How would I order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2} \ $ without approximating the irrational numbers? In fact, I would be interested in knowing a general way to solve such questions if there is one.
What I tried to so far, because they are all positive numbers, is to square $x,y,z$ but, obviously, the rational parts will not be equal so I cannot compare the radicals. Proving that $x<z$ is easy and so is $y<z$, but I'm stuck at $x < y \text{ or } x>y$.
On
Clearly $0 < x, y < 1$ and simple calculation shows $y = \sqrt{7-2\sqrt{10}}$ so the claim reduces to showing $y^2 - x^2 = 3 - 2\sqrt{10} + 2\sqrt{3} > 0$, i.e. that $\sqrt{10} < \sqrt{3} + \frac 32$. To show this, square both sides to see this is equivalent to $10 < \frac{21}{4} + 3 \sqrt{3}$, i.e. that $19 < 12 \sqrt{3}$. Squaring one final time, this is equivalent to $361 < 432$ which is obvious so we're done.
On
Suppose $ x \geqslant y $, then
$$ \begin{array} { r c l } \sqrt3 + \sqrt2 &\geqslant& \sqrt5 + 1 \\ (\sqrt3 + \sqrt2)^2 &\geqslant& (\sqrt5 + 1)^2 \\ 5 + 2 \sqrt6 &\geqslant& 6 + 2 \sqrt5 \\ 2( \sqrt6 - \sqrt5) &\geqslant& 1 \\ 2( \sqrt6 - \sqrt5)( \sqrt6 + \sqrt5) &\geqslant& ( \sqrt6 + \sqrt5) \\ 2(6 - 5) &\geqslant& \sqrt6 + \sqrt5 \\ 2 &\geqslant& \sqrt6 + \sqrt5 > \sqrt4 + \sqrt4 = 4 \\ \end{array} $$
A contradiction.
On
A visual demonstration of sorts:
Starting from $ \ x \ = \ \sqrt3 - 1 \ \ , \ $ we have
$$ x \ + \ 1 \ \ = \ \ \sqrt3 \ \ \Rightarrow \ \ x^2 \ + \ 2x \ + \ 1 \ \ = \ \ 3 \ \ . \ $$ So $ \ \sqrt3 - 1 \ $ is the larger of the two zeroes of $ \ x^2 + 2x - 2 \ \ . \ $ As for $ \ x \ = \ \sqrt5 \ - \ \sqrt2 \ \ , \ $ we may write
$$ x \ + \ \sqrt2 \ \ = \ \ \sqrt5 \ \ \Rightarrow \ \ x^2 \ + \ 2 \sqrt2 · x \ + \ 2 \ \ = \ \ 5 \ \ , $$
making $ \ \sqrt5 \ - \ \sqrt2 \ $ the larger of the two zeroes of the polynomial $ \ x^2 + 2 \sqrt2 · x - 3 \ \ . \ $ A graph of the two polynomials (the first in red, the second in blue) indicates the relative positions of these two zeroes.
If one mistrusts a graph (a not unreasonable attitude), we can show that the curve for $ \ x^2 + 2x - 2 \ $ lies "above" that of $ \ x^2 + 2 \sqrt2 · x - 3 \ \ $ [vertex at $ \ (-1 \ , \ -3) \ $ versus $ \ (-\sqrt2 \ , \ -5) \ \ , \ $ "higher" $ \ y-$ intercept; also, $$ x^2 + 2x - 2 \ > \ x^2 + 2 \sqrt2 x - 3 \ \ \Rightarrow \ \ 1 \ > \ 2·(\sqrt2 - 1)·x $$ $$ \Rightarrow \ \ \frac{1}{2·(\sqrt2 - 1)} \ = \ \frac12·(1 + \sqrt2) \ > \ x \ \ . \ ] $$ until the curves intersect at $ \ x \ = \ \frac{1}{2 \sqrt2 \ - \ 1} \ = \ \frac12 + \frac{\sqrt2}{2} \ \ , $ $ y \ = \ \frac32 \sqrt2 - \frac14 \ > \ 0 \ \ . \ $ Since this intersection is "above" the $ \ x-$axis, the positive $ \ x-$intercept of $ \ x^2 + 2x - 2 \ $ is smaller than that of $ \ x^2 + 2 \sqrt2 · x - 3 \ \ . $
One way is as follows. It can be proved that $f(x) = \sqrt{x}$ is an increasing function. So we have: $$x_1\lt x_2 \implies \sqrt{x_1}\lt \sqrt{x_2} \implies \sqrt{x_2} - \sqrt{x_1} \gt 0$$This implies that $x,y\gt 0$. Also $g(x) = x^2$ is an increasing function for $x\ge 0$, then: $$x_1\lt x_2 \implies x_1^2\lt x_2^2 \ \ \ \ x_1,x_2\in [0,\infty)$$ Now suppose $x\gt y$: $$x>y \implies x^2\gt y^2 \implies (\sqrt{3} -1)^2>(\sqrt{5} - \sqrt{2})^2 \implies 3+1-2\sqrt{3}\gt 5+2-2\sqrt{10} \implies -3-2(\sqrt{3}-\sqrt{10})\gt 0 \implies -3\gt 2(\sqrt{3}-\sqrt{10}) \implies 3<2(\sqrt{10}-\sqrt{3})\implies \frac{3}{2}\lt \sqrt{10}-\sqrt{3} \implies \frac{9}{4} \lt 10 +3-2\sqrt{30} \implies 2\sqrt{30}\lt 13 - \frac{9}{4} \implies 4\times30\lt \frac{43^2}{4^2} \implies 4^3\times30 \lt 43^2 \implies 1920\lt 1849$$ This is a contradiction and implies $x\lt y$. Other cases can be checked similarly.