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

**o&apartyrock** · April 29th 2009, 09:14 PM

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

**Grandad** · April 30th 2009, 06:07 AM

Hello o&apartyrock

Originally Posted by o&apartyrock

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

**aukie** · October 23rd 2009, 08:22 PM

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.)

**proscientia** · October 23rd 2009, 09:06 PM

Originally Posted by Grandad

$\Rightarrow p^2 = Nq^2 \Rightarrow p^2$ 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.

**aukie** · October 24th 2009, 07:08 AM

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 $N|q$ ?

**proscientia** · October 24th 2009, 02:31 PM

Originally Posted by aukie

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.

Originally Posted by aukie

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

I’m afraid I’m totally confused about that as well.

**tonio** · October 24th 2009, 03:10 PM

Originally Posted by aukie

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

**Bingk** · October 24th 2009, 06:53 PM

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.

**tonio** · October 24th 2009, 07:44 PM

Originally Posted by Bingk

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.

.

**Grandad** · October 25th 2009, 01:34 AM

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

**Bingk** · October 25th 2009, 08:43 AM

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.

**aukie** · October 25th 2009, 06:45 PM

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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