6 ms·

Observe that there is no prime number which is assigned a nonzero exponent by p_n but not p_m, and there is no prime number which is assigned a nonzero exponent

Observe that there is no prime number which is assigned a nonzero exponent by p_n but not p_m, and there is no prime number which is assigned a nonzero exponent by p_m but not p_n. Herein is the problem. This observation of yours needs to be proven and is in fact the whole point of the proof of the Fundamental Theorem of Arithmetic. That's the hard part. You'll also need to use the fact that every nonempty set of the positive integers has a least element. The reason your proof didn't seem difficult is because you glossed over the difficult parts and your proof isn't a proof.

thaumasiotes 10y ago
That sentence is followed by a proof. You can say the proof is wrong, but you're missing the point to say "it needs to be proven".
- yequalsx 10y ago
  If p_n assigned a positive exponent to any prime c while p_m assigned c a zero exponent, then the product of p_n would be congruent to 0 (mod c), but the product of p_m would not, and therefore the two products would not equal the same number. This is basically just a restatement of the Fundamental Theorem of Arithmetic. See Gower's Answer 3 in his post.
  - thaumasiotes 10y ago
    No, it's not, and I specifically quoted the part of Gower's Answer 3 that mentions it in my original comment. If you think this is a restatement of the fundamental theorem of arithmetic, maybe you think of the FTA as obvious after all. ;p
  - makomk 10y ago
    Nope, it's actually a restatement of a lemma traditionally used to prove the Fundamental Theorem of Arithmetic, and quite a nice one too in my opinion. (It's "obviously" true in the sense that any exception feels like it'd violate basic behaviour we'd expect from modulo arithmetic on primes, and with a bit of head-scratching it even seems to be possible to prove that.)
- cousin_it 10y ago
  Why can't the product of two primes A and B be congruent to 0 modulo some other prime C? It's trivial that A * B can't be equal to C, but not as easy to show that it can't be 2 * C, say.
  - thaumasiotes 10y ago
    This was answered in the same comment tree before you posted the question: https://news.ycombinator.com/item?id=11953790 https://news.ycombinator.com/item?id=11953790
    - cousin_it 10y ago
      Yes, I know the proof. The point is that your original proof sketch didn't acknowledge the need for that lemma.
- GFK_of_xmaspast 10y ago
  If it's lacking a correct proof it needs to be proven.
- rickhanlonii 10y ago
  To be fair, you were not rigorous enough to call that a proof and I think you know that. So there's really no point in arguing semantics here, it's just giving maths a bad look.
  - thaumasiotes 10y ago
    > To be fair, you were not rigorous enough to call that a proof and I think you know that. On the contrary. Referring to a well-known result without proving it in situ is very common. For example, I also assume without proving that if two integers a, b have the property that ka = kb, then a is equal to b. You don't see anyone complaining about that.
deleted 10y ago
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thaumasiotes 10y ago

That sentence is followed by a proof. You can say the proof is wrong, but you're missing the point to say "it needs to be proven".

yequalsx 10y ago
If p_n assigned a positive exponent to any prime c while p_m assigned c a zero exponent, then the product of p_n would be congruent to 0 (mod c), but the product of p_m would not, and therefore the two products would not equal the same number. This is basically just a restatement of the Fundamental Theorem of Arithmetic. See Gower's Answer 3 in his post.
- thaumasiotes 10y ago
  No, it's not, and I specifically quoted the part of Gower's Answer 3 that mentions it in my original comment. If you think this is a restatement of the fundamental theorem of arithmetic, maybe you think of the FTA as obvious after all. ;p
- makomk 10y ago
  Nope, it's actually a restatement of a lemma traditionally used to prove the Fundamental Theorem of Arithmetic, and quite a nice one too in my opinion. (It's "obviously" true in the sense that any exception feels like it'd violate basic behaviour we'd expect from modulo arithmetic on primes, and with a bit of head-scratching it even seems to be possible to prove that.)
cousin_it 10y ago
Why can't the product of two primes A and B be congruent to 0 modulo some other prime C? It's trivial that A * B can't be equal to C, but not as easy to show that it can't be 2 * C, say.
- thaumasiotes 10y ago
  This was answered in the same comment tree before you posted the question: https://news.ycombinator.com/item?id=11953790 https://news.ycombinator.com/item?id=11953790
  - cousin_it 10y ago
    Yes, I know the proof. The point is that your original proof sketch didn't acknowledge the need for that lemma.
GFK_of_xmaspast 10y ago
If it's lacking a correct proof it needs to be proven.
rickhanlonii 10y ago
To be fair, you were not rigorous enough to call that a proof and I think you know that. So there's really no point in arguing semantics here, it's just giving maths a bad look.
- thaumasiotes 10y ago
  > To be fair, you were not rigorous enough to call that a proof and I think you know that. On the contrary. Referring to a well-known result without proving it in situ is very common. For example, I also assume without proving that if two integers a, b have the property that ka = kb, then a is equal to b. You don't see anyone complaining about that.

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