For ## n\geq 2 ##, ## \sqrt[n]{n} ## is irrational?

[Math100]

Homework Statement

Establish the following fact:
For $n\geq 2$, $\sqrt[n]{n}$ is irrational.
[Hint: Use the fact that $2^{n}>n$.]

Relevant Equations

None.

Proof:

Suppose for the sake of contradiction that $\sqrt[n]{n}$ is rational for $n\geq 2$.
Then we have $\sqrt[n]{n}=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ such that
$gcd(a,b)=1$ where $b\neq 0$.
Thus $(\sqrt[n]{n})^{n}=(\frac{a}{b})^{n}$
$n=\frac{a^{n}}{b^{n}}$
$b^{n} n=a^{n}$.
Note that $gcd(a,b)=1$ implies $gcd(a^{n}, b^{n})=1$.
This means $b^{n}=1$ because $n\geq 2$ is an integer.
Now we have $a^{n}=n$.
Since $n\geq 2$, it follows that $a\neq 1$.
Thus $a^{n}\geq 2^{n}$ for $a\geq 2$.
But the fact that $2^{n}>n$ contradicts our assumption of $a^{n}\geq 2^{n}$ for
$a\geq 2$.
Therefore, for $n\geq 2$, $\sqrt[n]{n}$ is irrational.

 

[Mike S.]

The use of gcd seems so surprising to me. I wonder if you can simplify this...

1. $\sqrt[n]{n} = x$ with x being rational.
2. $n = x ^ n$
3. n is an integer, so x must be an integer
4. $1 ^ n$ is too small and $2 ^ n$ is too large.

 

[PeroK]

Homework Statement:: Establish the following fact:
For $n\geq 2$, $\sqrt[n]{n}$ is irrational.
[Hint: Use the fact that $2^{n}>n$.]
Relevant Equations:: None.

Proof:

Suppose for the sake of contradiction that $\sqrt[n]{n}$ is rational for $n\geq 2$.

This is the same mistake you always make. Okay, you might just have omitted the word "some". But based on previous homework, you always/often take the "opposite" instead of the negation.

You have to find some way to stop making this same mistake again.

 

[PeroK]

The use of gcd seems so surprising to me. I wonder if you can simplify this...

1. $\sqrt[n]{n} = x$ with x being rational.
2. $n = x ^ n$
3. n is an integer, so x must be an integer
4. $1 ^ n$ is too small and $2 ^ n$ is too large.

@Math100 This is so much easier to read than your rambling style. You need to change the way you write proofs.

 

[fresh_42]

Homework Statement:: Establish the following fact:
For $n\geq 2$, $\sqrt[n]{n}$ is irrational.
[Hint: Use the fact that $2^{n}>n$.]
Relevant Equations:: None.

Proof:

Suppose for the sake of contradiction that $\sqrt[n]{n}$ is rational for $n\geq 2$.
Then we have $\sqrt[n]{n}=\frac{a}{b}$ for some $a,b\in\mathbb{Z}$ such that
$gcd(a,b)=1$ where $b\neq 0$.
Thus $(\sqrt[n]{n})^{n}=(\frac{a}{b})^{n}$
$n=\frac{a^{n}}{b^{n}}$
$b^{n} n=a^{n}$.
Note that $gcd(a,b)=1$ implies $gcd(a^{n}, b^{n})=1$.
This means $b^{n}=1$ because $n\geq 2$ is an integer.
Now we have $a^{n}=n$.
Since $n\geq 2$, it follows that $a\neq 1$.
Thus $a^{n}\geq 2^{n}$ for $a\geq 2$.
But the fact that $2^{n}>n$ contradicts our assumption of $a^{n}\geq 2^{n}$ for
$a\geq 2$.
Therefore, for $n\geq 2$, $\sqrt[n]{n}$ is irrational.

This is correct. You should try to write it without contradiction, and generally avoid unnecessary statements. E.g. the line $b^n n=a^n$ set me on a completely wrong track. Your argument used the fraction formula, not the integer version. From $b^n n=a^n$ you can only conclude that $n$ divides $a$.

 

[nuuskur]

From the other thread we also have "$\sqrt[n]{a}$ is rational only if $\sqrt[n]{a}$ is an integer" for positive integers $a$. Exclude the possibility that $\sqrt[n]{n}$ is an integer and the result follows.

 

[fishturtle1]

The use of gcd seems so surprising to me. I wonder if you can simplify this...

1. $\sqrt[n]{n} = x$ with x being rational.
2. $n = x ^ n$
3. n is an integer, so x must be an integer
4. $1 ^ n$ is too small and $2 ^ n$ is too large.

How do you prove 3. without something 'Write $x =a/b$ with $\gcd(a, b) = 1$. Then $xb^n = a^n$. ...'

 

[epenguin]

@Math100 This is so much easier to read than your rambling style. You need to change the way you write proofs.

I have the impression he is being told and taught to write them that way, so sometimes pays more attention to supposedly required style than substance.

 

[Office_Shredder]

I thought the proof here was fine. It's basically the same as you would to to prove $\sqrt{2}$ is irrational, and I agree with fish that Mike's proof is only as good as the assumption in step 3.

There are plenty of posts that have actual mistakes, there's no point in clogging up this thread with a discussion about the style.

 

[WWGD]

It seems you can consider the function $$a^x-x $$ and show it has no zeros. For $$x=0$$ there's no equality ; $$ a^0=1>0$$. Then show $$a^x $$ grows faster than $$x$$.