5 ms·

If the proof is simple, give it.

by mjcohen 12y ago

ikeboy 12y ago
I've got something much simpler than what I was thinking not involving induction, so I'll give that instead. Assume n,m,x integer, and n^2/m^2=x. I claim x!=2. Proof: Write out both n and m's prime factorizations. Their squares' prime factorizations each contain an even number of all primes (zero is even). Therefore, for every prime, x's prime factorization contains an even number of it as well. Therefore x is not 2.
- gamegoblin 12y ago
  That is a proof by contradiction that the square root of any prime is irrational.
  - ikeboy 12y ago
    Where is the contradiction? I don't see any false assumptions that then result in contradiction.
    - gamegoblin 12y ago
      https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrationa... Your proof hinges on the not-directly-stated words "so, if x were 2, we would reach an impossibility" where impossibility == contradiction.
      - ikeboy 12y ago
        But that's not the proof I used. My proof was positive. They may be equivalent, but not if we count proof by contradiction as special. In response to edit: No, it doesn't. Having proven that x contains an even number of 2s as factors, it follows that x is not two. 2 does not have an even number of 2s as factors, x does, therefore they are different. No contradiction required.
        Retra 12y ago
        I can't see the difference between your proof and the given one, except that you didn't write yours out with the same degree of formality. If you wrote your proof using only applied axioms and theorems, it would be essentially identical. The key point here is that you are partitioning the possibilities in a binary fashion: either x has even number of prime factors or it does not. You've shown that it must not have an even number of prime factors. You haven't shown that it isn't 2. You implied that if it were true, that would contradict what you have shown. How you go from "X has an even number of prime factors" to "X can't be 2" is important.
        ikeboy 12y ago
        Take two cases. One where x has no factors of two, and one here it has at least 2. In the first case, x is not divisible by 2, while 2 is divisible by 2, therefore x is not 2. In the second case, x is divisible by 4, while 2 is not, therefore x is not 2. You're right that they are "essentially" the same proof, but excluding contradictions is weird, a bit like excluding the letter e. Rephrasing a proof to not use it may be possible in some cases.
        Retra 12y ago
        >Take two cases. This is where you are implicitly employing proof by contradiction. Why do you take two cases and not three? Or four? What assumption allows you to restrict your conclusions like that? Probably the Law of the Excluded Middle: (P or ~P). (This is equivalent to ~~p = p.) Once you assume or derive either of those, you have validated proof by contradiction.
        ikeboy 12y ago
        I proved that x has an even number of factors of 2. Why can't I then split that into scenarios when that even number is 0, and cases where it isn't? I don't see any restrictions here. Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof.

If the proof is simple, give it.

by mjcohen 12y ago

If the proof is simple, give it.

I've got something much simpler than what I was thinking not involving induction, so I'll give that instead. Assume n,m,x integer, and n^2/m^2=x. I claim x!=2. Proof: Write out both n and m's prime factorizations. Their squares' prime factorizations each contain an even number of all primes (zero is even). Therefore, for every prime, x's prime factorization contains an even number of it as well. Therefore x is not 2.

gamegoblin 12y ago
That is a proof by contradiction that the square root of any prime is irrational.
- ikeboy 12y ago
  Where is the contradiction? I don't see any false assumptions that then result in contradiction.
  - gamegoblin 12y ago
    https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrational https://proofwiki.org/wiki/Square_Root_of_Prime_is_Irrationa... Your proof hinges on the not-directly-stated words "so, if x were 2, we would reach an impossibility" where impossibility == contradiction.
    - ikeboy 12y ago
      But that's not the proof I used. My proof was positive. They may be equivalent, but not if we count proof by contradiction as special. In response to edit: No, it doesn't. Having proven that x contains an even number of 2s as factors, it follows that x is not two. 2 does not have an even number of 2s as factors, x does, therefore they are different. No contradiction required.
      - Retra 12y ago
        I can't see the difference between your proof and the given one, except that you didn't write yours out with the same degree of formality. If you wrote your proof using only applied axioms and theorems, it would be essentially identical. The key point here is that you are partitioning the possibilities in a binary fashion: either x has even number of prime factors or it does not. You've shown that it must not have an even number of prime factors. You haven't shown that it isn't 2. You implied that if it were true, that would contradict what you have shown. How you go from "X has an even number of prime factors" to "X can't be 2" is important.
        ikeboy 12y ago
        Take two cases. One where x has no factors of two, and one here it has at least 2. In the first case, x is not divisible by 2, while 2 is divisible by 2, therefore x is not 2. In the second case, x is divisible by 4, while 2 is not, therefore x is not 2. You're right that they are "essentially" the same proof, but excluding contradictions is weird, a bit like excluding the letter e. Rephrasing a proof to not use it may be possible in some cases.
        Retra 12y ago
        >Take two cases. This is where you are implicitly employing proof by contradiction. Why do you take two cases and not three? Or four? What assumption allows you to restrict your conclusions like that? Probably the Law of the Excluded Middle: (P or ~P). (This is equivalent to ~~p = p.) Once you assume or derive either of those, you have validated proof by contradiction.
        ikeboy 12y ago
        I proved that x has an even number of factors of 2. Why can't I then split that into scenarios when that even number is 0, and cases where it isn't? I don't see any restrictions here. Let's imagine I had shown that something was either one or two. You seem to be claiming I could not take positive proofs assuming each of them and turn it into a full proof.