Prove by contradiction that for every positive integer $k$, there is an integer $m$ such that $k\leq m^2\leq2k$.
Heres what I've done.
Take the negation of the statement above to attempt a contradiction. We have "There exist a positive integer $k$ such that for all integers $m$, $k>m^2$ or $m^2>2k$". We have the following cases.
Case 1: $k>m^2$ but $m^2\not>2k$
We have $k>m^2$ and $m^2\leq2k$, which implies that $m^2
. Case 2: $k\not>m^2$ but $m^2>2k$
We have $m^2>2k$ and $m^2\leq k$, which implies that $2k
implying $2k whch is impossibru as $k\geq 1$. Therefore contradiction. Case 3: $k>m^2$ and $m^2>2k$
We have $k>m^2$ and $m^2>2k$, which implies $2k
implying $2k whch is impossibru as $k\geq 1$. Therefore contradiction.
Actually thats the furthest I could go. Any hints or perhaps solutions?
Ok I Edited as I got new epihany from some ideas. Now I left with contradicting Case 1.