- #1

- 354

- 11

- 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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter e2m2a
- Start date

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[3]{64}## and ##\sqrt[3]{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.

- #1

- 354

- 11

- TL;DR Summary
- Is the odd root of an even number always an irrational number?

Mathematics news on Phys.org

- #2

Mentor

- 14,597

- 8,784

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

- #3

- 18,819

- 22,623

What is ##128##?

- #4

- 354

- 11

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

- #5

- 18,819

- 22,623

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.##

Last edited:

- #6

Mentor

- 14,597

- 8,784

Of course not consider every even or odd positive integer to the 7th power will be its own 7th root.e2m2a said:Is 2 the only exception?

- #7

Mentor

- 37,350

- 9,534

No.e2m2a said:Is the odd root of an even number always an irrational number?

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.

Last edited:

- #8

Science Advisor

- 8,140

- 571

- #9

- 18,819

- 22,623

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

- #10

Mentor

- 37,004

- 13,700

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.

<=>

An integer root of an integer is either an integer or irrational.

Share:

- Replies
- 31

- Views
- 2K

- Replies
- 6

- Views
- 1K

- Replies
- 3

- Views
- 1K

- Replies
- 1

- Views
- 1K

- Replies
- 23

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 7

- Views
- 1K

- Replies
- 2

- Views
- 710

- Replies
- 13

- Views
- 3K

- Replies
- 1

- Views
- 946