Prove that the irrational numbers do not form a field.?

Question

By providing a counterexample to one of the field axioms prove that the set of irrational numbers does not form a field.

Answer 1

I dont know anything about set theory or any fields and junk... but after reading Rob K, I think I have a counter example to whatever in the heck he said.

What about transcendental numbers? They are irrational and dont follow this squaring / arithmetic rule.

Answer 2

This should be pretty simple...to be a field, a set of numbers must be closed under all the arithmetic operations, that is, +,-,*,/. Obviously we can choose any square root of a nonsqare positive integer such as 2, 3, 5, 6, 7, 8, 10, ... and so on. Squaring such an irrational number obviously gives a natural number, and is not irrational. Since the multiplication of these two irrational numbers (which happens to be the same number, but that is irrelevant) is a rational number, the irrational numbers fail to be closed under multiplication and thus do not form a field.

Answer 3

The answers so far are correct, but even easier would be to remind that EVERY FIELD MUST contain zero and unit elements - the neutral elements of both binary operations - addition and multiplication. But 0 and 1 are rational - the set of irrational numbers does not contain them.

Note: the "smallest" possible field - Galois Field GF(2) contains exactly 2 elements GF(2) = {0; 1}

Answer 4

the set of Irrational numbers can not be expressed as a/b. 0 not belong to Irr. numbers, and 0 times any Irr. number is own number. So this axiom not hold and Irr. numbers not form a fieid

Answer 5

Let r be irrational. then -r is irrational. they add up to a rational number.