The discussion confirms that there are no natural number solutions \(m, n \in \mathbb{N}\) such that \(\sqrt{7} - \frac{m}{n} > \frac{1}{mn}\) while also ensuring \(\sqrt{7} - \frac{m}{n} > 0\). The proof utilizes inequalities and modular arithmetic, demonstrating that if \(0 < \sqrt{7} - \frac{m}{n} < \frac{1}{mn}\), it leads to contradictions regarding the parity of \(m\) and \(n\). Thus, the conclusion is that the initial condition cannot hold true for any natural numbers.
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[sp]Suppose that there exist $m,n$ such that $0 < \sqrt7 - \frac mn < \frac1{mn}$. Then $\frac mn < \sqrt7 < \frac{m^2+1}{mn}$, and (after squaring and multiplying through by $m^2n^2$) $$ m^4 < 7m^2n^2 < (m^2+1)^2 = m^4 + 2m^2 + 1,$$ $$0 < m^2(7n^2-m^2) < 2m^2+1,$$ $$ 0 < 7n^2 - m^2 < 2 + \tfrac1{m^2}.$$ Since $7n^2 - m^2$ is an integer, it must therefore be $1$ or $2$. But the square of an integer is congruent to $0$ or $1\pmod4$, so $7n^2-m^2 = 1\pmod4$ can never occur; and the only possible solution to $7n^2-m^2 = 2\pmod4$ is if $m$ and $n$ are both odd. But the square of an odd number is congruent to $1\pmod8$. So if $m$ and $n$ are both odd then $7n^2-m^2 = 6\ne2\pmod8$. Therefore there are no solutions. It follows that if $\sqrt 7 - \frac {m}{n}$ is greater than $0$ then it must be greater than $\frac1{mn}.$Albert said:$m,n\in N$ , and $\sqrt 7 - \dfrac {m}{n}>0$
prove :
$\sqrt 7 - \dfrac {m}{n}>\dfrac {1}{m\times n}$
nice solution !Opalg said:[sp]Suppose that there exist $m,n$ such that $0 < \sqrt7 - \frac mn < \frac1{mn}$. Then $\frac mn < \sqrt7 < \frac{m^2+1}{mn}$, and (after squaring and multiplying through by $m^2n^2$) $$ m^4 < 7m^2n^2 < (m^2+1)^2 = m^4 + 2m^2 + 1,$$ $$0 < m^2(7n^2-m^2) < 2m^2+1,$$ $$ 0 < 7n^2 - m^2 < 2 + \tfrac1{m^2}.$$ Since $7n^2 - m^2$ is an integer, it must therefore be $1$ or $2$. But the square of an integer is congruent to $0$ or $1\pmod4$, so $7n^2-m^2 = 1\pmod4$ can never occur; and the only possible solution to $7n^2-m^2 = 2\pmod4$ is if $m$ and $n$ are both odd. But the square of an odd number is congruent to $1\pmod8$. So if $m$ and $n$ are both odd then $7n^2-m^2 = 6\ne2\pmod8$. Therefore there are no solutions. It follows that if $\sqrt 7 - \frac {m}{n}$ is greater than $0$ then it must be greater than $\frac1{mn}.$
[Notice that if you drop the condition $\sqrt 7 - \frac {m}{n}>0$ then it is possible to have $\Bigl|\sqrt 7 - \frac {m}{n}\Bigr| < \frac1{mn}.$ For example, if $m=8$ and $n=3$ then $\Bigl|\sqrt 7 - \frac83\Bigr| \approx 0.0209 < \frac1{24} \approx 0.0417.$ But in that case, $\sqrt 7 - \frac 83$ is negative.][/sp]