<s>[INST] Proving the square root of 2 exists(Possible typo in Zorich's Mathematical Analysis) 
In the picture above, I don't understand how if $$s^2 \lt 2$$ then $$\left (s + \frac{2-s^2}{3s}\right)^2 \lt 2$$ holds. What insures that $(s + \frac{2-s^2}{3s}) \in X = \{x : x^2 \lt 2\}$ where $x$ is a positive real number ??
And when the identity is expanded should it not be $$\left(s+\frac\Delta{3s}\right)^2=s^2+2\cdot s \cdot\frac\Delta{3s}+\left(\frac\Delta{3s}\right)^2$$ instead of $$\left(s+\frac\Delta{3s}\right)^2=s^2+ 2 \cdot\frac\Delta{3s}+\left(\frac\Delta{3s}\right)^2$$
 Thanks in advance!
P.S. The book is Zorich's Mathematical Analysis I in case this is an error.
 [/INST] We know $1 < s^2 < 2$ so $0 < 2- s^2 = \triangle < 1$.  So $ \frac{\triangle}{3s} < \frac {\triangle}3 < \triangle < 1$ so $(\frac{\triangle}{3s})^2 < \frac{\triangle}{3s}$.
That's the groundwork.
So $(s + \frac {2-s^2}{3s})^2 = (s + \frac {\triangle}{3s})^2$
$= s^2 + 2s*\frac {\triangle}{3s} + (\frac {\triangle}{3s})^2$
$< s^2 + 2*\frac {\triangle}3 + \frac {\triangle}3$
$= s^2 + 3\frac {\triangle}3 = s^2 + \triangle$
$= s^2 + (2 - s^2) = 2$.
So $s^2 < (2+ \frac {2-s^2}{3s})^2 < 2$. 
And that's that.  
(the book had a typo but they didn't make any conclusions based on it.  They forgot to cancel out the lower $s$ and left it in for a few steps, and then, simply had it disappear.  It was probably a type-setting error and not an actual error of the author.)
Or if you want to do it without the gordang triangle.
$(s + \frac {2-s^2}{3s})^2 = s^2 + 2s*\frac {2-s^2}{3s} + \frac {(2-s^2)^2}{9s^2}$
$=s^2 + \frac {4 - 2s^2}3 + \frac {4 - 4s^2 +s^2}{9s^2}$
$=s^2 +\frac 43 - \frac {2s^2}3 + \frac 4{9s^2} -\frac 49 + \frac {s^2}9$
$= \frac {4s^2}9 + \frac 89 + \frac 4{9s^2}$
$=\frac 49(s^2 + \frac 1{s^2} + 2)$
$= \frac 49(2 -h + \frac 1{2-h} + 2)$
$< \frac 49(2 + 1/2 + 2) = \frac {18}9 = 2$.
</s>