Talk:Euclid's theorem/Archive 1

p+1 prime?

This article is terrible at the moment. p+1 is not necessarily prime. --Wikiwikiwildwest 07:19, 14 December 2006 (UTC) Erm, p'+1 that is. --Wikiwikiwildwest 07:30, 14 December 2006 (UTC)[reply]

Prime q

"If q is not prime then some prime factor p divides q." Doesn't this hold as well if q is prime? I mean, q divides q, right? So why the distinction whether q is prime or not? 88.65.186.193 (talk) 14:12, 25 April 2009 (UTC)[reply]

What Euclid, in the 3rd century BC, meant by "divides" (the translators say "measures") is not clear from the article. But even if you're right, the way it's presented is how Euclid wrote it. Michael Hardy (talk) 17:28, 19 March 2010 (UTC)[reply]

q prime?

I should point out that that q cannot possibly be prime, because P is the product of odd numbers and, therefore, an odd number itself; and any odd number plus one is an even number, and therefore not a prime number. -189.29.2.176 (talk) 15:04, 19 January 2010 (UTC)[reply]

P is not necessarily a product of odd primes; the number 2 can be included in the initial list, and in some cases in which 2 is included, q is prime. Michael Hardy (talk) 17:26, 19 March 2010 (UTC)[reply]

Deleted text

I made a deletion of some text from the first paragraph. Perhaps this was a strange bit of defacing. More likely, it was an attempt at adding some formatted text that went awry. Assuming this latter case, perhaps this deletion will prompt the author to return and try to again.--138.251.242.34 (talk) 19:16, 10 February 2010 (UTC)[reply]

Erdős proof

It's possible the mathematical language used is just beyond me, but stating that every integer can be written as r*s^2 is only true if you allow for non-integers or s=1. It would be highly helpful for at least my reading of the proof to know whether the variables represent integers or not. 194.16.178.140 (talk) 08:15, 9 March 2015 (UTC)[reply]

Proof using factorials

This proof is almost exactly the same as Euclid's proof. Should it be deleted for that reason? Brianbleakley (talk | contribs) 18:44, 17 August 2016 (UTC)[reply]

Yes, I agree. Maybe it could be mentioned as a variant of Euclid's proof, but then directly after it. Or simply deleted. Sapphorain (talk) 20:14, 17 August 2016 (UTC)[reply]

I moved that material to the end of the Euclid's proof section. Brianbleakley (talk | contribs) 19:17, 23 August 2016 (UTC)[reply]

Comment on the undoing by Sapphorain, 2 April 2017

1. The original wording by Saidak can be of interest on a special Saidak-page, not on the page titled "Euclid's theorem".
2. I cannot see why the 7-lines version of the proof should be more accessible for laymen than the short version. For:
2a. The central argument in the proof is described in essentially the same way in both versions: "For any n > 1, n and n + 1 have no common factors; they are coprime" versus "Because all prime divisors of a natural number n are different from the prime divisors of n+1, . . . ".
2b. The short version doesn't have the "induction for any k".
2c. Nor the superfluous and difficult to read and interprete "example" at the end.
2d. Using N₂, N₃, ..., but not N₁, is confusing.
2e. Why 'at least' in '1806 has at least four different prime factors' ?
Who refutes this arguments? Who has a better proposal for the text of the proof? -- Hesselp (talk) 21:49, 5 April 2017 (UTC)[reply]

1. It is more appropriate because it's clearer, not because it's Saidak's wording.

2. If you cannot see that I hope you are not a maths teacher.

(2c and 2e. This example is perfectly clear and correct. The fact that 1806 happens to have exactly four different prime factors cannot be inferred from the argument). Sapphorain (talk) 09:29, 6 April 2017 (UTC)[reply]

1-bis. The 7-lines version is clearer? In spite of my objections 2bcde ? Explain please.

2-bis. Your "If you cannot see that" is unclear to me. (I've been editor of a magazine on maths for secondary school students. For your information; I wouldn't see it as an argument in this discussion.)

2ce-bis.

i. The long formula in line 7 is complex, and misses appropriate spacing to facilitate reading.

ii. The example doesn't help to understand the preceding explanation; it's an extra puzzle.

iii. The number 1806 has four different prime factors, not at least four. You should introduce a variable in the formula to be able to use "at least" properly. -- Hesselp (talk) 12:22, 6 April 2017 (UTC)[reply]

The existing proof is clearer than what you propose, which is too short and abstract for a non mathematician, and which doesn't contain any example. The typography might not be perfect, be my guest if you want to improve it.

i. Then put appropriate spacing.

ii. But yes, it does help. An example always helps.

iii. Once again, the example is meant to illustrate the general argument, which tells us that, at this stage of the process, 1806 must have at least four different prime factors. The fact that it happens to actually have exactly four is totally irrelevant here. Sapphorain (talk) 16:11, 6 April 2017 (UTC)[reply]

Sapphorain, your judgement "clearer than what you propose" is motivated by "too short", too abstract" (without saying what could be added, and what could be better clarified; of course this depends on the level of the readers you have in mind), and by "lacking any example" (I agree).

On i: Can you see my "Second attempt" or "Second attempt plus" (see below) as an appropriate alternative?

On ii and iii: "An example always helps." Not here. I cannot see that the content of 'line 7' is of any help to an inexperienced reader. It isn't a concrete example of anything. I persist that your interpretation of this sentence is very hard to follow (for me)." -- Hesselp (talk) 22:23, 6 April 2017 (UTC)[reply]

Second attempt to improve the wording

Since each natural number (≥2) has at least one prime factor, and two successive numbers n and (n+1) don't have any prime factor in common, the product n×(n+1) has more different prime factors than the number n itself.
This implies that each term in the infinite sequence: 1, 2 (1×2), 6 (2×3), 42 (6×7), 1806 (42×43), (1806×1807), (1806×1807) × (1806×1807 + 1), · · · has more different prime factors than the preceding.
The sequence never ends, so the number of different primes never ceases to increase. -- Hesselp (talk) 10:18, 6 April 2017 (UTC)[reply]

Second attempt plus

Since each natural number (≥2) has at least one prime factor, and two successive numbers n and (n+1) don't have any prime factor in common, the product n×(n+1) has more different prime factors than the number n itself.
Consider the following special sequence of numbers, with their different prime factors within { } :
1×2 = 2 {2}, 2×3 = 6 {2, 3}, 6×7 = 42 {2,3, 7}, 42×43 = 1806 {2,3,7, 43}, 1806×1807 = ... {2,3,7,43, 13,139},
(1806×1807) × (1806×1807 + 1) {2,3,7,13,43,139, one or more new prime(s) }, et cetera
Observe that each term has more different prime factors than the preceding. The sequence never ends, so the number of different primes never ceases to increase. Q.e.d. -- Hesselp (talk) 22:20, 6 April 2017 (UTC)[reply]

(p_1p_2p_3p_4...)+1 is NOT necessarily prime

This seems to confuse some people. Should it be explicited somewhere? -- aureooms (talk) 22:23, 15 Mai 2017 (UTC) — Preceding unsigned comment added by 2A02:A03F:3D8B:AC00:8282:F490:1E9F:FCE5 (talk)

It seems clear to me. It says that the product (of all known primes) plus one is either prime or not. If it is prime it is a new prime. If not prime it has a prime factor not in the know list, so there must be a new prime. The product is or includes a prime not on the list, and so there is no finite list of all primes. Therefore there are an infinite number of primes.--JohnBlackburne^words_deeds 22:38, 14 May 2017 (UTC)[reply]