lean_workbook_plus_79054

SolutionWe will use $p$ -adic valuation. Let $m=v_p(a)$ and $n=v_p(b)$ for some prime $p$ . We are given $2kn \leq (2k+1)m$ and $(2j+1)m \leq (2j+2)n$ , for non-negative integers $k,j$ . (This follows from the identity $v_p(r^s)=sv_p(r)$ .) From the first inequality, we have $n-m \leq \frac{m}{2k}$ . Since $k$ can be any non-negative integer, the inequality is certainly true for $k=m$ , so we have $n-m \leq \frac{1}{2}$ . But $m$ and $n$ are non-negative integers, so we have $m \geq n$ . We can do the same thing with the second inequality: $m-n \leq \frac{n}{2j+1}$ . This inequality is true for $j=n$ , so $m-n \leq \frac{n}{2n+1} < \frac{1}{2}$ . Again, $m$ and $n$ are positive integers, so we also have $n\geq m$ . $m$ and $n$ must satisfy both $m\geq n$ and $n\geq m$ , so $m=n$ . This proves that $a$ and $b$ have equal $p$ -adic valuations for any prime $p$ , so they are themselves equal. $\square$