How do I prove that this inequality result holds for all positive integers How I can prove that this:
$$ \frac{1}{2} &lt; \frac{2}{3} &lt; \frac{3}{4} &lt; \frac{4}{5} &lt; \frac{5}{6}&lt;...&lt;\frac{n+1}{m+1}&lt;...$$
always holds when $n &lt; m$ and $n$ and $m$ are positive integers?
I started writing this $\frac{n}{m}&lt;\frac{n+1}{m+1}$ but then I don't know how to proceed.
Because $$\frac{n+1}{m+1}-\frac{n}{m}=\frac{m-n}{m(m+1)}&gt;0$$ for $m&gt;n&gt;0.$
This a high-school standard exercise on inequalities:
As all numbers are positive,
$$ \frac nm&lt;\frac{n+1}{m+1}\iff n(\not m+1)&lt;m(\not n+1)\iff n&lt;m. $$