In the first case, starting with p such that the highest power of 2 that divides p is an integer power of 2 (2^k for some integer k); then the highest power of 2 that divides p² is 2^2k; then the highest power of 2 that divides 2q² is also 2^2k; then the highest power of 2 that divides q is 2^(2k-1); therefore q must be a multiple of 2^(k-0.5); a noninteger power of 2.
This implies that there is a number 2^(0.5). It makes no claims as to whether or not this number is rational, or integer; it merely claims that such a number must exist. (Consider: if I had started instead with the equation x²-4=0, I would have ended up showing that a number of the form 4^(0.5) must exist—that number is rational, is indeed an integer).
Now, I think I can prove that an integer q which is a multiple of 2^(k-0.5) but which is not a multiple of 2^k, for integer k, does not exist; but I can only complete that proof by knowing in advance that 2^0.5 is irrational, so I can’t use it to prove the irrationality of 2^0.5. I can easily prove that a rational number of the form 4^(k-0.5) for integer k does exist; indeed, an infinite number of such numbers exist (examples include 2, 8, 32).
No matter how forcefully that first passage conveys the irrationality of √2, it does not prove it.
The paragraph, of course, was talking about integer powers of 2 that divide p. As in, the largest number 2^k such that 2^k divides p and k is an integer.
The largest real power of 2 that divides p is, of course, p itself, as 2^log_2(p) = p.
Looking over my post again, after a good night’s sleep, I see that it wasn’t as coherent as it appeared to me yesterday. Let me see if I can put my point a little more clearly.
The paragraph centers its claim of the irrationality of √2 on the idea that p² contains exactly twice as many powers of 2 as p does. But that is only true because √2 is irrational, making the demonstration a circular proof.
Consider. If √2 were rational, in the form of z/y for some coprime integers z and y, then it would be easy to find an integer that is not itself an integer power of 2, but whose square is an integer power of 2; z would be such a number.
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.
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.
But [p^2 contains exactly twice as many powers of 2 as p does] is only true because √2 is irrational, making the demonstration a circular proof.
Notice that you are claiming that all possible proofs of the statement “p^2 contains twice as many powers of 2 as p” require asserting without proof that sqrt(2) is irrational.
Why does the prime factorization of integers depend upon something that is, if not irrational, at least certainly not an integer? (Proof: 1^2 = 1, 2^2 = 4, and x ⇐ x^2 by induction.)
In the first case, starting with p such that the highest power of 2 that divides p is an integer power of 2 (2^k for some integer k); then the highest power of 2 that divides p² is 2^2k; then the highest power of 2 that divides 2q² is also 2^2k; then the highest power of 2 that divides q is 2^(2k-1); therefore q must be a multiple of 2^(k-0.5); a noninteger power of 2.
This implies that there is a number 2^(0.5). It makes no claims as to whether or not this number is rational, or integer; it merely claims that such a number must exist. (Consider: if I had started instead with the equation x²-4=0, I would have ended up showing that a number of the form 4^(0.5) must exist—that number is rational, is indeed an integer).
Now, I think I can prove that an integer q which is a multiple of 2^(k-0.5) but which is not a multiple of 2^k, for integer k, does not exist; but I can only complete that proof by knowing in advance that 2^0.5 is irrational, so I can’t use it to prove the irrationality of 2^0.5. I can easily prove that a rational number of the form 4^(k-0.5) for integer k does exist; indeed, an infinite number of such numbers exist (examples include 2, 8, 32).
No matter how forcefully that first passage conveys the irrationality of √2, it does not prove it.
The paragraph, of course, was talking about integer powers of 2 that divide p. As in, the largest number 2^k such that 2^k divides p and k is an integer.
The largest real power of 2 that divides p is, of course, p itself, as 2^log_2(p) = p.
Looking over my post again, after a good night’s sleep, I see that it wasn’t as coherent as it appeared to me yesterday. Let me see if I can put my point a little more clearly.
The paragraph centers its claim of the irrationality of √2 on the idea that p² contains exactly twice as many powers of 2 as p does. But that is only true because √2 is irrational, making the demonstration a circular proof.
Consider. If √2 were rational, in the form of z/y for some coprime integers z and y, then it would be easy to find an integer that is not itself an integer power of 2, but whose square is an integer power of 2; z would be such a number.
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.
Notice that you are claiming that all possible proofs of the statement “p^2 contains twice as many powers of 2 as p” require asserting without proof that sqrt(2) is irrational.
Why does the prime factorization of integers depend upon something that is, if not irrational, at least certainly not an integer? (Proof: 1^2 = 1, 2^2 = 4, and x ⇐ x^2 by induction.)