# Which of the following numbers are the smallest and largest: $\sqrt[5]{2}, \frac{4}{3}, \sqrt[6]{3}$?

Which of the following numbers are the smallest and largest: $\sqrt[5]{2}, \frac{4}{3}, \sqrt[6]{3}$ ?

I am not supposed to do any calculations with a calculator, any way to see this intuitively?

Thank you!

• intuitively?? I don't think so.. – user87543 Sep 11 '14 at 9:54
• Without any calculations ? – Claude Leibovici Sep 11 '14 at 9:55
• may be the question asks you not use the calculator for finding the exact number for all the three expressions. – Vikram Sep 11 '14 at 9:57
• I am not supposed to use a calculator, and all other questions before it has been more in line of quite simple algebraic manipulations. So I thought that there might be a way to "see" which of the numbers that are the largest and smallest... – Lukas Arvidsson Sep 11 '14 at 9:59
• May be you want to replace do any calculations by use calculator – user87543 Sep 11 '14 at 10:00

## 4 Answers

The trick to doing this without a calculator is that $x^n$ for $n>0$ is a (strictly) monotonically increasing function. Thus $x^n > y^n \Rightarrow x>y$.

This means we can eliminate the roots by raising to the 5th and 6th power. Comparing both numbers first to $4/3 = 2^2/3$ we get

$2<2^{10}/3^5 \Rightarrow \sqrt[5]{2}<4/3$

and

$3^7 < 2^{12} \Rightarrow \sqrt[6]{3}<4/3$

thus the largest number is $4/3$. To find the smallest we raise the two smaller numbers to the power $5\times6 = 30$ giving us $2^6$ and $3^5$ to compare. $2^6 = 64$ and it takes only a few multiplications to see that $3^5$ is indeed the larger so we get

$$\sqrt[5]{2} < \sqrt[6]{3} < \frac{4}{3}$$

• Thank you very much for your explanation Dan. Much appreciated! – Lukas Arvidsson Sep 11 '14 at 13:11
• You're welcome. :) – Dan Sep 11 '14 at 13:15
• Did you mean $$2<2^{10}/3^5 \Rightarrow \sqrt[5]{2}<4/3$$ and $$3^7 < 2^{12} \Rightarrow \sqrt[6]{3}<4/3 ?$$ – MartinG Sep 11 '14 at 14:21
• I did indeed. Edited the answer. – Dan Sep 11 '14 at 14:23

$2^6<3^5\Rightarrow ???$

$(\frac{4}{3})^2>(1.3)\cdot(1.3)=1.69$

$(\frac{4}{3})^4> (1.6)\cdot(1.6)=2.56$

$(\frac{4}{3})^6> (2.5)\cdot(1.6)=4$

Does this help you ??

• Not really. The OP needs bounds, not approximations. (Also, it looks very odd to approximate $1.69$ as $1.6$.) – TonyK Sep 11 '14 at 10:18
• @TonyK : I mean $4<(\frac{4}{3})^6$ – user87543 Sep 11 '14 at 10:37
• That's much better :-) – TonyK Sep 11 '14 at 10:42
• @TonyK : :) ${~~~}$ – user87543 Sep 11 '14 at 10:46
• Thank you very much for your answer! It does help me :) – Lukas Arvidsson Sep 11 '14 at 13:12

Hint: You can compare $\sqrt[5]{2}$ with $4/3$ by raising each number to its fifth power, $\sqrt[5]{2}$ to $\sqrt[6]{3}$ by raising each number to its $30^{\text{th}}$ power, and $4/3$ to $\sqrt[6]{3}$ by raising each number to its sixth power.

Raise all the numbers to the 6th power: $$(\sqrt[5]{2})^6 = \sqrt{2^{12/5}} < \sqrt{2^{12/4}} = \sqrt8 \lt \sqrt{9} = 3 \\(4/3)^6 = (1 +\tfrac{1}{3})^6 > 1 + 6 \times \tfrac{1}{3} = 3 \\ (\sqrt[6]{3})^6 = 3$$