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

• I
• e2m2a
In summary, the conversation discusses whether the odd root of an even number is always irrational. It is shown that this is not the case, as there are many examples of even numbers with odd roots, such as ##\sqrt{64}## and ##\sqrt{216}##. It is also proven that the only possible case for an even number to have an odd root is when the number is raised to an odd power. Additionally, it is mentioned that an integer power of a rational number that is not an integer is not an integer, which can be seen by writing the number as a fraction with coprime numerator and denominator.

#### e2m2a

TL;DR Summary
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.

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

• PeroK
What is ##128##?

jedishrfu said:
What about the even number 128? the 7th root is 2.
Is 2 the only exception?

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.##

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e2m2a said:
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.

e2m2a said:
Is the odd root of an even number always an irrational number?
No.
Consider ##\sqrt{64}, \sqrt{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.

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

mathman said:
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.

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.