTC)
 * "You will get a lot more traction using WP:secondary sources from reliable military historians. "
 * Shotgunning google search results isn't helpful. Kissinger "implying" is actually you inferring. (Hohum @ ) 17:44, 24 June 2024 (UTC)
 * I’ve given plenty of secondary sources no? What would “if either side had won” would otherwise imply? The Great Mule of Eupatoria (talk) 05:23, 25 June 2024 (UTC)
 * The leading sentences, and the tense of the statement make it very clear that he is speaking before the war has ended.
 * You are also ignoring half of this sentence: "You will get a lot more traction using WP:secondary sources from reliable military historians."
 * Don't expect another reply unless you provide useable sources that explicitly support your point. (Hohum @ ) 17:56, 25 June 2024 (UTC)
 * Frankly, the war being “an Israeli victory” can be disproven with common sense instead of some strange criteria about the species of sources considering you have been dismissing everything I provided even though it is beyond sufficient to explain why the “result” is something from an alternate history timeline. But here we go, hopefully this is good enough.
 * https://bootcampmilitaryfitnessinstitute.com/2021/11/23/what-was-the-yom-kippur-war-1973/
 * “Despite being surrounded, the Third Army managed to maintain its combat integrity east of the canal and keep up its defensive positions, to the surprise of many. According to Trevor N.
 * Dupuy, the Israelis, Soviets and Americans overestimated the vulnerability of the Third Army at the time. It was not on the verge of collapse, and he wrote that while a renewed Israeli offensive would probably overcome it, this was not a certainty.”
 * There are some military men who argued that the encirclement would have destroyed the third army, let’s look at what David Elazar, one of the generals during the war said according to this source
 * “According to David Elazar, Chief of Israeli headquarters staff, on 3
 * December 1973: "As for the third army, in spite of our encircling them they resisted and advanced to occupy in fact a wider area of land at the east.
 * Thus, we can not say that we defeated or conquered them””
 * And further
 * “Shortly before the ceasefire came into effect, an Israeli tank battalion advanced into Adabiya, and took it with support from the Israeli Navy. Some 1,500 Egyptian prisoners were taken, and about a hundred Egyptian soldiers assembled just south of Adabiya, where they held out against the Israelis. The Israelis also conducted their third and final incursion into Suez. They made some gains, but failed to break into the city centre. As a result, the city was partitioned down the main street, with the Egyptians holding the city centre and the Israelis controlling the outskirts, port installations and oil refinery, effectively surrounding the Egyptian defenders”
 * Showing an Israeli failure at the defeat too. This is also not mentioning the fact that Wikipedia completely ignores the effect of the war and its result which led Israel to negotiate after adamantly refusing 2-times pre war
 * The CIA says it’s inconclusive
 * Several sources and military analysts say it’s inconclusive
 * Israeli generals say they couldn’t defeat the Egyptians
 * Yet Wikipedia is adamant to simply this war with the most incorrect, misleading, and simplified result that ignores reality itself The Great Mule of Eupatoria (talk) 04:35, 26 June 2024 (UTC)
 * Yet Wikipedia is adamant to simply this war with the most incorrect, misleading, and simplified result that ignores reality itself The Great Mule of Eupatoria (talk) 04:35, 26 June 2024 (UTC)

Short-form (harv/sfn) reference formatting
This is a minor issue compared to the content questions, but the short-form references in the article are in a wide range of formats (plain-text vs templated, parenthesised vs without brackets). I propose to convert them all to parenthesised style and to use templates – i.e. to use sfnp and/or harvp – and to flag any references where the long-form bibliographic information about the source is missing. I'm happy to do the conversion myself, but, given this is a fairly sensitive article, I would like to check for disagreement / consensus here first. Best, Wham2001 (talk) 20:04, 4 July 2024 (UTC)

 * Since nobody has commented in a week and a half I am going to start on this, using sfnp/harvp. Given the article has 485 references it might take me a while, though... Wham2001 (talk) 19:14, 13 July 2024 (UTC)

title: Talk:Yom Kippur War


text:
Talk:Parliament of the United Kingdom

Andrew Bridgen and the Reclaim Party
Please update this article to include the single seat for the Reclaim party held by Andrew Bridgen 82.37.83.242 (talk) 13:22, 10 May 2023 (UTC)

Agreement
Yes i agree they should update it GothicGolem29 (talk) 14:38, 6 June 2023 (UTC)

History of broadcasts
The section on broadcast media is quite light on detail. I recall that there was a lot of resistance initially to the proceedings of the Commons being broadcast at all. At first broadcasting was only on radio, with TV cameras added later. Can such information be added to the article, please? --Viennese Waltz 11:40, 24 January 2024 (UTC)

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

title: Talk:Parliament of the United Kingdom


text:
Half Dome

Half Dome is a quartz monzonite batholith at the eastern end of Yosemite Valley in Yosemite National Park, California. It is a well-known rock formation in the park, named for its distinct shape. One side is a sheer face while the other three sides are smooth and round, making it appear like a dome cut in half. It stands at nearly 8,800 feet above sea level and is composed of quartz monzonite, an igneous rock that solidified several thousand feet within the Earth. At its core are the remains of a magma chamber that cooled slowly and crystallized beneath the Earth's surface. The solidified magma chamber was then exposed and cut in half by erosion, therefore leading to the geographic name Half Dome.

Geology
The impression from the valley floor that this is a round dome that has lost its northwest half, is just an illusion. From Washburn Point, Half Dome can be seen as a thin ridge of rock, an arête, that is oriented northeast–southwest, with its southeast side almost as steep as its northwest side except for the very top. Although the trend of this ridge, as well as that of Tenaya Canyon, is probably controlled by master joints, 80 percent of the northwest "half" of the original dome may well still be there.

Ascents
As late as the 1870s, Half Dome was described as "perfectly inaccessible" by Josiah Whitney of the California Geological Survey. The summit was finally reached by George G. Anderson in October 1875, via a route constructed by drilling and placing iron eye bolts into the smooth rock. Anderson had previously tried a variety of methods, including using pitch from nearby pine trees for extra friction.

Anderson subsequently went on to add ropes to his eye bolts, so that other people could climb. Among those who took advantage was the first woman to climb Half Dome in 1876, S. L. Dutcher, of San Francisco. In 1877 James Mason Hutchings along with Anderson led a climb which included Hutchings' daughter Cosie, his son Willie, his mother-in-law Florence Sproat (aged 65), and two other women.

Today, Half Dome may be ascended in several different ways. Thousands of hikers reach the top each year by following an 8.5 mi trail from the valley floor. After a rigorous 2 mi approach, including several hundred feet of rock stairs, the final pitch up the peak's steep but somewhat rounded east face is ascended with the aid of a pair of post-mounted steel cables originally constructed close to the Anderson route in 1919.

Alternatively, over a dozen rock climbing routes lead from the valley up Half Dome's vertical northwest face. The first technical ascent was in 1957 via a route pioneered by Royal Robbins, Mike Sherrick, and Jerry Gallwas, today known as the Regular Northwest Face. Their five-day epic was the first Grade VI climb in the United States. Their route has now been free climbed several times in a few hours' time. Other technical routes ascend the south face and the west shoulder.

Hiking the Cable Route
The Half Dome Cable Route hike runs from the valley floor to the top of the dome in 8.2 mi (via the Mist Trail), with 4800 ft of elevation gain. The length and difficulty of the trail used to keep it less crowded than other park trails, but trail traffic grew to as many as 1,000 people a day, and about 50,000 per year, before a permit system was introduced in 2010. The hike can be done from the valley floor in a single long day, but many people break it up by camping overnight in Little Yosemite Valley. The trail climbs past Vernal Fall and Nevada Fall, then continues into Little Yosemite Valley, then north to the base of the northeast ridge of Half Dome itself.

The final 400 ft ascent is steeply up the rock between two steel cables used as handholds. The cables are fixed with bolts in the rock and raised onto a series of metal poles in late May (the poles do not anchor the cables). The cables are taken down from the poles for the winter in early October, but they are still fixed to the rock surface and can be used. The National Park Service recommends against climbing the route when the cables are down or when the surface of the rock is wet and slippery. The Cable Route is rated class 3, while the same face away from the cables is rated class 5. The Cable Route can be crowded. Since 2010, all hikers who intend to ascend the Cable Route must first obtain permits when the cables are up between May and October. Permits are limited to 300 per day, and, more than 70,000 applications were received in the pre-season and daily permit lotteries. Permits are checked by a ranger on the trail, and hikers without permits are not allowed to hike beyond the base of the sub-dome or to the bottom of the cables. Hikers caught bypassing the rangers to visit either the sub-dome or main dome without a permit face fines of up to $5,000 and/or 6 months in jail.

Backpackers with an appropriate wilderness permit can receive a Half Dome permit when they pick up their wilderness permit with no additional reservation required. Rock climbers who reach the top of Half Dome without entering the subdome area can descend on the Half Dome Trail without a permit.

The top of Half Dome is a large, flat area where climbers can relax and enjoy their accomplishment. The summit offers views of the surrounding areas, including Little Yosemite Valley and the Valley Floor. A notable location to one side of Half Dome is the "Diving Board", where Ansel Adams took his photograph Monolith, the Face of Half Dome on April 10, 1927. Often confused with "the Visor," a small overhanging ledge at the summit, the Diving Board is on the shoulder of Half Dome.

From 1919 when the cables were erected through 2023, there have been 11 fatal falls from the cables. The latest fatality occurred on September 5, 2019.

Lightning strikes can be a risk while on or near the summit. On July 27, 1985, five hikers were struck by lightning, resulting in two fatalities.

The Cable Route was added to the National Register of Historic Places in 2012.

Notable ascents

 * 1875 (October 12) George G. Anderson via drilled spikes on the east slope.
 * 1875 (before November 10) John Muir
 * 1946 Salathe Route on southwest face (IV 5.7 A3), FA by John Salathe and Anton Nelson
 * 1957 Northwest Face (VI 5.8 A3), FA by Royal Robbins, Jerry Gallwas and Mike Sherrick. First Grade VI in North America.
 * 1963 Direct Northwest Face (VI 5.9 A5), FA by Royal Robbins and Dick McCracken
 * 1969 Tis-sa-ack (VI 5.9 A4), FA by Royal Robbins and Don Peterson.
 * 1973 First "clean ascent" of NW face by Dennis Hennek, Doug Robinson, and Galen Rowell, Hennek is on the cover of June 1974 National Geographic leading a nut protected traverse see Super Topo too

 * 1987 The Big Chill (VI 5.9 A4), FA by Jim Bridwell, Peter Mayfield, Sean Plunkett and Steve Bosque
 * 1989 Shadows (VI 5.10 A5), FA by Jim Bridwell, Charles Row, Cito Kirkpatrick, William Westbay
 * 1989 Kali Yuga (VI 5.10 A4+), FA by John Middendorf, Walt Shipley
 * 1997 Blue Shift (VI 5.11c a4) FA by Jay Smith and Karl McConachie.

Notable free climbs

 * 1964 Salathe Route (5.10), FFA by Frank Sacherer, Bob Kamps & Andy Lichtman
 * 1965 Snake Dike (5.7), FFA by Eric Beck, Jim Bridwell and Chris Fredericks
 * 1976 Regular Northwest Face, Higbee variation (VI 5.12d) by Art Higbee and Jim Erickson.
 * 1985 The Autobahn (5.11+/5.12a) by John Middendorf and Charles Cole.

 * 1988 Southern Belle (V 5.12d) by Dave Schultz and Scott Cosgrove
 * 2008 Regular Northwest Face, Higbee variation (VI 5.12a, 23 pitches), free solo climb by Alex Honnold.

In culture
Half Dome was originally called "Tis-sa-ack", meaning Cleft Rock in the language of the local Ahwahnechee people. Tis-sa-ack is also the name of the fourth route on the formation, ascended by Royal Robbins and Don Peterson over eight days in October 1969. Tis-sa-ack is the name of a mother from a native legend. The face seen in Half Dome is supposed to be hers. Tis-sa-ack is the name of a Mono Lake Paiute girl in the Yosemite Native American legend. John Muir interchangeably used "Tissiack," "South Dome," and "Half Dome" in his writings.

Others say Ahwahneechee Native Americans named Half Dome "Face of a Young Woman Stained with Tears" ("Tis-se’-yak") because of the colonies of brown-black lichens that form dark vertical drip-like stripes along drainage tracks in the rock faces.

Jack London's 1900 short story "Dutch Courage" tells of a fictional climb of Half Dome.

Monolith, the Face of Half Dome is a black and white photograph taken by Ansel Adams in 1927 that depicts the western face of Half Dome as seen from the "Diving Board". It was used by the Sierra Club as a visual aid for the environmental movement, and was a part of the portfolio Parmelian Prints of the High Sierras. The ashes of Ansel Adams were later scattered on Half Dome after he died.

Half Dome was used as the logo for a software company, Sierra On-Line.

In 1971, outdoor recreation and climbing equipment company The North Face created their company logo, based upon a stylized depiction of the Half Dome formation. It is still their logo 50 years later.

In 1988, Half Dome was featured on a 25 cent United States postage stamp. An image of Half Dome, along with John Muir and the California condor, appears on the California State Quarter, released in January 2005.

From 2010 until the introduction of REAL ID, California driver's licenses featured an illustration of Half Dome.

The design of the Wilshire Grand Center skyscraper in Los Angeles was inspired by Half Dome.

title: Half Dome


text:
Fundamental theorem of algebra

The fundamental theorem of algebra, also called d'Alembert's theorem or the d'Alembert–Gauss theorem, states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.

Equivalently (by definition), the theorem states that the field of complex numbers is algebraically closed.

The theorem is also stated as follows: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. The equivalence of the two statements can be proven through the use of successive polynomial division.

Despite its name, there is no purely algebraic proof of the theorem, since any proof must use some form of the analytic completeness of the real numbers, which is not an algebraic concept. Additionally, it is not fundamental for modern algebra; it was named when algebra was synonymous with the theory of equations.

History
Peter Roth (mathematician), in his book Arithmetica Philosophica (published in 1608, at Nürnberg, by Johann Lantzenberger), wrote that a polynomial equation of degree n (with real coefficients) may have n solutions. Albert Girard, in his book L'invention nouvelle en l'Algèbre (published in 1629), asserted that a polynomial equation of degree n has n solutions, but he did not state that they had to be real numbers. Furthermore, he added that his assertion holds "unless the equation is incomplete", by which he meant that no coefficient is equal to 0. However, when he explains in detail what he means, it is clear that he actually believes that his assertion is always true; for instance, he shows that the equation $$x^4 = 4x-3,$$ although incomplete, has four solutions (counting multiplicities): 1 (twice), $$-1+i\sqrt{2},$$ and $$-1-i\sqrt{2}.$$

As will be mentioned again below, it follows from the fundamental theorem of algebra that every non-constant polynomial with real coefficients can be written as a product of polynomials with real coefficients whose degrees are either 1 or 2. However, in 1702 Leibniz erroneously said that no polynomial of the type $x^{4} + a^{4}$ (with $a$ real and distinct from 0) can be written in such a way. Later, Nikolaus Bernoulli made the same assertion concerning the polynomial $x^{4} − 4x^{3} + 2x^{2} + 4x + 4$, but he got a letter from Euler in 1742 in which it was shown that this polynomial is equal to
 * $$\left (x^2-(2+\alpha)x+1+\sqrt{7}+\alpha \right ) \left (x^2-(2-\alpha)x+1+\sqrt{7}-\alpha \right ),$$

with $$\alpha = \sqrt{4+2\sqrt{7}}.$$ Also, Euler pointed out that
 * $$x^4+a^4= \left (x^2+a\sqrt{2}\cdot x+a^2 \right ) \left (x^2-a\sqrt{2}\cdot x+a^2 \right ).$$

A first attempt at proving the theorem was made by d'Alembert in 1746, but his proof was incomplete. Among other problems, it assumed implicitly a theorem (now known as Puiseux's theorem), which would not be proved until more than a century later and using the fundamental theorem of algebra. Other attempts were made by Euler (1749), de Foncenex (1759), Lagrange (1772), and Laplace (1795). These last four attempts assumed implicitly Girard's assertion; to be more precise, the existence of solutions was assumed and all that remained to be proved was that their form was a + bi for some real numbers a and b. In modern terms, Euler, de Foncenex, Lagrange, and Laplace were assuming the existence of a splitting field of the polynomial p(z).

At the end of the 18th century, two new proofs were published which did not assume the existence of roots, but neither of which was complete. One of them, due to James Wood and mainly algebraic, was published in 1798 and it was totally ignored. Wood's proof had an algebraic gap. The other one was published by Gauss in 1799 and it was mainly geometric, but it had a topological gap, only filled by Alexander Ostrowski in 1920, as discussed in Smale (1981).

The first rigorous proof was published by Argand, an amateur mathematician, in 1806 (and revisited in 1813); it was also here that, for the first time, the fundamental theorem of algebra was stated for polynomials with complex coefficients, rather than just real coefficients. Gauss produced two other proofs in 1816 and another incomplete version of his original proof in 1849.

The first textbook containing a proof of the theorem was Cauchy's Cours d'analyse de l'École Royale Polytechnique (1821). It contained Argand's proof, although Argand is not credited for it.

None of the proofs mentioned so far is constructive. It was Weierstrass who raised for the first time, in the middle of the 19th century, the problem of finding a constructive proof of the fundamental theorem of algebra. He presented his solution, which amounts in modern terms to a combination of the Durand–Kerner method with the homotopy continuation principle, in 1891. Another proof of this kind was obtained by Hellmuth Kneser in 1940 and simplified by his son Martin Kneser in 1981.

Without using countable choice, it is not possible to constructively prove the fundamental theorem of algebra for complex numbers based on the Dedekind real numbers (which are not constructively equivalent to the Cauchy real numbers without countable choice). However, Fred Richman proved a reformulated version of the theorem that does work.

Equivalent statements
There are several equivalent formulations of the theorem: The next two statements are equivalent to the previous ones, although they do not involve any nonreal complex number. These statements can be proved from previous factorizations by remarking that, if $n$ is a non-real root of a polynomial with real coefficients, its complex conjugate $$\overline r$$ is also a root, and $$(x-r)(x-\overline r)$$ is a polynomial of degree two with real coefficients (this is the complex conjugate root theorem). Conversely, if one has a factor of degree two, the quadratic formula gives a root.
 * Every univariate polynomial of positive degree with real coefficients has at least one complex root.
 * Every univariate polynomial of positive degree with complex coefficients has at least one complex root.
 * This implies immediately the previous assertion, as real numbers are also complex numbers. The converse results from the fact that one gets a polynomial with real coefficients by taking the product of a polynomial and its complex conjugate (obtained by replacing each coefficient with its complex conjugate). A root of this product is either a root of the given polynomial, or of its conjugate; in the latter case, the conjugate of this root is a root of the given polynomial.
 * Every univariate polynomial of positive degree $n$ with complex coefficients can be factorized as where $$c, r_1, \ldots, r_n$$ are complex numbers.
 * The $n$ complex numbers $$r_1, \ldots, r_n$$ are the roots of the polynomial. If a root appears in several factors, it is a multiple root, and the number of its occurrences is, by definition, the multiplicity of the root.
 * The proof that this statement results from the previous ones is done by recursion on $r$: when a root $$r_1$$ has been found, the polynomial division by $$x-r_1$$ provides a polynomial of degree $$n-1$$ whose roots are the other roots of the given polynomial.
 * Every univariate polynomial with real coefficients of degree larger than two has a factor of degree two with real coefficients.
 * Every univariate polynomial with real coefficients of positive degree can be factored as where $c$ is a real number and each $$p_i$$ is a monic polynomial of degree at most two with real coefficients. Moreover, one can suppose that the factors of degree two do not have any real root.

Proofs
All proofs below involve some mathematical analysis, or at least the topological concept of continuity of real or complex functions. Some also use differentiable or even analytic functions. This requirement has led to the remark that the Fundamental Theorem of Algebra is neither fundamental, nor a theorem of algebra.

Some proofs of the theorem only prove that any non-constant polynomial with real coefficients has some complex root. This lemma is enough to establish the general case because, given a non-constant polynomial $p$ with complex coefficients, the polynomial
 * $$q=p\overline{p},$$

has only real coefficients, and, if $z$ is a root of $q$, then either $z$ or its conjugate is a root of $p$. Here, $$\overline{p}$$ is the polynomial obtained by replacing each coefficient of $p$ with its complex conjugate; the roots of $$\overline{p}$$ are exactly the complex conjugates of the roots of $p$

Many non-algebraic proofs of the theorem use the fact (sometimes called the "growth lemma") that a polynomial function p(z) of degree n whose dominant coefficient is 1 behaves like zn when |z| is large enough. More precisely, there is some positive real number R such that

 * $$\tfrac{1}{2}|z^n|<|p(z)|<\tfrac{3}{2}|z^n|$$

when |z| > R.

Real-analytic proofs
Even without using complex numbers, it is possible to show that a real-valued polynomial p(x): p(0) ≠ 0 of degree n > 2 can always be divided by some quadratic polynomial with real coefficients. In other words, for some real-valued a and b, the coefficients of the linear remainder on dividing p(x) by x2 − ax − b simultaneously become zero.

 * $$p(x) = (x^2 - ax - b) q(x) + x\,R_{p(x)}(a, b) + S_{p(x)}(a, b),$$

where q(x) is a polynomial of degree n − 2. The coefficients Rp(x)(a, b) and Sp(x)(a, b) are independent of x and completely defined by the coefficients of p(x). In terms of representation, Rp(x)(a, b) and Sp(x)(a, b) are bivariate polynomials in a and b. In the flavor of Gauss's first (incomplete) proof of this theorem from 1799, the key is to show that for any sufficiently large negative value of b, all the roots of both Rp(x)(a, b) and Sp(x)(a, b) in the variable a are real-valued and alternating each other (interlacing property). Utilizing a Sturm-like chain that contain Rp(x)(a, b) and Sp(x)(a, b) as consecutive terms, interlacing in the variable a can be shown for all consecutive pairs in the chain whenever b has sufficiently large negative value. As Sp(a, b = 0) = p(0) has no roots, interlacing of Rp(x)(a, b) and Sp(x)(a, b) in the variable a fails at b = 0. Topological arguments can be applied on the interlacing property to show that the locus of the roots of Rp(x)(a, b) and Sp(x)(a, b) must intersect for some real-valued a and b < 0.

Complex-analytic proofs
Find a closed disk D of radius r centered at the origin such that |p(z)| > |p(0)| whenever |z| ≥ r. The minimum of |p(z)| on D, which must exist since D is compact, is therefore achieved at some point z0 in the interior of D, but not at any point of its boundary. The maximum modulus principle applied to 1/p(z) implies that p(z0) = 0. In other words, z0 is a zero of p(z).

A variation of this proof does not require the maximum modulus principle (in fact, a similar argument also gives a proof of the maximum modulus principle for holomorphic functions). Continuing from before the principle was invoked, if a := p(z0) ≠ 0, then, expanding p(z) in powers of z − z0, we can write

 * $$p(z) = a + c_k (z-z_0)^k + c_{k+1} (z-z_0)^{k+1} + \cdots + c_n (z-z_0)^n.$$

Here, the cj are simply the coefficients of the polynomial z → p(z + z0) after expansion, and k is the index of the first non-zero coefficient following the constant term. For z sufficiently close to z0 this function has behavior asymptotically similar to the simpler polynomial $$q(z) = a+c_k (z-z_0)^k$$. More precisely, the function

 * $$\left|\frac{p(z)-q(z)}{(z-z_0)^{k+1}}\right|\leq M$$

for some positive constant M in some neighborhood of z0. Therefore, if we define $$\theta_0 = (\arg(a)+\pi-\arg(c_k)) /k$$ and let $$z = z_0 + r e^{i \theta_0}$$ tracing a circle of radius r > 0 around z, then for any sufficiently small r (so that the bound M holds), we see that

 * $$\begin{align}

&\le \left|a +(-1)c_k r^k e^{i(\arg(a)-\arg(c_k))}\right| + M r^{k+1} \\[4pt] &= |a|-|c_k|r^k + M r^{k+1} \end{align}$$
 * p(z)| &\le |q(z)| + r^{k+1} \left|\frac{p(z)-q(z)}{r^{k+1}}\right|\\[4pt]

When r is sufficiently close to 0 this upper bound for |p(z)| is strictly smaller than |a|, contradicting the definition of z0. Geometrically, we have found an explicit direction θ0 such that if one approaches z0 from that direction one can obtain values p(z) smaller in absolute value than |p(z0)|.

Another analytic proof can be obtained along this line of thought observing that, since |p(z)| > |p(0)| outside D, the minimum of |p(z)| on the whole complex plane is achieved at z0. If |p(z0)| > 0, then 1/p is a bounded holomorphic function in the entire complex plane since, for each complex number z, |1/p(z)| ≤ |1/p(z0)|. Applying Liouville's theorem, which states that a bounded entire function must be constant, this would imply that 1/p is constant and therefore that p is constant. This gives a contradiction, and hence p(z0) = 0.

Yet another analytic proof uses the argument principle. Let R be a positive real number large enough so that every root of p(z) has absolute value smaller than R; such a number must exist because every non-constant polynomial function of degree n has at most n zeros. For each r > R, consider the number

 * $$\frac{1}{2\pi i}\int_{c(r)}\frac{p'(z)}{p(z)}\,dz,$$

where c(r) is the circle centered at 0 with radius r oriented counterclockwise; then the argument principle says that this number is the number N of zeros of p(z) in the open ball centered at 0 with radius r, which, since r > R, is the total number of zeros of p(z). On the other hand, the integral of n/z along c(r) divided by 2πi is equal to n. But the difference between the two numbers is

 * $$\frac{1}{2\pi i}\int_{c(r)}\left(\frac{p'(z)}{p(z)}-\frac{n}{z}\right)dz=\frac{1}{2\pi i}\int_{c(r)}\frac{zp'(z)-np(z)}{zp(z)}\,dz.$$

The numerator of the rational expression being integrated has degree at most n − 1 and the degree of the denominator is n + 1. Therefore, the number above tends to 0 as r → +∞. But the number is also equal to N − n and so N = n.

Another complex-analytic proof can be given by combining linear algebra with the Cauchy theorem. To establish that every complex polynomial of degree n > 0 has a zero, it suffices to show that every complex square matrix of size n > 0 has a (complex) eigenvalue. The proof of the latter statement is by contradiction.

Let A be a complex square matrix of size n > 0 and let In be the unit matrix of the same size. Assume A has no eigenvalues. Consider the resolvent function

 * $$ R(z)=(zI_n-A)^{-1},$$

which is a meromorphic function on the complex plane with values in the vector space of matrices. The eigenvalues of A are precisely the poles of R(z). Since, by assumption, A has no eigenvalues, the function R(z) is an entire function and Cauchy theorem implies that

 * $$ \int_{c(r)} R(z) \, dz =0.$$

On the other hand, R(z) expanded as a geometric series gives:

 * $$R(z)=z^{-1}(I_n-z^{-1}A)^{-1}=z^{-1}\sum_{k=0}^\infty \frac{1}{z^k}A^k\cdot$$

This formula is valid outside the closed disc of radius $$\|A\|$$ (the operator norm of A). Let $$r>\|A\|.$$ Then

 * $$\int_{c(r)}R(z)dz=\sum_{k=0}^{\infty}\int_{c(r)}\frac{dz}{z^{k+1}}A^k=2\pi iI_n$$

(in which only the summand k = 0 has a nonzero integral). This is a contradiction, and so A has an eigenvalue.

Finally, Rouché's theorem gives perhaps the shortest proof of the theorem.

Topological proofs

Suppose the minimum of |p(z)| on the whole complex plane is achieved at z0; it was seen at the proof which uses Liouville's theorem that such a number must exist. We can write p(z) as a polynomial in z − z0: there is some natural number k and there are some complex numbers ck, ck + 1, ..., cn such that ck ≠ 0 and:

 * $$p(z)=p(z_0)+c_k(z-z_0)^k+c_{k+1}(z-z_0)^{k+1}+ \cdots +c_n(z-z_0)^n.$$

If p(z0) is nonzero, it follows that if a is a kth root of −p(z0)/ck and if t is positive and sufficiently small, then |p(z0 + ta)| < |p(z0)|, which is impossible, since |p(z0)| is the minimum of |p| on D.

For another topological proof by contradiction, suppose that the polynomial p(z) has no roots, and consequently is never equal to 0. Think of the polynomial as a map from the complex plane into the complex plane. It maps any circle |z| = R into a closed loop, a curve P(R). We will consider what happens to the winding number of P(R) at the extremes when R is very large and when R = 0. When R is a sufficiently large number, then the leading term zn of p(z) dominates all other terms combined; in other words,

 * $$\left | z^n \right | > \left | a_{n-1} z^{n-1} + \cdots + a_0 \right |.$$

When z traverses the circle $$Re^{i\theta}$$ once counter-clockwise $$(0\leq \theta \leq 2\pi),$$ then $$z^n=R^ne^{in\theta}$$ winds n times counter-clockwise $$(0\leq \theta \leq 2\pi n)$$ around the origin (0,0), and P(R) likewise. At the other extreme, with |z| = 0, the curve P(0) is merely the single point p(0), which must be nonzero because p(z) is never zero. Thus p(0) must be distinct from the origin (0,0), which denotes 0 in the complex plane. The winding number of P(0) around the origin (0,0) is thus 0. Now changing R continuously will deform the loop continuously. At some R the winding number must change. But that can only happen if the curve P(R) includes the origin (0,0) for some R. But then for some z on that circle |z| = R we have p(z) = 0, contradicting our original assumption. Therefore, p(z) has at least one zero.

Algebraic proofs
These proofs of the Fundamental Theorem of Algebra must make use of the following two facts about real numbers that are not algebraic but require only a small amount of analysis (more precisely, the intermediate value theorem in both cases):
 * every polynomial with an odd degree and real coefficients has some real root;
 * every non-negative real number has a square root.

The second fact, together with the quadratic formula, implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if R is any real-closed field, then its extension C = R($\sqrt{−1}$) is algebraically closed.

By induction
As mentioned above, it suffices to check the statement "every non-constant polynomial p(z) with real coefficients has a complex root". This statement can be proved by induction on the greatest non-negative integer k such that 2k divides the degree n of p(z). Let a be the coefficient of zn in p(z) and let F be a splitting field of p(z) over C; in other words, the field F contains C and there are elements z1, z2, ..., zn in F such that

 * $$p(z)=a(z-z_1)(z-z_2) \cdots (z-z_n).$$

If k = 0, then n is odd, and therefore p(z) has a real root. Now, suppose that n = 2km (with m odd and k > 0) and that the theorem is already proved when the degree of the polynomial has the form 2k − 1m′ with m′ odd. For a real number t, define:

 * $$q_t(z)=\prod_{1\le i<j\le n}\left(z-z_i-z_j-tz_iz_j\right).$$

Then the coefficients of qt(z) are symmetric polynomials in the zi with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the elementary symmetric polynomials, that is, in −a1, a2, ..., (−1)nan. So qt(z) has in fact real coefficients. Furthermore, the degree of qt(z) is n(n − 1)/2 = 2k−1m(n − 1), and m(n − 1) is an odd number. So, using the induction hypothesis, qt has at least one complex root; in other words, zi + zj + tzizj is complex for two distinct elements i and j from {1, ..., n}. Since there are more real numbers than pairs (i, j), one can find distinct real numbers t and s such that zi + zj + tzizj and zi + zj + szizj are complex (for the same i and j). So, both zi + zj and zizj are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that zi and zj are complex numbers, since they are roots of the quadratic polynomial z2 − (zi + zj)z + zizj.

Joseph Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so "odd" can be replaced by "odd prime" and this holds for fields of all characteristics). For axiomatization of algebraically closed fields, this is the best possible, as there are counterexamples if a single prime is excluded. However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree n ∈ I has a root, where I is any fixed infinite set of odd numbers, then every polynomial f(x) of odd degree has a root (since (x2 + 1)kf(x) has a root, where k is chosen so that deg(f) + 2k ∈ I). Mohsen Aliabadi generalized Shipman's result in 2013, providing an independent proof that a sufficient condition for an arbitrary field (of any characteristic) to be algebraically closed is that it has a root for every polynomial of prime degree.

From Galois theory
Another algebraic proof of the fundamental theorem can be given using Galois theory. It suffices to show that C has no proper finite field extension. Let K/C be a finite extension. Since the normal closure of K over R still has a finite degree over C (or R), we may assume without loss of generality that K is a normal extension of R (hence it is a Galois extension, as every algebraic extension of a field of characteristic 0 is separable). Let G be the Galois group of this extension, and let H be a Sylow 2-subgroup of G, so that the order of H is a power of 2, and the index of H in G is odd. By the fundamental theorem of Galois theory, there exists a subextension L of K/R such that Gal(K/L) = H. As [L:R] = [G:H] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have L = R, thus [K:R] and [K:C] are powers of 2. Assuming by way of contradiction that [K:C] > 1, we conclude that the 2-group Gal(K/C) contains a subgroup of index 2, so there exists a subextension M of C of degree 2. However, C has no extension of degree 2, because every quadratic complex polynomial has a complex root, as mentioned above. This shows that [K:C] = 1, and therefore K = C, which completes the proof.

Geometric proofs
There exists still another way to approach the fundamental theorem of algebra, due to J. M. Almira and A. Romero: by Riemannian geometric arguments. The main idea here is to prove that the existence of a non-constant polynomial p(z) without zeros implies the existence of a flat Riemannian metric over the sphere S2. This leads to a contradiction since the sphere is not flat.

A Riemannian surface (M, g) is said to be flat if its Gaussian curvature, which we denote by Kg, is identically null. Now, the Gauss–Bonnet theorem, when applied to the sphere S2, claims that

 * $$\int_{\mathbf{S}^2}K_g=4\pi,$$

which proves that the sphere is not flat.

Let us now assume that n > 0 and

 * $$p(z) = a_0 + a_1 z + \cdots + a_n z^n \neq 0$$

for each complex number z. Let us define

 * $$p^*(z) = z^n p \left ( \tfrac{1}{z} \right ) = a_0 z^n + a_1 z^{n-1} + \cdots + a_n.$$

Obviously, p*(z) ≠ 0 for all z in C. Consider the polynomial f(z) = p(z)p*(z). Then f(z) ≠ 0 for each z in C. Furthermore,

 * $$f(\tfrac{1}{w}) = p \left (\tfrac{1}{w} \right )p^* \left (\tfrac{1}{w} \right ) = w^{-2n}p^*(w)p(w) = w^{-2n}f(w).$$

We can use this functional equation to prove that g, given by

 * $$g=\frac{1}{|f(w)|^{\frac{2}{n}}}\,|dw|^2 $$

for w in C, and

 * $$g=\frac{1}{\left |f\left (\tfrac{1}{w} \right ) \right |^{\frac{2}{n}}}\left |d\left (\tfrac{1}{w} \right ) \right |^2 $$

for w ∈ S2\{0}, is a well defined Riemannian metric over the sphere S2 (which we identify with the extended complex plane C ∪ {∞}).

Now, a simple computation shows that

 * $$\forall w\in\mathbf{C}: \qquad \frac{1}{|f(w)|^{\frac{1}{n}}} K_g=\frac{1}{n}\Delta \log|f(w)|=\frac{1}{n}\Delta \text{Re}(\log f(w))=0,$$

since the real part of an analytic function is harmonic. This proves that Kg = 0.

Corollaries
Since the fundamental theorem of algebra can be seen as the statement that the field of complex numbers is algebraically closed, it follows that any theorem concerning algebraically closed fields applies to the field of complex numbers. Here are a few more consequences of the theorem, which are either about the field of real numbers or the relationship between the field of real numbers and the field of complex numbers:

 * The field of complex numbers is the algebraic closure of the field of real numbers.
 * Every polynomial in one variable z with complex coefficients is the product of a complex constant and polynomials of the form z + a with a complex.
 * Every polynomial in one variable x with real coefficients can be uniquely written as the product of a constant, polynomials of the form x + a with a real, and polynomials of the form x2 + ax + b with a and b real and a2 − 4b < 0 (which is the same thing as saying that the polynomial x2 + ax + b has no real roots). (By the Abel–Ruffini theorem, the real numbers a and b are not necessarily expressible in terms of the coefficients of the polynomial, the basic arithmetic operations and the extraction of n-th roots.) This implies that the number of non-real complex roots is always even and remains even when counted with their multiplicity.
 * Every rational function in one variable x, with real coefficients, can be written as the sum of a polynomial function with rational functions of the form a/(x − b)n (where n is a natural number, and a and b are real numbers), and rational functions of the form (ax + b)/(x2 + cx + d)n (where n is a natural number, and a, b, c, and d are real numbers such that c2 − 4d < 0). A corollary of this is that every rational function in one variable and real coefficients has an elementary primitive.
 * Every algebraic extension of the real field is isomorphic either to the real field or to the complex field.

Bounds on the zeros of a polynomial
While the fundamental theorem of algebra states a general existence result, it is of some interest, both from the theoretical and from the practical point of view, to have information on the location of the zeros of a given polynomial. The simpler result in this direction is a bound on the modulus: all zeros ζ of a monic polynomial $$z^n+a_{n-1}z^{n-1}+\cdots+a_1z +a_0$$ satisfy an inequality |ζ| ≤ R∞, where

 * $$R_{\infty}:= 1+\max\{|a_0|,\ldots,|a_{n-1}|\}. $$

As stated, this is not yet an existence result but rather an example of what is called an a priori bound: it says that if there are solutions then they lie inside the closed disk of center the origin and radius R∞. However, once coupled with the fundamental theorem of algebra it says that the disk contains in fact at least one solution. More generally, a bound can be given directly in terms of any p-norm of the n-vector of coefficients $$a:=( a_0, a_1, \ldots, a_{n-1}),$$ that is |ζ| ≤ Rp, where Rp is precisely the q-norm of the 2-vector $$(1, \|a\|_p),$$ q being the conjugate exponent of p, $$\tfrac{1}{p} + \tfrac{1}{q} =1,$$ for any 1 ≤ p ≤ ∞. Thus, the modulus of any solution is also bounded by

 * $$ R_1:= \max\left \{ 1, \sum_{0\leq k<n} |a_k|\right \},$$
 * $$ R_p:= \left[ 1 + \left(\sum_{0\leq k<n}|a_k|^p\right )^{\frac{q}{p}}\right ]^{\frac{1}{q}},$$

for 1 < p < ∞, and in particular

 * $$ R_2:= \sqrt{\sum_{0\leq k\leq n} |a_k|^2 }$$

(where we define an to mean 1, which is reasonable since 1 is indeed the n-th coefficient of our polynomial). The case of a generic polynomial of degree n,

 * $$P(z):= a_n z^n+a_{n-1}z^{n-1}+\cdots+a_1z +a_0,$$

is of course reduced to the case of a monic, dividing all coefficients by an ≠ 0. Also, in case that 0 is not a root, i.e. a0 ≠ 0, bounds from below on the roots ζ follow immediately as bounds from above on $$\tfrac{1}{\zeta}$$, that is, the roots of

 * $$a_0 z^n+a_1z^{n-1}+\cdots+a_{n-1}z +a_n.$$

Finally, the distance $$|\zeta-\zeta_0|$$ from the roots ζ to any point $$\zeta_0$$ can be estimated from below and above, seeing $$\zeta-\zeta_0$$ as zeros of the polynomial $$P(z+\zeta_0)$$, whose coefficients are the Taylor expansion of P(z) at $$z=\zeta_0.$$

Let ζ be a root of the polynomial

 * $$z^n+a_{n-1}z^{n-1}+\cdots+a_1z +a_0;$$

in order to prove the inequality |ζ| ≤ Rp we can assume, of course, |ζ| > 1. Writing the equation as

 * $$-\zeta^n=a_{n-1}\zeta^{n-1}+\cdots+a_1\zeta+a_0,$$

and using the Hölder's inequality we find

 * $$|\zeta|^n\leq \|a\|_p \left \| \left (\zeta^{n-1},\ldots,\zeta, 1 \right ) \right \|_q.$$

Now, if p = 1, this is

 * $$|\zeta|^n\leq\|a\|_1\max \left \{|\zeta|^{n-1},\ldots,|\zeta|,1 \right \} =\|a\|_1|\zeta|^{n-1},$$

thus

 * $$|\zeta|\leq \max\{1, \|a\|_1\}.$$

In the case 1 < p ≤ ∞, taking into account the summation formula for a geometric progression, we have

 * $$|\zeta|^n\leq \|a\|_p \left(|\zeta|^{q(n-1)}+\cdots+|\zeta|^q +1\right)^{\frac{1}{q}}=\|a\|_p \left(\frac{|\zeta|^{qn}-1}{|\zeta|^q-1}\right)^{\frac{1}{q}}\leq\|a\|_p \left(\frac{|\zeta|^{qn}}{|\zeta|^q-1}\right)^{\frac{1}{q}},$$

thus

 * $$|\zeta|^{nq}\leq \|a\|_p^q \frac{|\zeta|^{qn}}{|\zeta|^q-1}$$

and simplifying,

 * $$|\zeta|^q\leq 1+\|a\|_p^q.$$

Therefore

 * $$|\zeta|\leq \left \| \left (1,\|a\|_p \right ) \right \|_q=R_p $$

holds, for all 1 ≤ p ≤ ∞.

Historic sources

 * (tr. Course on Analysis of the Royal Polytechnic Academy, part 1: Algebraic Analysis)
 * . English translation:
 * (tr. New proof of the theorem that every integral rational algebraic function of one variable can be resolved into real factors of the first or second degree).
 * – first proof.
 * – second proof.
 * – third proof.
 * – fourth proof.
 * (The Fundamental Theorem of Algebra and Intuitionism).
 * (tr. An extension of a work of Hellmuth Kneser on the Fundamental Theorem of Algebra).
 * (tr. On the first and fourth Gaussian proofs of the Fundamental Theorem of Algebra).
 * (tr. New proof of the theorem that every integral rational function of one variable can be represented as a product of linear functions of the same variable).
 * (tr. New proof of the theorem that every integral rational function of one variable can be represented as a product of linear functions of the same variable).

Recent literature

 * (tr. On the history of the fundamental theorem of algebra: theory of equations and integral calculus.)
 * (tr. The rational functions §80–88: the fundamental theorem).
 *
 * – English translation of Gauss's second proof.
 * (tr. On the history of the fundamental theorem of algebra: theory of equations and integral calculus.)
 * (tr. The rational functions §80–88: the fundamental theorem).
 *
 * – English translation of Gauss's second proof.
 *
 * – English translation of Gauss's second proof.
 *
 * – English translation of Gauss's second proof.
 * – English translation of Gauss's second proof.
 * – English translation of Gauss's second proof.

title: Fundamental theorem of algebra


text:
Talk:Twin prime conjecture

Removed text
I removed the following passage:
 * It's unknown if Gödel's incompleteness theorem has something to do in this conjecture. If the first statement of TPC is undecidable,it means no computer and no human can ever found either there is a biggest pair or it's endless.However, the problem does have an answer even if that's the case.

This text doesn't seem to have any clear meaning (at least, not clear and correct at the same time). --Trovatore 05:16, 2 March 2006 (UTC)

I removed this version too:
 * The statement there are an infinite number of primes p such that p + 2 is also primecould be an unprovable statement.If it is, that means no computer and no human can ever know either there is a biggest pair or it's endless.

Let me explain in more detail what's wrong with it:
 * There's no such thing as "an unprovable statement" full stop. (For example, any statement is provable if you add it as an axiom. Of course you may wind up proving everything in that case, but you will make the original statement provable.)
 * So something needs to be said about the formal theory in which TPC might be unprovable. Is it Peano arithmetic? Is it ZFC? What about adding large cardinal axioms?
 * But even if the text were changed to reflect the above, the stuff about no one ever being able to know whether TPC is true simply does not follow. Perhaps there are ways of knowing it that go beyond the formal theory from which TPC was shown unprovable.
 * (Of course, this still leaves out the technical possibility that TPC might be proven false from a theory we believe in, such as one of those listed in the second bullet point. Personally I just don't see that happening, but still, if something were added about independence/undecidability, the language probably shouldn't exclude the possibility outright.) --Trovatore 15:29, 2 March 2006 (UTC)

Are you saying that answering to the TPC question is simply very difficult? Some of the books I've read insist that knowing the answer could be impossible. (such as David Deutsch's "The Fabric of Reality" and Satoy's "The Music of Primes".)WAREL 16:36, 2 March 2006 (UTC)

 * I'm saying that simply showing that TPC is independent of some fixed formal theory, say PA or ZFC, is not sufficient to show that the answer is unknowable. It's conceivable, I suppose, that it is unknowable, but no techniques currently exist for proving that something is unknowable (and it's hard to imagine what such a technique could look like). --Trovatore 16:42, 2 March 2006 (UTC)

So, you're saying that we definitely could "know"(in general sense) the answer to the TPC sometime.Correct?WAREL 16:53, 2 March 2006 (UTC)

 * "Definitely could" is a tricky concept. I'm saying that no independence result, of the sort we currently understand, can rule out the possibility of knowing the answer. That doesn't necessarily mean we ever will know the answer, just that the techniques you seem to be talking about are not strong enough to rule out the possibility that we will. --Trovatore 17:44, 2 March 2006 (UTC)

Is it possible that this question is determined as we will not ever know the answer?WAREL 20:05, 2 March 2006 (UTC)

 * I'm reluctant to say it's impossible, but I don't see how we could come to know that we will never know it, short of God showing up and telling us we'll never know. --Trovatore 20:07, 2 March 2006 (UTC)

So,the answer is "it's possible". Isn't it? And also, it's possible that someone will show that we could never know the answer.Isn't it?WAREL 20:20, 2 March 2006 (UTC)

 * A very slippery word, "possible". I do not claim to know that no one will ever show we can't know the answer. That does not, in my estimation, justify making the claim that such a situation is "possible"; I think that would be severely misleading. --Trovatore 20:23, 2 March 2006 (UTC)

Why does it not justify making the claim that such a situation is "possible"?WAREL 01:55, 3 March 2006 (UTC)

 * Because it sounds like a positive affirmation, when it's actually nothing but an admission that I don't have a complete theory of knowledge, which is what I'd need (or something close to it, anyway) to completely rule out future knowledge that we'll never know the answer. --Trovatore 03:07, 3 March 2006 (UTC)

I have a question. You said it's possible that this question is determined that we will never know the answer. Why is that?WAREL 07:47, 6 March 2006 (UTC)

 * No, I did not say that. You keep trying to get me to say that. --Trovatore 14:48, 6 March 2006 (UTC)

Do you mean we're just too stupid yet to know the answer?WAREL 17:05, 6 March 2006 (UTC)

Du's proof
Does anyone have any information of someone talking about Du's proof?I wonder if it's valid or not.WAREL 07:08, 3 March 2006 (UTC)

 * (With due respect to the authors of that paper.) If we listed every time someone claimed to have proved the twin prime conjecture, this would be an awfully long and uninformative article. I am removing the reference to that proof, as the arxiv does not count as a peer-reviewed publication. Dmharvey 13:25, 3 March 2006 (UTC)

Is is not reviewed? How did you know that? WAREL 14:43, 3 March 2006 (UTC)

 * (1) I can't find anything on Mathematical Reviews Online to suggest that it has been published. (2) The arxiv, by definition, is not peer-reviewed, except for a brutal first pass which culls out any obvious nonsense. See Arxiv. (3) Also read peer review, especially the section entitled Criticisms of peer review. Dmharvey 14:59, 3 March 2006 (UTC)

What do you rely on when a mathematician comes up with a new result? I wrote in wikipedia perfect number section recently that odd perfect number N has at least 9 distinct prime factors, and at least 12 if 3 does not divide N (Nielsen 2006). I knew it from Arxiv.But I hear no one complaining about this result.Is it because it's trustable?How could you tell untrustable results from trustable results?WAREL 17:44, 3 March 2006 (UTC)

 * This is a very good question, and I can't give you any clear-cut answer. I saw your edits to perfect number and was tempted to revert, but did not, mostly because I don't know anything about work in that area, and I didn't have the energy to chase up any references you gave. I didn't even realise that your source was the arxiv. (You might consider reverting those edits yourself :-)) The situation with twin primes is quite special. The twin prime conjecture is one of those very famous statements which gets proved every second week. Therefore I set the bar much higher, and certainly the arxiv is not good enough. If someone very famous claimed to have proved the twin prime conjecture and had a pre-print on the arxiv, and if the mathematical world was buzzing with talk about it, then perhaps I would consider mentioning a rumour, but even then it's probably not good form. The thing about the arxiv is that it's not very difficult to have things posted there. Pretty much anyone can upload whatever they like;

there are moderators who glance over it briefly to make sure it's not pornography or random characters, and then there it is. One of the main reasons for its existence is to help people establish that they were the first ones to write something down. I don't even trust things I read in journals — but it's the best we've got as a community, and it seems to be the level of trust that Wikipedia generally considers acceptable. You might want to reconsider your approach to how much you trust things you read on the arXiv. After all, you don't trust things you read on Wikipedia, do you? :-) Dmharvey 18:15, 3 March 2006 (UTC)

So,what you're essentially saying is that Du is not famous enough ,right? But the time someone who's very famous says "I read it through ,and it's correct." is when you trust.I'm not blaming your attitude,though. I's just curious about what your "community" was. WAREL 18:31, 3 March 2006 (UTC)

 * Well, it's true that I have never heard of Du, but that doesn't mean that much, as you seem to be suggesting. As for my community, well I am a graduate student in number theory, I spend several tens of hours each week hanging around a mathematics department, surrounded by people who live and breathe mathematics, earn their daily living from mathematics, and some of whom are reasonably famous in their fields. This micro-community is part of a much larger macro-community of mathematicians around the world. Again, maybe that doesn't mean much to you, although it does mean a lot to me. I should also point out that there are different levels of trust. For example, I've carefully considered Euclid's proof that there are infinitely many prime numbers, and I believe it, and I don't think I need anyone any more to trust; I don't even need to trust Euclid himself any more. I also happen to believe that Fermat's last theorem is true, basically because the proof has been accepted by the community that I am part of,

even though I don't understand the proof completely myself. But I don't trust it as much as I trust Euclid's proof. My point is that in Wikipedia, as it's a collaborative project, we don't have the luxury of writing down whatever we personally believe in. There is a broad consensus that academic journals are generally good enough (although I'm sure that gets debated from time to time), but that things like the arXiv are probably not. It's possible that other editors (such as yourself) disagree with me about this last sentence. Dmharvey 18:49, 3 March 2006 (UTC)

 * Agree with Dmharvey here. WAREL, it is good if you have references for what you write, and it should be from either books or respectable journals. Bad information gives Wikipedia a reputation of unreliability, and that is not good. Oleg Alexandrov (talk) 19:26, 3 March 2006 (UTC)

Is someone even reading Du's thesis to see if it's true? If no one is reading just because he's not famous, it's not fair to say his proof is probably wrong,is it? WAREL 22:12, 3 March 2006 (UTC)

 * What about the proof outlined here? Or this one? Or this one? Or this one? And this person claims to have proved it, without any details provided. Have you checked all of those proofs? It's not fair to say their proofs are probably wrong, is it? Are you aware of this proof, which originally appeared on the arXiv, and recently (2004) generated a lot of quite serious mathematical interest.... but turned out to be wrong? Dmharvey 22:36, 3 March 2006 (UTC)

So, are you saying that Du's paper is not worth reading? If so,is it unworth for you or the entire mathematical community? WAREL 22:45, 3 March 2006 (UTC)

 * No, I'm not saying that it's not worth reading. It may well be. If I hear from people I trust that it's worth reading and might be correct, then I may well read it. But otherwise I'm not planning to read it, because reading things (especially the kind of reading one must do to verify a proof) is time-consuming, and I have lots of other things to do (like have interesting conversations on wikipedia :-)). What I'm trying to say is that I don't currently have any evidence that anyone has read it and checked it, let alone that anyone whose mathematical skills I trust has read it and checked it. If it appeared in a professional journal, that would be some indication that such a thing has happened. It wouldn't mean that the proof is correct — sometimes even journals publish things that turn out to be wrong — but it would sure increase my interest. Dmharvey 22:51, 3 March 2006 (UTC)

Are all people you trust saying that Du's paper is unworth reading?WAREL 23:01, 3 March 2006 (UTC)

 * I haven't asked any of them. Dmharvey 23:01, 3 March 2006 (UTC)

So I'll wait to write about it until then.WAREL 23:05, 3 March 2006 (UTC)

 * Indeed, the job of Wikipedians is not to proofread any paper which shows up at ArXive. Our job is to include only well-established and published material on which there is consensus that it is correct. WAREL, in case you did not read that, you may want to take a look at Wikipedia's policy about No original research. Oleg Alexandrov (talk) 23:32, 3 March 2006 (UTC)

No, I just asked if there "is" a consensus.WAREL 00:01, 4 March 2006 (UTC)

Is it alright with everybody to have some sentences of Du's result once one of the people who Dmharvey trust said it's worth reading?WAREL 04:16, 4 March 2006 (UTC)

 * That's fine with me. As soon as someone I trust tells me it's worth reading, I'll let you know. By the way WAREL, your English is pretty good. Some problems with grammar sometimes, but not bad. Where did you learn English? Dmharvey 12:42, 4 March 2006 (UTC)

 * That would be a nice conclusion to this fruitful discussion, were it not for the fact that WAREL keeps inserting the link to Du's paper elsewhere . I will thus share my findings.

 * The paper is not worth reading, and in fact, it is not worth the paper on which I printed it. Leaving aside the insanely obfuscated notation which the authors set up, the proof of the main lemma 4.5 is invalid, it is a meaningless manipulation with symbols. The nonsense actually begins in definition 2.3 and the ensuing discussion leading to eq. 2.19, but until lemma 4.5 it does not bite. The rest of the paper seems to be more or less correct, but it only consists of a few elementary properties of dubious value, and more to the point, it does not get us any closer to a proof of the twin prime conjecture. -- EJ 01:54, 5 March 2006 (UTC)

In fact, it looks like they also proved Goldbach's conjecture, one week earlier:. What's fascinating about these two papers is how incredibly similar they are; the lemma numbering matches up beautifully. Dmharvey 02:51, 5 March 2006 (UTC)

 * LOL! I can't wait when they will extend the method to cover Riemann Hypothesis as well. -- EJ 03:26, 5 March 2006 (UTC)

Look carefully,though. They didn't say they'd proved Goldbach's theorem.WAREL 05:13, 5 March 2006 (UTC)

 * Okay, now I have looked carefully. The abstract says "We proved that any even number not less than 6 can be expressed as the sum of two old primes, 2n=p_i+p_j." I don't know what an "old prime" is, but seeing as the only other time the word "old" appears in the article is in the word "Goldbach" and "oldest unsolved problem", I think it's safe to say that it's a typographical error, and is supposed to be "odd". Then, theorem 1.1 (the main "theorem" of the paper) is stated as follows: "There is at least one pair of double primes for any even number $$2n=p_i+p_j \geq 6$$. Again, I don't know what a "pair of double primes is", but I take it to mean a pair $$p_i, p_j$$ of primes numbers such that $$p_i + p_j = 2n$$." This is in fact what is suggested by their statement that "[The goldbach conjectures states that] for any even number 2n there exists a pair of double primes $$(p_i, p_j)$$ such that $$2n=p_i+p_j$$. Additionally, in the example for n = 23 given just

before definition 2.2, they state that there are four double prime pairs, namely (3, 43), (5, 41), (17, 29), (23, 23), which agrees with this meaning. I'm guessing that the strange English grammar is a literal translation of some Chinese grammar that hasn't worked too smoothly. I don't know enough Chinese to say. So yes, it certainly looks to me like they are claiming they have proved the Goldbach conjecture. Dmharvey 12:24, 5 March 2006 (UTC)

 * One could also point out that the statement and proof of corollary 6.2 is almost free of such grammar errors, and makes clear what they are up to. -- EJ 17:14, 5 March 2006 (UTC)

WAREL, I can't help but notice that you still did not provide any evidence that Du's proof is notable, let alone accepted. You agreed above that you will wait with such additions until people who Dmharvey trusts say that it's worth reading. Can I ask why don't you keep your word? - EJ 17:14, 5 March 2006 (UTC)

Their English is indeed terrible. Have you found any mathematical mistakes that make the tpc theorem not working?WAREL 20:30, 5 March 2006 (UTC)

 * Did you read my post above? Once again: everything from definition 2.3 till the end of section 2 is complete nonsense. Given a set of integers of size t, there is no way of computing the number of its elements which are not congruent to 0 or -2 modulo p, knowing t and p alone. It can be as low as 0, and it can be as large as t. -- EJ 21:08, 5 March 2006 (UTC)

anon external links
An anon using a Rogers account keeps pushing links to the work of Martin C. Miner, owner of a Canadian-based real estate website, relating to probabilistic arguments in favor of TPC. Now, I'm in favor of mentioning such arguments; it's my understanding that there are excellent heuristic reasons to believe TPC is true, if not quite as obvious as the ones for Goldbach's conjecture (the latter you can practically work out in your head). But I don't see any reason to think the best arguments along this line come from this non-peer-reviewed website. Moreover I'm suspicious that the anon may be Miner.

It would be good if someone could write something sourceable about the probabilistic arguments. --Trovatore 18:46, 11 March 2006 (UTC)

 * Well, someone already did. We have a whole section on the Hardy-Littlewood conjecture. -- EJ 19:20, 11 March 2006 (UTC)
 * Ah, I see now. I should have read the article more carefully. --Trovatore 19:27, 11 March 2006 (UTC)

The name is Winer and this is Winer speaking now. One of my students was trying to place my articles on wikipedia after he noticed they were removed. He was shocked at the arrogance he encountered in trying to replace them. Quite frankly, so am I. These are 'external links' not legal testimony. If this is how mathematicians work, then there's no wonder you're all stuck in the stone ages. —The preceding unsigned comment was added by 38.116.204.67 (talk • contribs) 18:40, 13 March 2006   (UTC)

 * Wikipedia's position on original research is clear. Unless your proofs (or commentary on those proofs) are published in a peer-reviewed journal, they don't exist for the purpose of Wikipedia.  This is not necessarily how mathematicians work -- although a quick study of one of the links shows you've defined "random" sufficiently differently than the standard definition that one would suspect other terms being redefined as well -- but it's how a non-expert-based encyclopedia works.  — Arthur Rubin | (talk) 19:14, 13 March 2006 (UTC)

 * I also notice that "you" (the author of the website referred to) are proud of your links from dmoz.org and Google directory. You should note that the Google directory is a copy of dmoz.org, and the Open Directory Project standards do not require notability or accuracy. — Arthur Rubin | (talk) 19:54, 13 March 2006 (UTC)

 * Thanks for your input. In the year 2350 when they discover that I was right... I'm sure history will take note of your position. —The preceding unsigned comment was added by 72.136.90.95 (talk • contribs) 03:56, 14 March 2006   (UTC)

Merge with Twin prime?
Please discuss at Talk:Twin prime --Trovatore 21:18, 13 March 2006 (UTC)

Since several people have supported this, and none opposed it, I have done the merge. JamesBWatson (talk) 14:59, 13 January 2010 (UTC)

A valid proof of the twin prime conjecture or just another conjecture?
Euclid’s proof of the potential infinitude of the prime numbers declares that, for the first n prime numbers p1, p2, p3, …, and pn, there is at least one prime number among the natural numbers inclusive between pn + 1 and p1p2p3…pn + 1 since the latter is either prime to all the pi (1 &le i &le n) or is divisible by a prime number greater than pn.

It is clear that the argument applies just as well to p1p2p3…pn – 1. Thus, one could readily assert that there exists “infinitely many” natural number n such that, for the first n prime numbers, both p1p2p3…pn –1 and p1p2p3…pn + 1 are prime numbers. For examples, for n = 2: (2)(3) – 1 = 5 and (2)(3) + 1 = 7; for n = 3: (2)(3)(5) – 1 = 29 and (2)(3)(5) + 1 = 31; for n = 5: (2)(3)(5)(7)(11) – 1 = 2309 and (2)(3)(5)(7)(11) + 1 = 2311; etc. [in general, there are “infinitely many” prime numbers of the form: a power of 2 times a product of powers of odd prime numbers plus or minus 1 &mdash for examples, (22)(3) &plusmn 1 = 11 or 13; (2)(33)(5) &plusmn 1 = 269 or 271; etc.].

The “arguable issue” is whether this contention is already a valid proof of the twin prime conjecture or just another “conjecture” considering that it could not also be actually completely verified nor disproven by a counterexample --- that is, one cannot establish that there exists some natural number y such that, for all natural number z > y, both p1p2p3…pz – 1 and p1p2p3…pz + 1 could not be prime numbers.

Please read my discussion text Prime number: About the incomplete totality of the set of all prime numbers in the Wikipedia article Prime number. [BenCawaling@Yahoo.com] BenCawaling 10:34, 16 April 2006 (UTC)

 * This is not a proof. It's plausible -- but I don't think it's even a sketch of a proof.  — Arthur Rubin | (talk) 13:58, 16 April 2006 (UTC)

 * That is my general idea --- any argument purported to be a "proof" of the twin prime conjecture would just be another conjecture. BenCawaling 07:53, 18 April 2006 (UTC)

 * I don't generally like to be blunt, but there are exceptions. That last statement is abject nonsense: if a proof was found it would be a theorem, not a conjecture, just like any other theorem in number theory. In addition, your third paragraph shows that you are very confused. You would benefit from realizing this, and go and refresh your memory as to what a proof is. Meanwhile, such posts are not appropriate for the discussion page of an article on mathematics - this is an encyclopedia which presents accepted results, not alt.philosophy. Elroch 09:31, 18 April 2006 (UTC)

Impossible Statement
"In 1940, Paul Erdős showed that there is a constant c < 1 and infinitely many primes p such that p′ - p < c ln p where p′ denotes the next prime after p."

All primes are integers. Therepore the difference of two consecutive primes could never be less than one, as this statement implies. Is there a type somewhere in there, or is this just plain untrue? -- He Who Is[ Talk ] 23:16, 26 July 2006 (UTC)

 * I got caught with this one already! It is not just p' - p < c, look carefully and you will see it is p' - p < c ln p.  I skipped over the natural logarithm of p every time I read it until someone pointed it out to me! Gondooley 23:42, 26 July 2006 (UTC)

 * I have changed the layout to make it more clear. (I hope! Let me know what you think!) Gondooley 23:46, 26 July 2006 (UTC)

Looks much better. I'm just in awe now at how I actually never noticed that. Thanks. -- He Who Is[ Talk ] 18:29, 27 July 2006 (UTC)

 * No problem. It is humbling isn't it! :-) Gondooley 00:23, 28 July 2006 (UTC)

Goldbach
If this proof were false, then there would be a finite number of primes such that p - p' = 2, and goldbach's conjecture would be rendered false. If this theorem were prooved correct it would be a significant step to prooving Goldbach's conjecture.

The converse is also true. If Goldbach's major or minor conjecture were proved correct, this conjecture would automatically be prooved alongside it.

I would not be surprised if Goldbach's conjecture actually derives from this one, or was produced while attempting to solve it.

Corrected twin primes constant
I corrected 0.66016118158468695739278121100145 to 0.6601618158468695739278121100145. (Am I geeky, or what?) One of the digits had somehow gotten doubled. I also corrected it on the French Wikipedia, but will leave other languages for those more familiar with them than I am. The incorrect number has propagated to numerous non-wiki websites. Keith Lynch 21:51, 15 October 2006 (UTC)

Significance?
Is the twin prime conjecture important? Does it relate to other results in number theory, in math, in computer science, in theoretical physics, ...? All that I see here is a mention of sieve theory --- would it be a stretch to relate the TPC to integer factoring/discrete logarithm algorithms, and hence to encryption?

Since this problem is so famous and so accessible to the non-mathematician, an encyclopedia needs some explanation of why mathematicians study it. If the only reason is "it's old and simple but surprisingly hard", then we'll have to say that. Joshua R. Davis 15:53, 5 February 2007 (UTC)

 * I don't know a relation to integer factoring, discrete logarithm algorithms, or encryption. An explanation of why it is studied must be based on reliable sources. Maybe it's partly because "it's old and simple but surprisingly hard", but unsourced speculation from editors would be original research. I don't think an encyclopedia "needs" to explain it, although it might be nice. The twin prime conjecture by itself doesn't seem to me to have great significance for other things, but maybe techniques used in a proof could be used for many other things - and maybe the first proof (if one is found) will just be a special case of a much more general theorem. PrimeHunter 16:16, 5 February 2007 (UTC)

 * I agree with the need for sources, and I don't intend to invent applications for the TPC. That's why I was asking for them here. But Make technical articles accessible and Manual of Style (mathematics) (see also WikiProject Mathematics/Motivation for discussion) encourage motivation, "practical applications", etc. Cheers. Joshua R. Davis 16:46, 5 February 2007 (UTC)

Wiener-Ikehara theorem
I moved the following fragment which was added to the article by User:Karl-H here.

 * Using Wiener-Ikehara theorem, the 'Hardy-Littlewood' conjecture is equivalent to the asymptotic behavior:

$$ \psi _{2} (x) = \sum_{n=1}^{x} \Lambda (n) \Lambda (n+2)\sim  C_2 x $$

Or equivalently the series:

$$ g(s)= \sum_{n=1}^{\infty} \Lambda (n) \Lambda (n+2)n^{-s}$$

has a pole at s=1 with residue $$ C_2 $$

It uses unexplained notation, it doesn't cite a reference, and it's not clear at all how this equivalent formulation is in any way useful. -- Jitse Niesen (talk) 03:10, 24 March 2007 (UTC)

You can take a look at the paper ' Distributional Wiener-Ikehara theorem and twin primes' by Jacob Korevaar, published ( i downloaded the e-print but it's published) at Indag. Mathematica N.S 16 (1) 37-49 in case you have a doubt i can send you the .pdf --Karl-H 15:01, 24 March 2007 (UTC)

 * Then you should have referenced Korevaar, explained that Λ is the von Mangoldt function, and made an effort to embed this remark in the overall narrative. Moreover, it seems to me that the paper just states that the conjecture is equivalent to
 * $$ \psi _{2} (x) = \sum_{n=1}^{x} \Lambda (n) \Lambda (n+2)\sim  C_2 x $$
 * That's apparently supposed to be obvious. The Wiener-Ikehara theorem comes in later. I do hope that anything you add to the article is correct. -- Jitse Niesen (talk) 13:18, 29 March 2007 (UTC)

My Conjecture
There exists a twin prime pair between n and 2n, for all integers n > 6.

In June 2007, I made this conjecture after a long search, first by hand, and then by computer, revealed no counterexamples to my conjecture.

PhiEaglesfan712 - June 26, 2007
 * Talk pages are for discussing improvements to the article and not for general discussion of the article topic. If no reliable source has mentioned your conjecture then it's unsuitable for the article. That being said, twin primes appear to be fairly common and your conjecture seems likely to be true. I just verified it for n < 10100 in a few minutes with PARI/GP. PrimeHunter 22:09, 26 June 2007 (UTC)

 * Nice conjecture. I'm not aware of any prior paper.  I suggest you publish it.  Giftlite 22:31, 26 June 2007 (UTC)

 * It's easy for amateur mathematicians to make plausible prime conjectures which nobody can currently prove or disprove. There are probably thousands of such conjectures on the Internet (I have seen hundreds). I think it would be very hard to get a serious math journal to publish this conjecture. It's also quite possible that it follows from other published conjectures about twin prime occurrences. PrimeHunter 23:23, 26 June 2007 (UTC)

Your conjecture is well-known. It follows, for instance, from the Hardy-Littlewood conjectures with a reasonable error term by an argument identical to how one shows that Bertrand's Postulate follows from the Prime Number Theorem (with an error term). -MM 10/3/08 —Preceding unsigned comment added by 130.74.180.190 (talk) 15:47, 3 October 2008 (UTC)

Claim to be from Euclid
I have removed the claim that this conjecture was proposed by Euclid. This statement was added a long time ago by 68.5.186.133. The sole other contribution made by this user is an edit to Euclid that is evidently nonsense. Eric119 (talk) 04:50, 8 April 2008 (UTC)
 * I agree with the removal. In Daniel Goldston (or somebody identifying as him) says: "No one really knows if Euclid made the twin prime conjecture. He does have a proof that there are infinitely many primes, and he or other Greeks could easily have thought of this problem, but the first published statement seems to be due to de Polignac in 1849". PrimeHunter (talk) 12:59, 8 April 2008 (UTC)

History
The article lacks a reasonable History section. Is this the oldest unsolved conjecture in mathematics? Albmont (talk) 13:39, 20 October 2008 (UTC)

title: Talk:Twin prime conjecture