<s>[INST] How to order $x = \sqrt{3}-1, y = \sqrt{5}-\sqrt{2}, z = 1+\sqrt{2}$ ascendingly? 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 $x < y \text{ or } x>y$.
 [/INST] 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.
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