# Why is square root of 2 less than cube root of 3

Square root of 2=1.41421..... Cube root of 3=1.44224..... 4th root of 4=1.41421...... 5th root of 5=1.37972...... 6th root of 6=1.34800...... 7th root of 7=1.32046...... . . nth root of n=.......... Similarly thereafter nth root of n will converge towards 1.

Q.2 why is square root 2 equals 4th root of 4? Intuitively sqrt 2 is diagnol of 1 by 1 square, and sqrt 3 is diagnol of 1 by 1 parallelogram, explain geometrially?

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$\sqrt{2} < \sqrt[3]{3}$ because $2\sqrt{2} < 3$. $\sqrt{2} = \sqrt[4]{4}$ because $4 = 2^2$. And no, $\sqrt[3]{3}$ is not the diagonal of a $1$ by $1$ parallelogram. – Daniel Fischer Jun 26 '13 at 7:17
What is the first question? – lab bhattacharjee Jun 26 '13 at 7:19
Consider the graph $x^{\frac{1}{x} }$, which has a maximum point at $x=e$, and is strictly increasing on $[1, e]$ and strictly decreasing on $[e, \infty)$. This is easily shown via calculus. – Calvin Lin Jun 26 '13 at 7:21
oh sorry, i edited that cubrt 3 diagnol. Can any ond explain that what other roots can be look like in geometry. – Waqar Ahmad Jun 26 '13 at 7:21
Take the volume of a unit n-ball (the n-dimensional analog to the 2D unit circle or 3D unit sphere). The volume as a function of dimension behaves similarly - it actually peaks at $n\approx 5.256$. Thus, the peak is either at $n=5$ or $n=6$, and after that decreases. Math works very funny when you're comparing units from different dimensions. – Foo Barrigno Jun 26 '13 at 11:11

Let us find extreme values of $f(x)=x^{\frac1x}$ where $x$ is real positive

So, $\ln f(x)=\frac{\ln x}x$

$$\frac{d \ln f(x)}{dx}=\frac{1-\ln x}{x^2}$$

So, for the extreme values of $\ln f(x),1-\ln x=0\iff x=e$

$$\frac{d^2 \ln f(x)}{dx^2}=\frac{-\frac1x\cdot x^2-(1-\ln x)(2x)}{x^4}$$ which is $<0$ at $x=e$

and $$\frac{d \ln f(x)}{dx}>0\text{ or } < \text{ according as }1-\ln x>0\iff x<e\text{ or }1-\ln x<0\iff x>e$$

So, $x^{\frac1x}$ will increase when $x\in(0,e)$ and decreasing when $x\in(e,\infty)$ and we know $2<e<3$

As $e$ lies between $2,3$ the above derivation can not help us to compare $x^{\frac1x}$ at $x=2,3$ or any two values between which $e$ lies

So we can solve it as follows:

$\sqrt2 <\sqrt[3]3$ i.e., $2^{\frac12}<3^{\frac13}\iff (2^{\frac12})^6<(3^{\frac13})^6\iff 2^3<3^2$ which is true

We have taken $6$th power as lcm$(2,3)=6$

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If one has points on either side of an extremum, one cannot generally compare them. So while your calculation gives a nice insight in the behaviour of $x^{1/x}$, it is not sufficient to allow for the conclusion $\sqrt2 < \sqrt[3]3$. – Lord_Farin Jun 26 '13 at 7:36
@Lord_Farin, thanks for your observation.Please have a look into the edited answer – lab bhattacharjee Jun 26 '13 at 8:59
That works, thanks for editing; as the point was minor anyway, you already got my upvote before the edit. Cheers! – Lord_Farin Jun 26 '13 at 9:09