Prove by contradiction that for every positive integer $k$, there is an integer $m$ such that $k\leq m^2\leq2k$ 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 $m^2<k\leq2k$.
Case 2: $k\not>m^2$ but $m^2>2k$
We have $m^2>2k$ and $m^2\leq k$, which implies that $2k<m^2\leq k$ implying $2k<k$ 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 $2k<m^2<k$ implying $2k<k$ 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.
 A: Your case 1 states "There exists a positive integer $k$ such that for all integers $m$, $m^2 < k \leq 2k$."  The second inequality is meaningless so we get rid of that.  This leaves
"There exists a positive integer $k$ such that for all integers $m$, $m^2 < k$."
I will show this is not possible.  Okay, so take ANY integer $k$.  If $k \leq 0$, then $m^2 \geq 0$ and therefore $m^2 < k$ is false.  So, assume $k > 0$.  Now, choose $m = k+1$.  Then $m^2 = k^2 + 2k + 1 = k + (k^2 + k + 1) > k$.
A: Here's an alternative solution. We take advantage of the gap between consecutive squares. Suppose that for some $k$ there exists no squares between $k$ and $2k$. Let $m$ be the largest integer such that $m^2 < k$. Then we must have
$$m^2 < k < 2k < (m+1)^2$$
In particular this means that the gap between $(m+1)^2$ and $m^2$ is greater than $k$.
$$(m+1)^2 - m^2 = 2m+1 > k$$
Clearly $m\neq 1$. Then $m\ge 2$. But that means
$$2m \le m^2 < k \implies 2m + 1 \le k$$
There is our desired contradiction.
A: Hint $\rm\ f(x) = x^2$ is increasing, $\rm\:f(x + 1) < 2\,f(x)\:$ for $\rm x\ge 3\:$
thus $\rm\:f(m) < k\le f(m\!+\!1)\:\Rightarrow\:f(m\!+\!1) < 2\,f(m) < 2k$
A: ok because we have not constraint,we can say that for $k>0$; $k<2*k$,now    if $m<0$ then $m^2>2*m$;else $ 2*m<=m^2$;now let us return to problem statment, suppose that  $k>m^2$ always hold,this is not  always because  we can easily find value for $k$ and $m$,for which $k <=m^2$ ;because we used fact that for $k>0$ ;$k<2*k$;we  have to show that $k>m^2$ and $m^2>2*k$ or $k>m^2$ and $m^2<2*k$ at the same time;but first is false,and second we can always find value for $k$ and $m$ ,such that second condition does not  hold,so i think my argument is correct
