How do I sort a number of surd expressions?

Question

Suppose I have the a number of expressions, some of which resolve to rational number and others to irrational numbers. $\sqrt{56}$, $7.5$, $5 + \sqrt{6}$, $10-\sqrt{6}$, $11-\sqrt{12}$ and $\sqrt{12} + \sqrt{17}$. My aim is to sort the surds in ascending order.

Answer 1

First, compare $$ 5 + \sqrt{6} \text{ with } 10 - \sqrt{6} $$ This is effectively the same as comparing $2.5$ and $\sqrt{6}$.

$2.5 = \sqrt{2.5 \times 2.5} = \sqrt{6.25}$

$\sqrt{6.25} > \sqrt{6}$

Therefore $5+\sqrt{6} > 10 - \sqrt{6}$

So the ordering at this point is $$ 10 - \sqrt{6} < 5+\sqrt{6} $$ with all the other items left unordered.

Next I've compared $11 - \sqrt{12}$ with $5+\sqrt{6}$

This is the same as comparing $(6 - \sqrt{12})^2=48-12\sqrt{12}$ and $6$ (because I subtract $5$ from both sides). In this case, the comparison that needs to be made is between $48-6$ and $12\sqrt{12}$. $42^2 = 1764$ and $12\times12 = 144$. It is clear that $5+\sqrt{6}$ is thus greater than $11 - \sqrt{12}$.

Something similar is then done for $10-\sqrt{6}$ and $11-\sqrt{12}$ after subtracting ten, we are left with $-\sqrt{6}$ and $1-\sqrt{12}$. $6$ and $(1-\sqrt{12})^2 = 13-2\sqrt{12}$. The comparison is therefore between $13-6=7$ and $2\sqrt{12} = \sqrt{48}$. Because $\sqrt{49} > \sqrt{48}$ $11-\sqrt{12} > 10-\sqrt{6}$.

The ordering at this point is therefore $$ 10 - \sqrt{6} < 11-\sqrt{12} < 5+\sqrt{6} $$

The next stage of the puzzle could be to see where $\sqrt{56}$ fits. To compare this with $5+\sqrt{6}$ we can square both expressions, so the comparison is between $56$ and $(5+\sqrt{6})^2 = 31+10\sqrt{6}$. Therefore, we are interested in whether $25 > 10\sqrt{6}$. We can rewrite that inequality as $\sqrt{625} > \sqrt{600}$. Thus, $\sqrt{56}$ is greater than $5+\sqrt{6}$. So the ordering is:

$$ 10 - \sqrt{6} < 11-\sqrt{12} < 5+\sqrt{6} < \sqrt{56} $$

We can see where $7.5$ fits with limited effort. Because $7.5 = \sqrt{56.25}$ it is thus bigger than $\sqrt{56}$ so the ordering should now be: $$ 10 - \sqrt{6} < 11-\sqrt{12} < 5+\sqrt{6} < \sqrt{56} < 7.5 $$

The last step of the solution is to work out where $\sqrt{12} + \sqrt{17}$ should go.

If we compare this to $\sqrt{56}$ we can square both expressions before comparing them. Thus, we compare $(\sqrt{12}$ and $\sqrt{17})^2 = 29 + \sqrt{816}$ and $56$. Thus we compare $56-29 = 27$ and $\sqrt{816}$. So $\sqrt{729}$ and $\sqrt{816}$. We thus deduce that $\sqrt{56} < \sqrt{12} + \sqrt{17}$. To compare to the next biggest number $\sqrt{56.25}$ we just repeat the above computation, but with 56.25 instead of 56. It turns out that $27.25^2 = 742.5625$ is still not bigger than $\sqrt{816}$ so $\sqrt{12} + \sqrt{17}$ is bigger than 56.25.

The final ordering is thus: $$ 10 - \sqrt{6} < 11-\sqrt{12} < 5+\sqrt{6} < \sqrt{56} < 7.5 < \sqrt{12}+\sqrt{17} $$