# Roots of integers

An integer is either a perfect square or its square root is irrational. Said a different way, when you compute the square root of an integer, there are either no figures to the right of the decimal or there are an infinite number of figures to right of the decimal and they don’t repeat. There’s no middle ground. You can’t hope, for example, that the decimal expansion might stop or repeat after a hundred or so terms.

This theorem came up recently when I was talking to one of my kids about her math class, so I decided to look up the proof. It’s easier than I expected, not much harder than the familiar proof that the square root of 2 is irrational. Here goes.

Suppose a/b is a fraction in lowest terms, i.e. a and b are relatively prime, and a/b a solution to xn = c where n > 0 is an integer and c is an integer. Then

(a/b)n = an / bn = c

and so

an / b = c bn-1.

Now the right side of the equation above is an integer, so the left side must be an integer as well. But b is relatively prime to a, and so b is relatively prime to an. The only way an / b could be an integer is for b to equal 1 or -1. And so a/b must be an integer.

This proof shows that what we said about square roots extends to cube roots and in fact to all integer roots. For example, the fifth root of an integer is either an integer or an irrational number.

Update: Note that there’s another proof in the comments, one that I believe is easier to follow.

## 13 thoughts on “Roots of integers”

1. I. J. Kennedy

Another way to get the same result is to assume a/b is fraction in lowest terms and is not an integer (i.e. b ≠ 1), and consider powers (a/b)n. Clearly an and bn are relatively prime, and the denominator bn ≠ 1, so an/bn is not an integer. In other words, no (non-integer) fraction, when raised to a power, can produce an integer.

3. How do you know that an and bn are relatively prime? The work amounts to the messy parts in the post. This version is only simpler because it hides the hard bits in a lemma.

4. I. J. Kennedy

Perhaps you’re right, but I’m not sure. In any case we know a^n and b^n are relatively prime by the Fundamental Theorem of Arithmetic. If a factors to certain primes, then multiplying a times itself isn’t going to yield any new primes. So the factorizations of a^n and b^n have no primes in common, because a and b don’t.

5. I remember being amazed when I first saw this proof, at how easy it was. I wondered then (and still do) why we never had it in school. We never did any proofs except in geometry and trig.

6. Denise, I agree that is seems odd that proofs in high school are limited to geometry. (And from what I understand, proofs have largely been removed from geometry.) Number theory seems like a reasonable place to learn proofs.

7. sherifffruitfly

“Denise, I agree that is seems odd that proofs in high school are limited to geometry.”

The intellectual horsepower level of education majors doesn’t permit anything more than that.

8. The statement is that an integer is either (1) a perfect square or (2) its square root is irrational. What about the 2nd part?

9. Caio Braz

Weiwei, we’ve got the two cases covered:
Part 1 is ok, if n is an integer, if its square root is an integer, its ok!

Part 2: if some integer has a rational (i.e. a number in form a/b, with a and b integers) square root, then, some fraction (a/b)^2 will be this integer.

The proof stands that it’s impossible to a fraction yield this result, so if the square root is not an integer, it can’t be a rational, so, it’s irrational

10. @caio Indeed! Thanks for the clarification :)

11. S.B. Easwaran

Mr Kennedy’s proof is neat. If a and b are co-prime, a^n and b^n will be co-prime because it’s the same primes that the decomposition of a and b yield that will be n-plicated (on the lines of duplicated, triplicated etc) in a^n and b^n. Since a and b have no prime factors in common, a^n and b^n cannot have any prime factors in common.

What I particularly like is how Mr Kennedy’s proof generalises the irrationality idea for all exponents and all nth roots.

Regards,

12. S.B. Easwaran

And Mr Cook, I really like how you put it: “Said a different way, when you compute the square root of an integer, there are either no figures to the right of the decimal or there are an infinite number of figures to right of the decimal and they don’t repeat. There’s no middle ground.”

Corking! That’s a gem in the communication of ideas, especially for students who are still bending their minds to the idea of irrationality.

And, may I ask, why restrict that sentence to square roots alone? One might put it this way: “Said a different way, when you compute the N-TH ROOT of an integer, there are either no figures to the right of the decimal or there are an infinite number of figures to right of the decimal and they don’t repeat. There’s no middle ground.” (CAPS FOR EMPHASIS)

Regards,

13. John Linehan

As a teacher of English for many decades who changed to teaching Maths, and ended up teaching teachers how to teach maths, I believe that part of the problem of teaching mathematical concepts is the divergence in meaning of words in Maths and in common usage.

The meaning we all ascribe to ‘irrational’ and many other words in maths and the sciences has various meanings when we use them in conversation, but has a specific meaning in maths.

Sometimes I’ve thought that it would have helped if we made up entirely new words whenever we believe that their common usage might make them difficult for learners to understand.