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Math Help - Prove that the square root of a non-square number is irrational

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    Prove that the square root of a non-square number is irrational

    Prove: If integer N>1 is not a perfect square, then sqrt(N) is irrational.
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    Irrational numbers

    Hello o&apartyrock
    Quote Originally Posted by o&apartyrock View Post
    Prove: If integer N>1 is not a perfect square, then sqrt(N) is irrational.
    Assume that \sqrt{N} is rational; in other words

    \exists\, p, q \in\mathbb{N},\, \sqrt{N} = \frac{p}{q}

    and assume further that hcf(p,q)=1; in other words, that the fraction \frac{p}{q} is in its lowest terms.

    Then N = \frac{p^2}{q^2}

    \Rightarrow p^2 = Nq^2 \Rightarrow p^2 has a factor N, since hcf(p,q) = 1

    \Rightarrow p has a factor N (For if not, then N is the product of the squares of one or more of the prime factors of p, and is therefore a perfect square. Contradiction.)

    \Rightarrow p = Nr, r\in\mathbb{N}

    \Rightarrow p^2 = N^2r^2 = Nq^2

    \Rightarrow q^2 = Nr^2

    \Rightarrow q has a factor N

    \Rightarrow hcf(p,q) \ge N > 1

    Contradiction.

    Therefore \sqrt{N} cannot be expressed in the form \frac{p}{q}, and is therefore irrational.

    Grandad
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    Why is this line true?

    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.)
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    Quote Originally Posted by Grandad View Post
    \Rightarrow p^2 = Nq^2 \Rightarrow p^2 has a factor N, since hcf(p,q) = 1

    \Rightarrow p has a factor N (For if not, then N is the product of the squares of one or more of the prime factors of p, and is therefore a perfect square. Contradiction.)
    Almost, but not quite. For example, 6^2 has a factor 12 but 6 does not have a factor 12. You should take into account not only that N is not a perfect square but also that \frac{p^2}N=q^2 is a perfect square (in this case \frac{6^2}{12}=3 is not a perfect square).


    This is how I would write out the proof. Let N=p_1^{n_1}p_2^{n_2}\cdots p_k^{n_k} where the p_i are distinct primes. Then at least one of the n_i must be odd since N is not a perfect square, say n_1 is odd.

    If \sqrt N=\frac ab where a,b\in\mathbb Z, b\ne0 and \gcd(a,b)=1, then

    Nb^2\ =\ a^2

    \Rightarrow\ N\mid a^2

    \Rightarrow\ p_1\mid a^2 (since p_1\mid N)

    \Rightarrow\ p_1\mid a (since p_1 is prime)

    \Rightarrow\ p_1 divides a^2 an even number of times

    This is a contradiction because p_1 divides N an odd number of times and p_1\nmid b^2 and so p_1 divides Nb^2 an odd number of times.
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    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 p^2 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 p^2 = Nq^2 and gcd(p,q)=1.

    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 p^2 = Nq^2 and gcd(p,q)=1, \Rightarrow N|q^2, but how does this imply N|q?
    Last edited by aukie; October 24th 2009 at 12:28 PM.
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    Quote Originally Posted by aukie View Post
    But we are given some extra constraints here, namely p^2 = Nq^2 and gcd(p,q)=1.
    Thanks for the reminder! I had completely forgotten about p and q having to be coprime.

    Quote Originally Posted by aukie View Post
    Given p^2 = Nq^2 and gcd(p,q)=1, \Rightarrow N|q^2, but how does this imply N|q?
    Iím afraid Iím totally confused about that as well.
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    Quote Originally Posted by aukie View Post
    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 p^2 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 p^2 = Nq^2 and gcd(p,q)=1.

    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 p^2 = Nq^2 and gcd(p,q)=1, \Rightarrow N|q^2, but how does this imply N|q?

    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 p^2=Nq^2 at the moment of argumenting that N \mid p^2 \Longrightarrow N \mid p, 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 p^2=Nq^2 and this can't be since it appears at the first power in N...!.

    Now, p^2=Nq^2\,\,\,but\,\,\,N \nmid p \Longrightarrow \exists prime\,\,\,r\,\,\,s.t.\,\,r \mid N \wedge r \nmid p

    But this prime factor appears on the right hand in p^2=Nq^2 and thus it must appear in the left hand: r \mid p^2 \Longrightarrow r \mid p and we get our contradiction.

    Tonio
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    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.
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    Quote Originally Posted by Bingk View Post
    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

    \color{red}\mbox{And who said q cannot be} \pm 1?

    tonio

    ... 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.
    .
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    Putting right my error!

    Hello everyone -

    Since I wrote this line, six months ago,
    \Rightarrow p^2 = Nq^2 \Rightarrow p^2 has a factor N, since hcf(p,q) = 1

    \Rightarrow p has a factor N (For if not, then N is the product of the squares of one or more of the prime factors of p, and is therefore a perfect square. Contradiction.)
    I (like many others) have been puzzling over what I meant by it!

    I was wrong: the conclusion " p has a factor N" is not valid, and I apologise for the confusion this may have caused. I was attempting to modify the well known proof that \sqrt2 is irrational, and I was guilty of over-simplification.

    What I should have said is that p^2=Nq^2 implies that there is a prime factor, n say, of N,
    for which the quotient \frac{p^2}{N} is a multiple of n. (For if not, then N is the product of even powers of one or more of the prime factors of p, and is therefore a perfect square. Contradiction.)

    The proof then follows similar lines to my original 'proof', using n rather than N:

    q^2=\frac{p^2}{N}=rn, for some integer r

    \Rightarrow q^2 has a prime factor n

    \Rightarrow q has a prime factor n

    But n|N and N|p \Rightarrow n|p\Rightarrow \gcd(p,q)\ge n, and this contradicts the initial hypothesis that \gcd(p,q)=1.

    Further comment

    I can fully understand where the problems arise as far as the part played by q is concerned, and the confusion caused by the fact that q is co-prime with p. The fact is, of course, that \gcd(p,q)=1 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 p^2 = Nq^2 is clear enough, and it obviously means that N is a factor of p^2. So forget for the time being that q^2 is the other factor of p^2, and concentrate on N. We'll use the fact that p and q are co-prime all in good time.

    Since N 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 n (and which proscientia called p_1).

    The fact that p^2, with its array of even-powered prime factors, has N as a factor with its odd-powered prime factor n means that when we simplify (by cancelling) the fraction \frac{p^2}{N} there will inevitably be at least one n left uncancelled in the numerator. That's where I get my statement from that \frac{p^2}{N} is a multiple of n.

    It's only at the end that we need to invoke the condition that \gcd(p,q)=1. It's when we try to do so too early that the confusion arises.

    I hope that helps to clear things up.

    Grandad
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    No one said that q cannot be \pm 1 ... 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.
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    very nicely done tonio! its seems like you are a fan of apostol.

    grandad is right the number of possible contradictions one can arrive at is intriging.
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