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# The Proof is Trivial! Watch

1. (Original post by ukdragon37)
I just got to reading Lemma 5, and you are committing this fallacy:

An infinitude of X implies an infinitude of Y, satisfying property P with Y being a subset of X.

Which is not true in general without further proof. (It's like saying, there is an infinite number of natural numbers, so there is an infinite number of natural numbers smaller than 10).

In fact this reminds me of a very old Cambridge maths interview question

Spoiler:
Show

Prove there are an infinitude of primes that end in 7.

and quoting the infinitude of primes result is not an acceptable answer!

In the end this Lemma 5 might well be as difficult as TPC.
When you say Y is a subset of X that is not what is happening here at all, since the statement X that there are an infinite number of numbers divisible by 6 must then show that there are an infinite number of numbers not of the form 6n+1, which leads to our conclusion. I dont see the fallacy
2. (Original post by Nebula)
What I would like to see is you using this method to show, say, 95 is not of the form 6ab+-a+-b. That is, how can you get a missing value of 6ab+-a+-b, from restrictions on T?
I am certain that you can you use programs such as wolfram alpha to do this but my claim was not to have a formula but to have an expression that acts as a seive discarding all those integers of the form 6k+-1 that are not prime. I have mathematically proved that if 95 is of the form 6k+-1 then it must have a K of one of those forms mentioned above. If you want you can go ahead and put this into wolfram alpha and test for yourself.
3. (Original post by theuser77)
I am certain that you can you use programs such as wolfram alpha to do this but my claim was not to have a formula but to have an expression that acts as a seive discarding all those integers of the form 6k+-1 that are not prime. I have mathematically proved that if 95 is of the form 6k+-1 then it must have a K of one of those forms mentioned above. If you want you can go ahead and put this into wolfram alpha and test for yourself.
Referring ofc to this specific scenario since we know that 95 is not prime.
4. (Original post by ukdragon37)
Let's look at your statements closely:

"Euclid was able to prove that there is an infinitude of primes." Ok.

"We have already proven that all primes are of the form, ...., furthermore we have shown that an integer of the form is prime if and only if Q is not of the forms stated in Lemma 5." Ok.

"Hence we can conclude that there are an infinite number of Q’s that are required in order to satisfy our requirement that is prime if and only if Q is not of the forms stated in Lemma 5." Not ok. This only holds if the two previous statements can be combined. You are committing the fallacy that there is an infinite number of primes (X) so there is an infinite number of primes of your desired form (Y) such that it will result in your set of Qs being infinite.
OMG im so srry that was from an old copy of the solution, that i have updated massively, so does that mean you have not seen what i have said about when T is divisible by 6. Please completely ignore that euclid stuff let me reupload the updated solution
5. ii.
6. (Original post by theuser77)
OMG im so srry that was from an old copy of the solution, that i have updated massively, so does that mean you have not seen what i have said about when T is divisible by 6. Please completely ignore that euclid stuff let me reupload the updated solution
Go ahead, my evening is free anyways. Do you mean this doc?
7. You still fall into the exact same trap.
8. yes, here i dont talk about euclid i think
(Original post by ukdragon37)
Go ahead, my evening is free anyways. Do you mean this doc?
9. (Original post by Noble.)
You still fall into the exact same trap.
That is not true You have stated that the fallacy is that " there is an infinite number of X which implies an infinite number of Y such that Y is a subset of X" i think but in this case the statement X that there is an infinite number of values ofN not divisble by 6 and the statement Y that there are infinite amount of T that are divisible by 6 are exactly the same are they not
10. (Original post by Noble.)
You still fall into the exact same trap.
(Original post by theuser77)
yes, here i dont talk about euclid i think
Well my primary bone to pick in this one would be that you said:

"Hence we can conclude that when T is divisible by 6, the expressions above cannot exist since leaves a remainder of 1 or -1 upon division by 6", and then say that there is an infinitude of numbers not of the forms you gave because there is an infinitude of numbers divisible by 6.

But T cannot be divisible by 6! You have defined by fiat that T is of the form 6k +- 1, so to assert it is possible for T to be divisible by 6 is a contradiction and you can conclude anything from that. Example: Let T be a natural in the form 10 - k for natural k. Hence k < 10. Suppose T is greater than 10, then clearly there are an infinite number of them. Hence clearly there are an infinite number of possible k.

Is T divisible by 6 or not? If it is, it can't be of the form 6k +- 1. If it is not, then the rest of the proof for the lemma does not follow.
11. Again I think its my explanations that are letting me down here.
Ok so here is what i meant.
consider the expression
a(t)+b
when T is divisible by 6, it cannot be of the form 6b+-1, hence the expressons shown above are impossible., but when T is leaves a remainder of+-1 depending on which expression we are using then T can be expressed in the form 6b-1 and hence such a form is possible. But there is an infinite number of t's that are divible by 6 and hence there are an infinite number of expressions not of the form 6ab+a+b. is that sufficient?
oh yeah and where t is a positive integer.
(Original post by ukdragon37)
Well my primary bone to pick in this one would be that you said:

"Hence we can conclude that when T is divisible by 6, the expressions above cannot exist since leaves a remainder of 1 or -1 upon division by 6", and then say that there is an infinitude of numbers not of the forms you gave because there is an infinitude of numbers divisible by 6.

But T cannot be divisible by 6! You have defined by fiat that T is of the form 6k +- 1, so to assert it is possible for T to be divisible by 6 is a contradiction and you can conclude anything from that. Example: Let T be a natural in the form 10 - k for natural k. Hence k < 10. Suppose T is greater than 10, then clearly there are an infinite number of them. Hence clearly there are an infinite number of possible k.

Is T divisible by 6 or not? If it is, it can't be of the form 6k +- 1. If it is not, then the rest of the proof for the lemma does not follow.
12. (Original post by Noble.)
You still fall into the exact same trap.
(Original post by theuser77)
....
Also I just noticed you have the constraint that Q has to be of the form x^2 - y^2 = 24Q. So I would agree this means again you are falling into my trap where asserting there is an infinite number of Q of the desired form in Lemma 5 (X) implies there is an infinite number of Qs of the form x^2 - y^2 = 24Q (Y).
13. (Original post by theuser77)
Again I think its my explanations that are letting me down here.
Ok so here is what i meant.
consider the expression
a(t)+b
when T is divisible by 6, it cannot be of the form 6b+-1, hence the expressons shown above are impossible., but when T is leaves a remainder of+-1 depending on which expression we are using then T can be expressed in the form 6b-1 and hence such a form is possible. But there is an infinite number of t's that are divible by 6 and hence there are an infinite number of expressions not of the form 6ab+a+b. is that sufficient?
oh yeah and where t is a positive integer.
So you are picking T is divisible by 6, there is an infinitude of T, hence an infinitude of Q not satisfying those forms since those forms are impossible when T is divisible by 6. One problem is my last post above, where this does not imply the infinitude of a subclass of Q which is of the form x^2 - y^2 = 24Q. I'll consider some more.
14. (Original post by theuser77)
Again I think its my explanations that are letting me down here.
Ok so here is what i meant.
consider the expression
a(t)+b
when T is divisible by 6, it cannot be of the form 6b+-1, hence the expressons shown above are impossible., but when T is leaves a remainder of+-1 depending on which expression we are using then T can be expressed in the form 6b-1 and hence such a form is possible. But there is an infinite number of t's that are divible by 6 and hence there are an infinite number of expressions not of the form 6ab+a+b. is that sufficient?
oh yeah and where t is a positive integer.
I'm going to be blunt because you continually seem to argue that you're not "conveying your ideas clearly" - the issue is with the maths. Once you start going down the route of infinite primes and finding an infinite number of solutions you fall into the trap ukdragon37 pointed out. Also, no that isn't sufficient, it's the same trap if you're trying to conclude there are infinite twin-primes from it.
15. (Original post by ukdragon37)
Also I just noticed you have the constraint that Q has to be of the form x^2 - y^2 = 24Q. So I would agree this means again you are falling into my trap where asserting there is an infinite number of Q of the desired form in Lemma 5 (X) implies there is an infinite number of Qs of the form x^2 - y^2 = 24Q (Y).
I didnt say that Q has to be of any form other than the 4 forms i have stated clearly, if you substitute such a Q into the equation, with the only requiremnt of thos above, then we will yield twin prime solution.
that is what i have proved, but with added knowledge that there are an infinite number of Q not of the four forms we can find twin prime solution infinitely often. Btw please critisms of the work and not just vague things surrounding prime numbers. i appreaciate what you guys have been doing as i do believe we are simply fixing this soloutiona as it is certainly not irreperable.
16. (Original post by theuser77)
I didnt say that Q has to be of any form other than the 4 forms i have stated clearly, if you substitute such a Q into the equation, with the only requiremnt of thos above, then we will yield twin prime solution.
that is what i have proved, but with added knowledge that there are an infinite number of Q not of the four forms we can find twin prime solution infinitely often. Btw please critisms of the work and not just vague things surrounding prime numbers. i appreaciate what you guys have been doing as i do believe we are simply fixing this soloutiona as it is certainly not irreperable.
You're just rewording what you've done thinking it changes anything, it doesn't. Saying "but with the added knowledge..." when it's an integral part of your attempted proof doesn't change the fact it doesn't work.
17. (Original post by Noble.)
You're just rewording what you've done thinking it changes anything, it doesn't. Saying "but with the added knowledge..." when it's an integral part of your attempted proof doesn't change the fact it doesn't work.
Yes but your point that i have placed constraints on Q by the equation are invalid since Q can be of any form provided that it is not of the forms in lemma 4. I just think you are thinking about this illogicaly. Ok here are the steps in what my proof is claiming to prove and pls tell me which step is wrong and why:
1. There are infinitely many primes of the form 6k+1 or 6k-1 (proven)
2. twin primes differ by 2 and are of the form 6k+1 and 6k-1 (proven)
up to this point there have been no constraints on what k can be
3. where 6Q+1 and 6Q-1 are equated to x and y , then x^2-y^2=24Q
where Q can be any integer.
4. when Q is not of the forms 6ab+a+b, 6ab+a-b,6ab-a+b or 6ab-a-b
then Q is prime. (proven)
5. Subbing all values of Q not of the forms mentioned in 4 into the equation:
24Q=x^2-y^2, will give solutions of the form 6Q+1 and 6Q-1, but since Q is not of the forms mentioned in 4 the 6Q+1 and 6Q-1 will be a prime pair since they differ by 2.
6.By showing that there are an infinite number of Q's not of the forms mentioned in 4 , we can sub each of those Q's into the equation to give unique primes of the form 6Q+1 and 6Q-1.
7. we can show that there are an infinite number of Q's not of the forms mentioned in 4 by factorise them to reveal they are the same as
a(6b+-1)+-b
so now we can consider ALL numbers of the form 6T+-1. these numbers can never be disivible by 6 and this is obvious.
now consider all numbers of the form
aS+-b. If S is of the form 6b+-1, then 6ab+-a+-b will be the expression that will arise as a result. But if S is not of the form 6b+-1 then a different expression will arise. Henec for all such values not of the form 6b+-1 we have a value of S that ensure that we not have an expression of the 6ab+-a+-b
18. (Original post by theuser77)
I didnt say that Q has to be of any form other than the 4 forms i have stated clearly, if you substitute such a Q into the equation, with the only requiremnt of thos above, then we will yield twin prime solution.
that is what i have proved, but with added knowledge that there are an infinite number of Q not of the four forms we can find twin prime solution infinitely often. Btw please critisms of the work and not just vague things surrounding prime numbers. i appreaciate what you guys have been doing as i do believe we are simply fixing this soloutiona as it is certainly not irreperable.
I'm just going to keep picking holes as that seems to be the best way to get across to you:

"Therefore we can conclude that an integer R of the form 6k +- 1 is prime if and only if k is not of the forms S" where I write S for short of the forms you disallow.

Consider R = 47 = 6*8 - 1, which is prime. But k = 8 is of one of the forms in S, since 8 = 6(1 * 1) + 1 + 1.

Note: I presume you are not allowed to let either a = 0 or b = 0, because then one can express any number n as a form in S simply by letting a = n and b = 0 and n = 6 * 0 + a - 0.

EDIT: Will give the second counterexample some thought
19. (Original post by ukdragon37)
I'm just going to keep picking holes as that seems to be the best way to get across to you:

"Therefore we can conclude that an integer R of the form 6k +- 1 is prime if and only if k is not of the forms S" where I write S for short of the forms you disallow.

Consider R = 47 = 6*8 - 1, which is prime. But k = 8 is of one of the forms in S, since 8 = 6(1 * 1) + 1 + 1.

Note: I presume you are not allowed to let either a = 0 or b = 0, because then one can express any number n as a form in S simply by letting a = n and b = 0 and n = 6 * 0 + a - 0.

EDIT: Will give the second counterexample some thought
I gave him a counterexample to this a couple of pages back, he ignored it and decided to attach a PDF version of his proof instead. Maybe now we've provided two counterexamples he'll LaTeX it up and attach that.

(Original post by Noble.)
Not true. Take and then

is not false prime, it's prime.
and obviously 129 = 6(4*5)+4+5
20. (Original post by Noble.)
I gave him a counterexample to this a couple of pages back, he ignored it and decided to attach a PDF version of his proof instead.

and obviously 129 = 6(4*5)+4+5
My intuition is that a very large class of numbers can be expressed in the forms he described as undesirable, especially if he allows a, b in Z. However I wonder if one can find a composite k such that it is not in one of the forms. Alas, I've ran out of procrastination time for tonight.

But I agree this isn't really heading in any useful direction, since there is no proof that the desirable solutions don't just generate, say, cycles of primes within a finite class.

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