Is the odd root of an even number always an irrational number?

[e2m2a]

TL;DR

Is the odd root of an even number always an irrational number?

Is the odd root of an even number always an irrational number? For example the 7th root or the 11th root, etc. of an even number.

 

[jedishrfu]

What about the even number 128? the 7th root is 2.

 

[fresh_42]

What is $128$?

 

[e2m2a]

What about the even number 128? the 7th root is 2.

Is 2 the only exception?

 

[fresh_42]

You have the equation $$N:=2n=\left(\dfrac{r}{s}\right)^{2m+1}$$ to solve. This means $$2ns^{2m+1}=r^{2m+1}$$ so $r$ has to be even. Since we can assume that $r/s$ is a canceled fraction, we also may conclude that all prime factors of $n$ occur in $r$. This yields $$s^{2m+1}=\dfrac{r^{2m+1}}{2n}=\dfrac{r}{2n}r^{2m}.$$ By assumption, no prime factor of $s$ divides $r,$ so $s=1$ and all solutions are $$N=r^{2m+1}$$ with even $r.$

 

[jedishrfu]

Is 2 the only exception?

Of course not consider every even or odd positive integer to the 7th power will be its own 7th root.

 

[Mark44]

Is the odd root of an even number always an irrational number?

No.
Consider $\sqrt[3]{64}, \sqrt[3]{216}$ and many others, including 5th, 7th, and higher odd roots.

Notice that $\sqrt[2n+1]{m^{2n+1}} = m$, where m and n are positive integers, and m could be even or odd.

 

[mathman]

Looks trivial - no. Take any even number to odd power. It will be even and its odd root will be number you started with.

 

[fresh_42]

Looks trivial - no. Take any even number to odd power. It will be even and its odd root will be number you started with.

And I have proven that this is the only possible case.

 

[mfb]

Nothing special about odd or even here.

An integer power of a rational number that's not an integer is not an integer. This is easy to see if the number is written as a/b with coprime a,b. No power of a will ever cancel with b.
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An integer root of an integer is either an integer or irrational.