# How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$

How can you determine which one of these numbers is bigger (without calculating):

$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$

• Also, there are methods of taking cube roots and especially square roots by hand, less known nowadays. In this case, only one digit is sufficient--and doing it this way is almost as fast. Commented Nov 28, 2014 at 3:02
• Commented Nov 28, 2014 at 16:44

Raise them both to the power of $6$.

Since they are both positive, their order will be preserved and you will get:

$$\left({\dfrac{1}{2}}\right)^2=\frac{1}{4} > \frac{1}{27}=\left({\dfrac{1}{3}}\right)^3$$

• Very nice, simple, and yet often overlooked tactic on problems like these (that occur on standardized exams in the US at least where time is of the essence). Commented Nov 27, 2014 at 23:25
• @JohnD: Thanks. Thought a comment would do, but then realized that a proper answer might be useful... Commented Nov 27, 2014 at 23:27
• @barakmanos but you don't get to the last equation Commented Dec 2, 2014 at 11:22
• @user2637293: What does "don't get to the last equation" mean? Commented Dec 2, 2014 at 11:48
• @user2637293: I'm sorry, but you are not making yourself clear. In my answer above, each side of the equation is raised to the power of $6$. Since both sides are positive, the order (direction of the $>$ sign) is preserved. Commented Dec 2, 2014 at 12:59

No need to do any calculations at all: since we are talking about numbers between $0$ and $1$, a cube root is larger than a square root: $$\Bigl(\frac12\Bigr)^{1/3}>\Bigl(\frac12\Bigr)^{1/2}>\Bigl(\frac13\Bigr)^{1/2}\ .$$

• Yeah, this quiestion would be more interesting if the indices were switched. Commented Nov 28, 2014 at 10:24

When is $x^y > y^x$ ? When $x^{1/x} > y^{1/y}$. Let's look at the function $x^{1/x}$. Differentiating, we find it has a maximum at $x=e$. Since $1/2$ and $1/3$ are both less than $e$, the one that's nearer wins. So $(1/2)^2 > (1/3)^3$, so $(1/2)^{1/3} > (1/3)^{1/2}$.

But more to the point, this shows that $e^\pi > \pi^e$, which might be a lot harder without a calculator.

$$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}$$ $$\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$ Now it is obvious that $$\sqrt[3]2<\sqrt3$$ Thus $$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$

$((\frac{1}{2})^{\frac{1}{3}})^6=(\frac{1}{2})^2=\frac{1}{4}$

$((\frac{1}{3})^{\frac{1}{2}})^6=(\frac{1}{3})^3=\frac{1}{27}$

So as it is obvious from the above relations, $((\frac{1}{2})^{\frac{1}{3}})^6>((\frac{1}{3})^{\frac{1}{2}})^6$, so we can say $(\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}}$

If you don't want to depend on the "trick" of raising to the sixth power, you can compare the logs: $\frac 13 \log \frac 12=\frac {- \log 2}3$ and $\frac 12 \log \frac 13=\frac {-\log 3}2$ Now $\frac 12 \gt \frac 13$ and $\log 3 \gt \log 2$, so $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} \lt \left(\frac{1}{2}\right)^{\frac{1}{3}}$

Since $3^3=27 > 4=2^2$, we have $$\left(\frac{1}{2}\right)^2 > \left(\frac{1}{3}\right)^3$$ Take square roots and get $$\left(\frac{1}{2}\right) > \left(\frac{1}{3}\right)^{3/2}$$ Take cube roots and get $$\left(\frac{1}{2}\right)^{1/3} > \left(\frac{1}{3}\right)^{1/2}$$

For $x ≥ 0$ and $y > 0$, $x^y$ grows with $x$ for fixed $y$; it grows with $y$ for fixed $x > 1$, and decreases with $y$ for fixed $x < 1$. This becomes much more obvious if you take the logarithm. Special case as David said: For $0 < x < 1$, a cube root is greater than a square root.

So increasing $x$ from $1/3$ to $1/2$ increases the result, and decreasing the exponent from $1/2$ to $1/3$ also increases the result since $x < 1$. No need to calculate anything.

Checking whether $\left(\frac{1}{3}\right)^{1/3} < \left(\frac{1}{2}\right)^{1/2}$ would have been more difficult. Raising to the sixth power would give $1/9 < 1/8$ which is a lot closer than $1/4 > 1/{27}$.

What's wrong with just logging? Logarithm is a monotone increasing function, so the inequality sign stays the same.

First log, the multiply both sides by 2 and 3. LHS becomes $\log (\frac{1}{2})^2$, right $\log(\frac{1}{3})^3$. Now exponentiate. LHS is $\frac{1}{2} \cdot \frac{1}{2} \cdot 1$, RHS is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$. Each term on the LHSis greater than each term on the RHS, hence the inequality follows.