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How to prove theorems - August 26, 2010
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Spiked Math Comic - How to prove theorems



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25 Comments

Um, no. As constructed, N could be divisible by primes smaller than p_{n}. The correct conclusion is "By construction, N is not divisible by any of the p_{i}, therefore there is a p_{n+1} not on the finite list. QED (by straight induction)"

Uhm, I'm confused, say I take n=2, then we have the primes p_1=2 and p_2=3. This means that N=6, which is divisible by both 2 and 3. Or are you including 1 in the list of primes?

Don't be hard on me if I make a simple mistake, just trying to learn :)

N=P_1+P_2+1=7
you forgot the +1

ps. I'm an idiot and theres supposed to be a * instead of a + between P_1 and P_2

As constructed, N could be divisible by primes smaller than p_{n}.

Nope, division by any of the p_{n} primes will give a remainder of one.
p_{n}*a = 0 (mod p_{n})
therefore
N-1 = 0 (mod p_{n})
and so
N = 1 (mod p_{n})
QED

(pardon my use of equal signs)

GAH!!! replace p_{n} in that proof by p_{i}...

Oh man..That takes me back..
And yay...I'm first!

Oops...guess I'm not first!

Uhm, I'm confused, say I take n=2, then we have the primes p_1=2 and p_2=3. This means that N=6, which is divisible by both 2 and 3. Or are you including 1 in the list of primes?
unless the 1 was added after, N=2*3+1 = 7, which is not divisible either by 2 nore 3.

As constructed, N could be divisible by primes smaller than p_{n}.
I don't think so. Since all the prime smaller than p_{n} are in the product, N/p_{i} would be equal to p_1*p_2*...*p_{i-1}*p_{i+1}*...*p{n} + 1/p_{i}, which can't be a whole number, since the first part is whole, and p_{i} > 1.

Great, it was a simple mistake... I was assuming p_1+...+p_n+1 instead of multiplying the primes. Thanks :)

what? how is this comic funny?

The "funny" comes in the gray/brown box above the paper with the proof. I always found it rather odd why so many Math proofs use reduction to the absurd.

leave mr troll alone ;)

now otoh... I have been known to be terrible with proofs, not in the sense that I am not good with them, but in that I tend to "skip over" "simple" steps and then get it counted against me :/ I really should stop assuming people think like I do (I can see the logical jump from one of my steps to the next, even if there would be a few steps in between that the instructor would like)

There's that hoary joke about the math prof saying, "Now, from this it is obvious that. . . .
"Is it obvious?"
Shuffles papers.
Runs out of room.
45 minutes pass.
Prof runs back into room, shouts, "Yes! I was right! It IS obvious!" and continues with proof.

It's amusing how many proofs start out assuming you're wrong, proving you're wrong about that, and getting two wrongs to make a right.

That is a good point. Perhaps an alternate joke would be, "Step 1, assume I am wrong. Step 2, conclude that such an assumption is clearly absurd."

Just a short note on everyone saying that p_{n+1} cannot be smaller than p_n: of course it can, since it has nowhere been assumed that we know all primes smaller than p_n already. Suppose for instance that we only know two prime numbers yet, 5 and 7. We then get 2 as a new prime number (and three as well, for that matter).

The theorem is not whether we can find all the primes. It's about whether there is an upper limit on number of primes. So we can safely assume that there are only n primes(initial assumption), whether we can find them or not(or what their values are), is a different issue which doesn't make any difference.

Sorry. I misinterpreted your point(I thought you were discrediting the proof). You're right in that p_{n+1} need not be > p_{i} for all i.

2*3*5*7*11*13+1=30031
30031 is not prime.

30031 is not a prime sure it is 59 * 509.....So it comes under the second case that it is divisible by prime 59 which is greater than 13...

The point is not that it's a new prime that is necessarily greater than the previous primes, but rather that it's a NEW prime, possibly more than one. And, from this prime plus the already discovered primes, we can discover another one or more, and so on.

So his proof is flawed; it's not too hard to fix it.

His proof is not flawed, his (and everybody's) language is.
Just try to translate everything he said to math language and you will see.

How about this( picking up from the definition of N from the proof).
N certainly has a prime factorization which is a list of numbers of length at least 1.
None of the p_{i} are on this list (N=~1 mod (p_{i]))
Therefore any finite list of primes is incomplete; there are infinitely many primes.

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