# Elementary number Theory / squarefree numbers [duplicate]

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.

• 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. – Zang MingJie Oct 5 '18 at 12:34
• See also this answer which explicitly discusses the squarefree aspect. – 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 $$\gcd(a,b)=1$$.