Adesh said: Here is my solution: If we have two real numbers \(\displaystyle a\) and \(\displaystyle 1\), then by Archimedean property we have $$ a \lt 1 \times n_1 \\ a \lt n_1$$ Again, given two real numbers \(\displaystyle a\) and \(\displaystyle 1\), by Archimedean property we have $$ 1 \lt n_2 a \\ 1/n_2 \lt a $$ Let's take \(\displaystyle n = ~max~(n_1, n_2)\) we have $$ 1/n \leq 1/n_2 \lt a \lt n_1 \leq n \\ 1/n \lt a \lt n$$ Hence proved. Is my solution correct? Click to expand... It's not important, but you should have included something like "by Archimedean principle there exist natural numbers $n_1$ and $n_2$ such that ..."