Can you help me to prove "Show that for any square-free integer $n > 1, \sqrt{n}$ is an irrational number."?

I'm asked to show that any square-free number $n, n^{\frac{1}{2}}$ is irrational.

I applied proof by contradiction here and assumed $n^{\frac{1}{2}}$ is a rational number. Therefore $n^{\frac{1}{2}}$ is equal to, lets say, $\frac{a}{b}$ where a and b are natural numbers. Since $b^{2} \times n = a^{2}$, I know that $n$ must contain a square number but I cannot prove it yet. I hope you can help me.

  • $\begingroup$ first step: prove that it is equivalent to “for any prime integer”, then it should be very easy, just as same as prove 2^(1/2) is irrational. $\endgroup$ – Zang MingJie Oct 5 '18 at 12:34
  • $\begingroup$ See also this answer which explicitly discusses the squarefree aspect. $\endgroup$ – Gone Oct 5 '18 at 13:15

You are almost there. Suppose $\sqrt{n}=\frac{a}{b}$ when $\gcd(a,b)=1$. You got it right, $b^2n=a^2$. Easy to see $a\ne 1$ because otherwise both $b$ and $n$ would be $1$ and that is a contradiction to $n>1$. So $a>1$ and hence there is a prime $p$ which divides $a$. Then $p^2$ divides $a^2$ and hence $p^2$ divides $b^2n$. But $n$ is square free so $p^2$ can't divide $n$. From here you can get that $p|b^2$ (because otherwise $p^2$ would have divided $n$) and that implies $p|b$. So $p|a$ and $p|b$ which contradicts $\gcd(a,b)=1$.

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