How to prove that nth root of n is irrational

Question

Can someone please tell me if this is correct or not?

Assume $n^{1/n}$ is rational, i.e $n^{1/n} = a/b$; $a, b$ integers and $b > 1, \gcd(a,b)=1$, and $a^n = nb^n$.

If b exists and $b>1$, it must have at least one prime factor $p$ (Fundamental Theorem of Arithmetic). It follows that $p|b$ and $p|a^n$.

But $p|a^n$ implies $p|a$ (by Theorem: if $p$ prime and $p|uv$, then $p|u$ or $p|v$. Let $v=u^{n-1} \Rightarrow p|u^n$ implies $p|u$).

Therefore $p$ is a common factor of $a, b$ and we have a contradiction.

PS: I am very new at proofs, so please be kind and only reply with elementary proofs/concepts.

Answer 1

I think lulu's comment suggests a more subtle error: even for $n\geq 2$, it is a priori possible that $n^{1/n}$ is an integer, in which case $b=1$ and your argument is flawed. But using calculus (or other methods), one can show the function $f:[1,\infty)\to \mathbb{R}$, $f(x) = x^{1/x}$ has a maximum of approximately $1.445$ at $x=e$. In other words, for $x>1$ $f(x)$ is strictly between $1$ and $2$ and then your argument holds.

Answer 2

That proves that $b$ has no prime factors. Which means $b = 1$ and that $n^{\frac 1n} = a\in \mathbb Z$ and $a^n =n$ for some integer $a$.

Now we must prove that is impossible.

If $a = 1$ then $a^n = 1$ and $n=1$ and indeed $1^{\frac 11} = 1$ with no contradiction. (the statement you are trying to prove is only true if $n > 1$.)

If $a > 1$ how can we prove $a^n \ne n$?

Intuitively I would try to make an argument that $a^n = a\cdot a\cdot .... \cdot a \ge a+a+a+.... + a> 1+1+1+.... + 1 = n$ but that would require a bit of justification and maybe a prove by induction.

(How do I know $a\cdot a \cdot a ...\ge a+a+a+...$? I mean it's "obvious" but what are the details... Well for just two value $a\cdot a \ge 2a =a+a$ and we can bootstrap up by induction... if $a^k \ge k\times a$ then $a^{k+1} = a\cdot a^k \ge 2a^k = a^k + a^k \ge (a+a+....+a) +(a+a+.... +a)>(a+a+....+a)+ a= (k+1)a$ ....that'd work if we put it all together...)

Alternatively we can say that $a> 1$ means that if $m = a-1$ then $m \ge 1$ so $a^n = (1+m)^n$ and be the binomial thereom $(1+m)^n = \sum_{k=0}^n {n\choose k}m^k > \sum_{k=0}^n 1 = n+1 > n$.

That'll do it.

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Seeing the analytic proofs of Integrand and Reimann'sPointyNose proves the $a^n = n; n > 1$ can not have integer $a$ solutions (and further that $1 < a < 2$) I'm a bit unhappy with my handwavy do it in blocks approach.

But an alternative (without any calculus or Mean Value Theorem):

Claim: If $n \in \mathbb N$ and $b > 0$ then $(1+b)^n \ge 1+nb$.

Pf: By binomial theorem $(1+b)^n = 1 + nb + \sum_{k=2}^n{n\choose k}b^k \ge 1+nb$.

And we know if $0 < a < 1$ then $a^n < 1$ and if $a=1$ then $a^n =1$ so if $a^n = n$ the $a > 1$.

So $n = a^n = (1 + (a-1))^n \ge 1+n(a-1) = 1+an-n$ so $an < 2n-1$ and $a < 2-\frac 1n$.

So $1 < a < 2$.....

.... which at this point I rewrite the entire proof frome scratch:

===== REDO FROM GROUND UP ====

If $n \in \mathbb N$ and $n > 1$ then if $1 < n^{\frac 1n} < 2$ and $n^{\frac 1n}$ is irrational.

Pf: Let $r= n^{\frac 1n}$ so $r^n = n$.

Then $r > 1$ (if $0< r\le 1$ then $r^n \le 1 < n$).

Let $m=r-1 > 0$ so $r = 1+m$ and $n=r^n =(1+m)^n\le 1+nm$ so $m < 1$ and $1<r=m+1 < 2$

If $r$ were rational so that $r=\frac ab; a,b \in \mathbb Z; \gcd(a,b)=1; b> 0$, then $r>0$ means $a > 0$ and $1< r< 2$ means $r$ is not an integer and $b \ne 1$.

Then $a^n = nb^n$ and any prime factor of $b$ would be a factor of $a$. But that contradicts $\gcd(a,b) = 1$.

So $r$ is irrational.

Answer 3

As pointed out, you must show that ${n^{\frac{1}{n}}}$ is not an integer, hence ${b>1}$ for this proof to be complete. @Integrand gave an awesome proof for this. Another way you could prove this is by using the Intermediate Value Theorem.

Firstly, if ${n^{\frac{1}{n}}=k}$ for some integer ${k}$, then ${n=k^n}$. Replace $k$ with a variable, ${x}$, and consider the function

$${f(x)=x^n-n}$$

Notice that ${f(x)=0}$ gives you solutions for ${n=k^n}$. We only care about ${x\geq 0}$ (since ${n > 0}$). For the next part of the proof, we assume ${n\geq 2}$.

Notice that ${f'(x)=nx^{n-1}>0\ \forall\ x > 0}$. So ${f}$ is increasing on ${(0,\infty)}$. Next, notice that

$${f(1)=1-n < 0\ \forall\ n \geq 2}$$

and

$${f(2)=2^{n}-n > 0\ \forall\ n \geq 2}$$

So by the Intermediate Value Theorem, ${\exists\ c \in (1,2)}$ such that ${f(c)=0 \Leftrightarrow c^n=n}$. But ${c}$ is not an integer since it's between ${1}$ and ${2}$. To finish the proof, we have to argue this is the only positive root. Well, since ${f(x)}$ was shown to be strictly increasing on the domain we care about, indeed, it has to be. And we are done. We have shown ${n^{\frac{1}{n}}}$ cannot be an integer for ${n\geq 2}$.