Prove: If integer N>1 is not a perfect square, then sqrt(N) is irrational.
Hello o&apartyrockAssume that is rational; in other words
and assume further that ; in other words, that the fraction is in its lowest terms.
Then
has a factor , since
has a factor (For if not, then is the product of the squares of one or more of the prime factors of , and is therefore a perfect square. Contradiction.)
has a factor
Contradiction.
Therefore cannot be expressed in the form , and is therefore irrational.
Grandad
Almost, but not quite. For example, has a factor but does not have a factor You should take into account not only that is not a perfect square but also that is a perfect square (in this case is not a perfect square).
This is how I would write out the proof. Let where the are distinct primes. Then at least one of the must be odd since is not a perfect square, say is odd.
If where and then
(since
(since is prime)
divides an even number of times
This is a contradiction because divides an odd number of times and and so divides an odd number of times.
Hmm... now you have me slightly more confused! The proof you have provided, proscientia is quite elegant.
I still think Grandad's proof is valid though. Of course "if N divides then N divides p" is not generally true as your counter example of N=12 and p=6 shows. But we are given some extra constraints here, namely and .
But GrandDad's proof appears in Mathematical Analyis 2nd ed by Apostol Theorem theorem 1.10. So it must be correct. and my question still remains:
Given and , but how does this imply ?
Apostol's argument's correct because he's assuming something that hasn't been assumed here, namely: N has no square factors or, as said in number theory, N is square-free, and this is all the difference; otherwise his argument would be wrong, or at least lacking, since he doesn't remind at the moment of argumenting that , but he needs not to: if N doesn't divides p then some prime factor of N doesn't divide p (this can be also said in the case 12 divides 6^2 but 12 doesn't divide 6, since 12 has prime factors 2,2,3 and 6 only 2,3...but 12 is not square free), but this factor appears in and this can't be since it appears at the first power in N...!.
Now,
But this prime factor appears on the right hand in and thus it must appear in the left hand: and we get our contradiction.
Tonio
I'm so confused ... you assume that sqrt(N) is a reduced fraction, and so N = p^2/q^2 ... but since (p,q)=1, that implies that N is a rational number, and so we have a contradiction since N is supposed to be an integer greater than 1 ... cuz after that, we are treating N as if it's a rational number, and in that case, we can't really use prime decomposition on it, nor can we use the definition of divisibility ....
Could we do this also:
N is an integer greater than 1 and is not a perfect square.
Assume that sqrt(N) is a rational number say p/q with (p,q)=1 (reduced fraction).
=> N = p^2/q^2 => q^2|p^2 (since N is an integer) => q^2 = 1 (since (p,q)=1) => N = p^2. Contradiction since N is not a perfect square ... so sqrt(N) is not a rational. Since N is positive, sqrt(N) is a real number and since it is not a rational, it must be irrational.
Hello everyone -
Since I wrote this line, six months ago,I (like many others) have been puzzling over what I meant by it!has a factor , since
has a factor (For if not, then is the product of the squares of one or more of the prime factors of , and is therefore a perfect square. Contradiction.)
I was wrong: the conclusion " has a factor " is not valid, and I apologise for the confusion this may have caused. I was attempting to modify the well known proof that is irrational, and I was guilty of over-simplification.
What I should have said is that implies that there is a prime factor, say, of , for which the quotient is a multiple of . (For if not, then is the product of even powers of one or more of the prime factors of , and is therefore a perfect square. Contradiction.)
The proof then follows similar lines to my original 'proof', using rather than :
, for some integer
has a prime factor
has a prime factor
But and , and this contradicts the initial hypothesis that .
Further comment
I can fully understand where the problems arise as far as the part played by is concerned, and the confusion caused by the fact that is co-prime with . The fact is, of course, that tends to bring with it a whole lot of contradictions - that's the whole nature of the problem!
So at the risk of adding further confusion, may I attempt to amplify my argument, and so atone in some measure for my original error?
The statement is clear enough, and it obviously means that is a factor of . So forget for the time being that is the other factor of , and concentrate on . We'll use the fact that and are co-prime all in good time.
Since is not a perfect square, it therefore has, among its prime factors, at least one with an odd power. It is this prime number that I am calling (and which proscientia called ).
The fact that , with its array of even-powered prime factors, has as a factor with its odd-powered prime factor means that when we simplify (by cancelling) the fraction there will inevitably be at least one left uncancelled in the numerator. That's where I get my statement from that is a multiple of .
It's only at the end that we need to invoke the condition that . It's when we try to do so too early that the confusion arises.
I hope that helps to clear things up.
Grandad
No one said that q cannot be ... my problem is that someone should have said that q^2 MUST be 1 ...
p^2=N*q^2 with (p,q)=1, should make you reach the conclusion that q^2 = 1, not that there is a contradiction with (p,q)=1. The reason you reach that contradiction is because you're assuming q>1, because you're assuming that there is a prime r such that r|q (and consequently r|p), but there is no prime that divides 1 (or -1). That case is being overlooked, and in an indirect way, you're basically saying that q^2 cant equal 1.
I think proscientia's proof is okay, since although it's doesn't state that b^2=1, it correctly states that p1 does not divide b^2.