# How do I sort a number of surd expressions?

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.

• Note that you can always check your work by computing each quantity numerically. Of course that's not how you are meant to address the problem, but it provides an excellent check once you have solved it by other means.
– lulu
Sep 9, 2020 at 17:04
• That's very true! Sep 9, 2020 at 17:07

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: $$10 - \sqrt{6} < 11-\sqrt{12} < 5+\sqrt{6} < \sqrt{56} < 7.5 < \sqrt{12}+\sqrt{17}$$

• As a quick remark, note that $10-\sqrt 6 > 5+\sqrt 6\iff 5>2\sqrt 6$ but the latter inequality is clearly true (squaring shows that it is equivalent to $25>24$).
– lulu
Sep 9, 2020 at 17:08