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Kindly comments on Rationality Quotes September 2012 - Less Wrong
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The proof assumes unique prime factorization. Once we factor p as 2^a 3^b 5^c etc., then we know that p^2 factors as 2^(2a) 3^(2b) 5^(2c) etc. This is (implicitly) how we know that p^2 contains exactly twice as many powers of 2 as p.
If √2 is rational, then √2 can be written as z/y for some integers z and y, where z and y are coprime. Then, 2=z²/y².
Consider the hypothetical integer z. It is equal to √2*y. Since y and z are coprime, y cannot contain a factor of √2. Thus, z does not contain a factor of 2; the highest integer power of 2 that is a factor of z is 2^0.
On the other hand, z² does have a factor of 2; it is equal to 2*y² (since y has no factor of √2, y² therefore has no factor of 2).
Therefore, to claim that p² contains exactly twice as many powers of 2 as p is exactly equivalent to claiming that √2 is irrational.
Even if √2 is rational, it is not an integer, and (especially in your second paragraph) you are trying to do things with it that only make sense with integers.
I don't think that continuing this conversation can be productive, since you seem to be objecting to standard techniques. A rigorous textbook in number theory can probably explain the proof that √2 is irrational more thoroughly, and you can follow back the lemmas and see exactly where your confusion lies.
Isn't the point of math that all mathematical truths are logically equivalent? (In beore Gödel.)
Depends on the axioms you're using.
Well, there's an informal notion that if two theorems T1 and T2 are both true, yet T1 <=> T2 is much easier to prove than either T1 and T2, then the two are equivalent. (There's also the formal notion that two axioms are equivalent if assuming either one lets you prove the other, but I don't think that's especially relevant here.)
That's not close to being the most confused part of the comment you're replying to.