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

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

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    $\begingroup$ 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. $\endgroup$ – imallett Nov 28 '14 at 3:02
  • $\begingroup$ Related: math.stackexchange.com/questions/517555/… $\endgroup$ – Henry Nov 28 '14 at 16:44
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Raise them both to the power of $6$.

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

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

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    $\begingroup$ 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). $\endgroup$ – JohnD Nov 27 '14 at 23:25
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    $\begingroup$ @JohnD: Thanks. Thought a comment would do, but then realized that a proper answer might be useful... $\endgroup$ – barak manos Nov 27 '14 at 23:27
  • $\begingroup$ @barakmanos but you don't get to the last equation $\endgroup$ – user2637293 Dec 2 '14 at 11:22
  • $\begingroup$ @user2637293: What does "don't get to the last equation" mean? $\endgroup$ – barak manos Dec 2 '14 at 11:48
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    $\begingroup$ @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. $\endgroup$ – barak manos Dec 2 '14 at 12:59
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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: $$\Bigl(\frac12\Bigr)^{1/3}>\Bigl(\frac12\Bigr)^{1/2}>\Bigl(\frac13\Bigr)^{1/2}\ .$$

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  • $\begingroup$ Yeah, this quiestion would be more interesting if the indices were switched. $\endgroup$ – Rawling Nov 28 '14 at 10:24
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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.

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

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

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 $(\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}}$

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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 $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} \lt \left(\frac{1}{2}\right)^{\frac{1}{3}}$

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

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

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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 $1/4 > 1/{27}$.

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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.

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