# Another proof that *n*^{th} roots of integers are irrational, except for perfect nth powers.

### (Based on the Rational Root Theorem)

**Scott E. Brodie**

6/2/06

We begin by recalling two helpful propositions from Euclid.

First, VII.24:

If two numbers are relatively prime to any number, then their product is also relatively prime to the same.

**Proof**: Suppose *a* is relatively prime to both *b* and *c*. Since *a* and *b* are relatively prime, there exist integers (perhaps negative) *m* and *n* such that *ma* + *nb* = 1*ja* + *kc* = 1*j*, *k*.

Multiplying these two equations together,

(ma + nb)(ja + kc) | = 1 |

= maja + makc +nbja + nbkc | |

= (maj + mkc + nbj)a + (nk)bc | |

= 1 |

so *a* and *bc* are relatively prime.

Repeating the argument verifies that if *a* is relatively prime to *b*, then *a* is relatively prime to *b ^{n}* for any positive integer

*n*.

Second, VII.30:

If two numbers, multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original factors.

It is no more work to prove a simple generalization: If integers *a* and *b* are relatively prime, and *a* divides the product *bc*, then *a* divides *c*. [See the aforementioned "Remark" for the proof].

With these tools in hand, we can now prove the *Rational Root Theorem*, from which the general result on the irrationality of *n*-th roots follows as a simple corollary:

### Rational Root Theorem

Let *P*(*x*) be a polynomial with integer coefficients, say

*P*(*x*) = *a*_{n}*x*^{n} + *a*_{n-1}*x*^{n-1} + ... + *a*_{1}*x* + *a*_{0}

and suppose that *r* = *c*/*d* is a rational root of *P* [that is,*P*(*r*) = 0]*c* and *d* are relatively prime). Then *c* divides *a*_{0} and *d* divides *a*_{n}.

**Proof**: Inserting the argument *x* = *c*/*d* into the expression for *P*(*x*) yields

0 = *a*_{n} *c*^{n}/*d*^{n} + *a*_{n-1} *c*^{n-1}/*d*^{n-1} + ... + *a*_{1} *c*/*d* + *a*_{0}

Multiplying through by *d*_{n} and isolating the first term yields

-*a*_{n}*c*^{n} = *a*_{n-1}*c*^{n-1}d + ... + *a*_{1}c*d*^{n-1} + *a*_{0}*d*^{n}

Since *d* is a factor of every term on the right hand side of this equation, *d* must divide *a*_{n}*c*^{n}. But *c* and *d* are relatively prime, so *d* and *c*^{n} are relatively prime, and it follows from the generalized version of VII.30 that *d* divides *a*_{n}.

Isolating the last term instead of the first, we see that

*a*_{n}*c*^{n} + *a*_{n-1}*c*^{n-1}d + ... + *a*_{1}c*d*^{n-1} = - *a*_{0}*d*^{n}

As before, since *c* is a factor of every term in the left side of this equation, *c* must divide *a*_{0}*d*^{n}. Since *c* and *d* are relatively prime, *c* and *d*^{n} are relatively prime, and we conclude that *c* must divide *a*_{0}.

Now consider the equation for the *n*^{th} root of an integer *t*: *x*^{n} - *t* = 0.

If *r* = *c*/*d* is a rational *n*^{th} root of *t* expressed in lowest terms, the Rational Root Theorem states that *d* divides 1, the coefficient of *x*^{n}. That is, that *d* must equal 1, and *r* = *c* must be an integer, and *t* must be itself a perfect *n*^{th} power.

Of course, the same arguments could be applied to just the simple equation *x*^{n} = *t*. (Indeed, such an argument is the essence of Proofs 2 and 4 of the irrationality of the square root of 2.) Somehow, though, the role of VII.24 and VII.30 is better seen in the more general case. And, for essentially the same effort, we get the useful Rational Root Theorem in the bargain.

(Note that Gauss' Lemma is a specialization of the above theorem to *monic polynomials*.)

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