# Is there a number $a$ such that $a^2$ is irrational, but $a^4$ is rational?

This problem is from Spivak Calculus, now you immediately think $a^2 = \sqrt{2}$ and $a^4 = 2$, so the answer to the problem is yes. But, the way Spivak has developed his book:

1. So far we only have the algebraic axioms (associativity, commutativity, units, inverses, distributivity).
2. Significantly, we don't have the completeness axiom yet.
3. We have already proved that if there is a number $x$ such that $x^2 = 2$ then $x$ can't be rational.

The important thing is that we haven't yet proved that such $x$ indeed exists (to prove this we need the completeness axiom, which we still don't have), only that if it happens to exist it should be irrational, thus I can't say that $\sqrt{2}$ is a solution to this problem just yet. Am I right or am I being just too picky?

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Take y = $\sqrt x$ where $x$ is as above in pt. 3, so we need to show existence of such y. – Ram Jan 30 '13 at 4:51
What is meant by number? – jspecter Jan 30 '13 at 4:55
Offhand it seems that the most you can say at this point is that if there is a number $a$ such that $a^4=2$, it has the desired property. That’s clearly correct and does not go beyond the axioms that you’ve been given thus far. I suspect that you’re supposed to do exactly what you’ve done: worry about existence and recognize that $\sqrt[4]{2}$ is such a number $-$ if it exists. – Brian M. Scott Jan 30 '13 at 4:55
@Ram Without completeness (or some property that will imply completeness), for all we know all real numbers might be rational, so we can't assume that such a $y$ exists. – Brett Frankel Jan 30 '13 at 4:56
The best you can get then is that if the fourth root of 2 exists then it satisfies the hypotheses. – Sam DeHority Jan 30 '13 at 4:59

You're right. So far you don't even have enough axioms to prove that your number system is unique up to isomorphism -- in fact, any field, including $\mathbb{Q}, \mathbb{R}, \mathbb{C}, F_2,$ etc all satisfy them. You can neither prove nor disprove existence of such an $\alpha$.