4
$\begingroup$

How can I know which one of the following numbers is the greatest: $$2^{1/2},3^{1/3},4^{1/4},5^{1/5},{6^{1/6}}$$ That can be also written as: $$\sqrt[2]{2},\sqrt[3]{3},\sqrt[4]{4},\sqrt[5]{5},\sqrt[6]{6}$$ Is there a fast way to determine what the greatest number is without a calculator? $n$th root algorithm is not fast so I can't use it. How can we solve this and similiar problems only by hand?

$\endgroup$
10
$\begingroup$

$$\begin{array}{rrcl} & n^{1/n} &>& (n+1)^{1/(n+1)} \\ \iff& n^{n+1} &>& (n+1)^n \\ \iff& n &>& \displaystyle \left(1+\frac1n\right)^n \\ \impliedby& n &>& e \end{array}$$

Since $3 > e$, we can be sure that $3^{1/3} > 4^{1/4} > 5^{1/5} > 6^{1/6}$.

As for $2$, notice that $\left(1+\dfrac12\right)^2 = 2.25 > 2$, so $2^{1/2} < 3^{1/3}$.

So the greatest number is $3^{1/3}$.

$\endgroup$
  • $\begingroup$ Sorry, but how do you go from $n^{n+1}$ to $n$ and from $(n+1)^n$ to $\left(1+\frac{1}{n}\right)^n$? $\endgroup$ – Hanlon Jun 23 '18 at 14:27
  • $\begingroup$ I divide both sides by $n^n$ $\endgroup$ – Kenny Lau Jun 23 '18 at 14:28
  • $\begingroup$ Ok, I get it for the left-hand side but not for the right one. $\endgroup$ – Hanlon Jun 23 '18 at 14:29
  • $\begingroup$ $(n+1)^n \div n^n = ((n+1) \div n)^n$ $\endgroup$ – Kenny Lau Jun 23 '18 at 14:30
  • $\begingroup$ Ok. Thanks. Only one more thing: you said that since $3>e$, then it must be that $3^{1/3}>4^{1/4}>5^{1/5}>6^{1/6}$. How did you conclude that i.e. what are the omitted steps in proving that? $\endgroup$ – Hanlon Jun 23 '18 at 14:35
6
$\begingroup$

Hint: The function $f(x)=x^{1/x}$ is decreasing for $x \ge e$. It only remains to compare $2^{1/2}$ and $3^{1/3}$. For that, raise them to the $6$-th power.

$\endgroup$
4
$\begingroup$

Their logarithms are $$\frac{\ln n}{n}$$ for $n=2,\ldots,6$. The function $f(x)=(\ln x)/x$ is increasing on $(1,e)$, has a maximum at $e$, and is decreasing on $(e,\infty)$. To find the largest in your list, just compare $f(2)$ and $f(3)$. To find the smallest in your list, just compare $f(2)$ and $f(6)$.

$\endgroup$
3
$\begingroup$

To compare fractions, you would put them over the same denominator, and you could apply a similar method in this case, using $(x^a)^b=x^{a b}$.

For example, you can write $2^{1/2} = (2^3)^{1/6} = 8^{1/6}$ and $3^{1/3} = (3^2)^{1/6} = 9^{1/6}$, which makes it clear that $3^{1/3} > 2^{1/2}$.

You can compare all five in one go by writing them all in the form $n^{1/60}$, but whether you can calculate, e.g., $5^{12}$ without a calculator is another matter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.