# How would you do this ? (1 Viewer)

#### An0nyM0usUseR455

##### New Member
I got the RHS of the inequality just dont understand why 1 is less than nth root of n. Should it not be less than or equal to?

#### tywebb

##### dangerman
Yes I think it should be less than or equal to, equal in the case of n=1.

If you allow n to be a real number you can get it to be less than 1 but the maximum is $\bg_white \sqrt[e]{e}$ so for positive integers the maximum is $\bg_white \sqrt[3]{3}\approx1.44$ and minimum is 1. So a tighter bound for positive integers is $\bg_white 1\le\sqrt[n]{n}\le\sqrt[3]{3}$

You can see it more clearly in the graph of $\bg_white y=x^\frac{1}{x}$

Range for real $\bg_white x>0$ is $\bg_white 0<\sqrt[x]{x}\le\sqrt[e]{e}$

Range for positive integers $\bg_white n$ is $\bg_white 1\le\sqrt[n]{n}\le \sqrt[3]{3}$

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#### tywebb

##### dangerman
So to do the actual question:

a.

$\bg_white \text{Base Step: }$
$\bg_white 2^1=2>1\text{ but in this case I would also check }2^2=4>2\text{ and you will see why in the induction step}$

$\bg_white \text{Induction Step:}$
$\bg_white \text{If for }k>1, 2^k>k\text{ then }$
$\bg_white 2^{k+1}=2^k\cdot2>k\cdot2=k+1+k-1>k+1\text{ since }k>1$

Hence by the principle of mathematical induction, for all positive integers $\bg_white n, 2^n>n$

b.

$\bg_white 2^n>n\ge1\therefore1=\sqrt[1]{1}\le\sqrt[n]{n}<\sqrt[n]{2^n}=2\text{ i.e., }1\le\sqrt[n]{n}<2$

#### An0nyM0usUseR455

##### New Member
So to do the actual question:

a.

$\bg_white \text{Base Step: }$
$\bg_white 2^1=2>1\text{ but in this case I would also check }2^2=4>2\text{ and you will see why in the induction step}$

$\bg_white \text{Induction Step:}$
$\bg_white \text{If for }k>1, 2^k>k\text{ then }$
$\bg_white 2^{k+1}=2^k\cdot2>k\cdot2=k+1+k-1>k+1\text{ since }k>1$

Hence by the principle of mathematical induction, for all positive integers $\bg_white n, 2^n>n$

b.

$\bg_white 2^n>n\ge1\therefore1=\sqrt[1]{1}\le\sqrt[n]{n}<\sqrt[n]{2^n}=2\text{ i.e., }1\le\sqrt[n]{n}<2$
Thanks bro appreciate it !!!

I applied the same steps as u, was just confused why there was no equal sign but I am glad you agree with me !!!

Have a great night