<s>[INST] Spivak's Calculus 4th ed: Absolute value definition p. 11 After defining the absolute value $|a|$ of $a$ as
$$
|a| = 
\begin{cases}
a, & a\geq 0\\
-a, & a\leq 0,
\end{cases}
$$
the text provides some examples. While $|-3| = 3$ and $|7| = 7$ are simple enough (by direct application of the above definition), I do not understand the latter two examples, which are
$$
|1 + \sqrt{2} - \sqrt{3}| = 1 + \sqrt{2} - \sqrt{3}
$$
and
$$
|1 + \sqrt{2} - \sqrt{10}| = \sqrt{10} - \sqrt{2} - 1.
$$
Without calculating the actual values of $\sqrt{2}$, $\sqrt{3}$, $\sqrt{10}$ and so on, how can one arrive at these equations?
EDIT: As @Bernard points out, the idea is to just compare the squares. Both of these absolute values use the sum $1 + \sqrt{2}$, we can think of its square, $3 + 2\sqrt{2}$ (which is less than $3 + 2\cdot2 = 3 + 4)$. In the first, we are essentially doing $3 + 2\sqrt{2} - 3$, which is a positive value, so we can just leave it as is. For the second, $3 + 2\sqrt{2} < 3 + 4 < 10$, so we have to re-order the $\sqrt{10}$ to come first, and then subtract the $1$ and $\sqrt{2}$ afterwards. It's more calculating than I'd like (and I'm not sure exactly what Spivak was trying to accomplish by throwing in these examples), but I am satisfied for now.
 [/INST] For the first absolute value, it is enough to compare the squares: $(1+\sqrt2)^2=3+2\sqrt 2>(\sqrt 3)^2$.
 For the second  absolute value, $3+2\sqrt 2<3+4<10$.
</s>