If c > 1, prove that its n^th root is greater than 1 too

[issacnewton]

HelloI am trying to prove that if $c > 1$, then $c^{1/n} > 1$ , where $n\in \mathbb{N}$.
Now I present my proof here. Since $c>1$, it follows that $c > 0$. Now
$n^{\mbox{th}}$ root theorem says that if $a>0$,then there exists a unique positive
$n^{\mbox{th}}$ root of $a$. So $c^{1/n} > 0 $. Since we have to prove that $c^{1/n} > 1$ ,
I assume the negative of this goal. Assume $c^{1/n} \leqslant 1$. If $c^{1/n} = 1$, it
immediately follows that $c=1$, which leads to contradiction. So assume that
$c^{1/n} < 1$. Since $c^{1/n} > 0 $, we have that $0 < c^{1/n} < 1$. If $n=1$, we get
$c<1$ which is contradiction. Now I use a property of real numbers which I have
already proved. If $0<b<1$, and $n>1$, it follows that $b^n < b$. So for $n>1$,
letting $b = c^{1/n}$ , we get that $ (c^{1/n})^n < c^{1/n}$, which is
$c < c^{1/n}$. But since we assumed that $c^{1/n} < 1$, it follows that $c<1$ , which
contradicts our premise that $c>1$ . Hence $c^{1/n} \nleqslant 1$. So we prove that
$c^{1/n} > 1$.
Is this proof OK ?Furthermore, given $a>1$, I have proved that $a^n > a$ for $n>1$. So plugging, $a = c^{1/n}$,
we can arrive at conclusion that $(c^{1/n})^n > c^{1/n}$. So we can conclude that
$1< c^{1/n} < c $ if $n>1$

(Emo)

 

[Plato2]

I am trying to prove that if $c > 1$, then $c^{1/n} > 1$ , where $n\in \mathbb{N}$.
Now I present my proof here. Since $c>1$, it follows that $c > 0$. Now
$n^{\mbox{th}}$ root theorem says that if $a>0$,then there exists a unique positive
$n^{\mbox{th}}$ root of $a$. So $c^{1/n} > 0 $.

Have you tried to use induction ?

 

[issacnewton]

No, I have not tried induction. It could be used here. But is this proof correct ?