Absolute value of a square root

Question

Can anyone explain to me why:

$$|\sqrt{5} - 5| = 5 - \sqrt{5}$$

Obviously I understand why the $5$ turns positive but I don't really get why the square root of $5$ turns negative?

Answer 1

Note that $\sqrt{5} < 5$ so $\sqrt{5}-5< 0$. Call $x = \sqrt{5}-5$ and recall that $|x| = \begin{cases}x \quad \,\,\,\,\mathrm{if}\, x \geq 0 \\ -x \quad \mathrm{if} \,x<0\end{cases}$ and here $x<0$ so $|x| = -x=-(\sqrt{5}-5) = 5 - \sqrt{5}$.

Answer 2

Well, this first comes down to $$|a-b| = |-(b-a)| = |-1|\cdot|b-a| = |b-a|$$ We thus get that $|\sqrt{5} - 5| = |5 - \sqrt{5}|$

We now note that if $a-b > 0$ then $|a-b| = a-b$, and since $5-\sqrt{5}$ is clearly greater than zero we have $$|\sqrt{5} - 5| = |5 - \sqrt{5}| = 5 - \sqrt{5}$$

Answer 3

The purpose of the absolute value is to (1) leave a number alone if it is to the right of zero on the number line, and (2) if a number is to the left of zero on the number line, move it an equal distance to the right of zero on the number line. For example, $|5|=5$ and $|-5|=5$. However, an expression may be to the left of zero on the number line but not have a negative sign to "drop". For example, $\sin(\frac{3\pi}{2})$ is to the left of zero on the number line (it's value is $-1$) but there is no negative sign in $\sin(\frac{3\pi}{2})$ to drop off. So instead, we put a negative sign in front of it as in $$ \left|\sin(\frac{3\pi}{2})\right|= -\sin(\frac{3\pi}{2}). $$ This works because multiplication by $-1$ reflects a quantity across zero on the number line. For example, 5 is to the right of zero and $-5$ is to the left of zero. Likewise, $-5$ is to the left of zero and $-(-5)$ is to the right of zero. So when you have a negative quantity and you want to put it an equal distance to the right of zero (i.e. positive), you just multiply the negative quantity by $-1$. This reflects it across to the other side of zero. If $\square$ is to the left of zero (i.e. negative) then $-\square$ is an equal distance to right of zero (i.e. positive). That is, if $\square <0$ then $|\square|=-\square$.