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How to compare the size of following numbers without using the calculator?

$a=\sqrt{2}+\sqrt{6}+\sqrt{7},$

$b=\sqrt{3}+\sqrt{4}+\sqrt{8},$

$c=\sqrt{5}+\sqrt{5}+\sqrt{5}$

guest
  • 33

2 Answers2

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Looking at the three expressions, one easily notices that the sum of the radicants is $15$ in all three cases. This together with the concavity of the square root ($\sqrt{\lambda a+(1-\lambda)b}\ge\lambda\sqrt{a}+(1-\lambda)\sqrt{b}$ for all $0\le\lambda\le1$) helps to compare the numbers. Note that the concavity of the square root is strict, so in the above equation, if $a\neq b$ and $\lambda\notin\{0,1\}$, the left hand side is strictly greater.

Clearly $c$ is the largest of these numbers, because here all three radicants are equal. So what remains is to compare $a$ with $b$.

It is also obvious that for a and b, additionally the product of the first two terms is equal. Therefore it is useful to square the numbers: $$\begin{aligned} a^2 &= 2 + 6 + 7 + 2 \sqrt{12} + 2 \sqrt{14} + 2 \sqrt{42}\\ b^2 &= 3 + 4 + 8 + 2 \sqrt{12} + 2 \sqrt{24} + 2 \sqrt{32} \end{aligned}$$ Obviously the leading terms are equal, so ultimately we only have to compare $a' := 2 \sqrt{14} + 2 \sqrt{42} = \sqrt{8}(\sqrt{7} + \sqrt{21})$ with $b' := 2\sqrt{24}+2\sqrt{32} = \sqrt{8}(\sqrt{12}+\sqrt{16})$. Here we see that again, the sum of the radicants is the same, but for $a'$ their difference is larger, and therefore due to concavity $a'<b'$.

So putting everything together, we get $$a < b < c.$$

celtschk
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One method, which isn't particularly nice, is to use rough estimations. For example, $\sqrt2$ we know must be bigger than 1, and smaller than 1.5 (since $1.5^2$ is $\frac{9}{4}$ which is 2.25). Since it's between 1 and 1.5, I averaged it to be 1.25.

I then did this with the rest of the roots, to get the approximate values $\sqrt2=1.25,\sqrt3=1.75,\sqrt5=2.25,\sqrt6=2.25,\sqrt7=2.75,\sqrt8=2.75$

And of course $\sqrt4$ = 2. Then you simply substitute these values in, and it turns out that the result you get from this method isn't too far off the real answers, from this method I got

$a=6.25,b=6.5,c=6.75$

I checked with a calculator, these aren't too far off and the order is correct.

As I said though, this is not such a nice method but it works.

lagrange103
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  • It works in this case! I can't give a counterexample, because you don't say how you approximate $\sqrt n$ in general. But it looks very dodgy to me. – TonyK Sep 14 '14 at 10:15
  • Agreed - extremely dodgy. Well at it's core it is really guess and check. https://www.khanacademy.org/math/pre-algebra/exponents-radicals/radical-radicals/v/approximating-square-roots – lagrange103 Sep 14 '14 at 10:35
  • Perhaps if you start by figuring out the closest perfect square both above and below n, then you 'squeeze' it until you get closer – lagrange103 Sep 14 '14 at 10:36