is this true: given an integer n, sqrt(n) is rational only if sqrt(n) is an integer
Yes—assuming we mean the real square root, this is true.
If \(n\) is an integer and \(\sqrt n\) is rational, write
\[ \sqrt n=\frac{a}{b} \]
where \(a,b\) are integers with no common factor and (b>0). Squaring gives
\[ n=\frac{a^2}{b^2} \]
so
\[ a^2=nb^2. \]
That means \(b^2\) divides \(a^2\). But since \(a\) and \(b\) have no common factor, \(a^2\) and \(b^2\) also have no common factor. Therefore \(b^2=1\), so \(b=1\).
So \(\sqrt n=a\), an integer.
Therefore:
\[ \boxed{\sqrt n \in \mathbb{Q} \implies \sqrt n \in \mathbb{Z}} \]
for integer \(n\ge 0\). Negative \(n\) has no real square root, so it is not a counterexample in the real numbers.
