For any $a \in \mathbb R$ and any $n \in \mathbb N^+$ there exists $q \in \mathbb Q$ such that $|a-q|< \frac{1}{n}$.

Question

For any $a \in \mathbb R$ and any $n \in \mathbb N^+$ there exists $q \in \mathbb Q$ such that $|a-q|< \frac{1}{n}$.

I think i can prove this is false, let $a=2,n=2,q=1/2$ so $|2-\frac{1}{2}|< \frac{1}{2}$ which is wrong. Is there a formal way to prove this?

Answer 1

As pointed out in the comments, the theorem says "there exists", which means that providing one counterexample doesn't mean that no such $q$ exists. To solve the problem, consider $k=\lfloor na\rfloor$ and let $q=\frac{k}{n}$. Then $$ |a-\frac{k}{n}|=|\frac{na-\lfloor na\rfloor}{n}|<\frac{1}{n} $$