In Z/3, 0^2=0, 1^2 = 1 and 2^2 = 1, so there is no element whose square is 2. Now suppose sqrt(2) is rational. Then p/q maps to a = (p mod 3) / (q mod 3). But since reduction mod 3 respects all the arithmetic operations, a^2 = 2 in Z/3, a contradiction.

From this it is easy to see that n^(1/m) is irrational if there exists at least one prime p such that n is not a perfect mth power in Z/p.

This way of proving things is perhaps not as "pretty" as some of the more elementary methods, but makes a nice connection to modular arithmetic, I think.

--Stuart Anderson

It's reasonably elementary, I would say. But the problem is below. While the next paragraph is correct, finding a suitable p does not appear to be simple even for m = 2. Or should I think better?

>From this it is easy to see that n^(1/m) is irrational if
>there exists at least one prime p such that n is not a
>perfect mth power in Z/p.

However, for prime p, exactly half the nonzero elements of Z/p are squares, so there is a priori a 50% chance of success for each prime. It'should not be a long search, therefore, but still it would be much nicer if some pattern were available to guide one.

It is also true that for any exponent m, the elements of the form a^m form a multiplicitave subgroup in Z/p, so there cannot be more than (p-1)/2 such elements. Again, one has an a priori chance of at least 50% that a given prime p will work.

At the moment, I cannot see a general method for finding primes that work. I will give it'some more thought.

--Stuart Anderson