Kindly help to prove this problem.
Prove that for every positive rational number $s$ satisfying the condition $s^2 > 2$ one can always find a smaller rational number $s - k (k > 0)$ for which $(s - k) (s - k) > 2.$
Request! I had very big problem in dealing the above type of problems. Once any one can help me out, I am sure I can do such similar problems. Please help me.
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Hint: what happens if you consider the sequence $s-\frac{1}{n}$ for $n\in\mathbb{N}$?
We may assume $\sqrt{2}<s\leq2$. Solving the equation $x^2-2=0$ by Newton's method and starting with $s_0:=s$ gives
$$s_1={1\over2}\left(s+{2\over s}\right)\ .$$
This implies
$$s_1-\sqrt{2}={\bigl(s-\sqrt{2}\bigr)^2\over 2s}={\bigl(s-\sqrt{2}\bigr)\over 2s}\cdot\bigl(s-\sqrt{2}\bigr)\ .$$
Here both factors on the right hand side are positive, and the first one is $<{1\over2}$.
Very explicitly (and constructively), you can argue as follows:
Let $s\in\mathbb{Q}_+$, such that $s^2>2$. Let $k=\frac{s^2-2}{2s}>0$, then $$(s-k)^2 = s^2-2sk+k^2>s^2-2sk = 2.$$
