The real numbers have no zero divisors, so it's a pretty safe bet that both are at least integral domains. To determine if there is an inverse, don't start from scratch. Multiply (a+sqrt(2)*b)*(a-sqrt(2)*b) and try to figure out where to go from there.

Ok, my guess was really just a hint. a^2-2*b^2 is unlikely to be 1. Now think about (a+b*sqrt(2))*k*(a-b*sqrt(2))=1 where k is some constant in your system. Can you solve for k if you are working in the rationals?

Well, no. Using that value of k, I'd say the inverse of a+sqrt(2) is (a-sqrt(2))/(a^2-2), isn't it? Your main worry over the rationals is whether a^2-2*b^2 could be zero.

I thought we were getting towards being able to agree that the inverse of (a+b*sqrt(2)) is (a-b*sqrt(2))/(a^2-2*b^2)? In so far as that is defined. Is it defined over the rationals? Let's not go backwards here.

You are still missing the whole point! [itex](a+ b)(a- b)= a^2- b^2[/itex]. I'm sure you learned that long ago. So what do you get if you multiply both numerator and denominator of
[tex]\frac{1}{a+ b\sqrt{2}}[/tex]
by
[tex]a- b\sqrt{2}[/tex]

I don't see why people in this thread have an aversion to "starting from scratch". Linear algebra is a very useful technique in abstract algebra, and both of the questions

What, when multiplied by (a + b[itex]\sqrt{2}[/itex]), gives 1?

What, when multiplied by (a + b[itex]\sqrt{2}[/itex]), gives 0?

are linear systems of equations in two unknowns. (treating a and b as "knowns")

Rationalizing denominators is a good trick, but far from required for this problem.

Rationalizing is when you get rid of the square root in the denominator? I was wrong. You can rationalize the denominator with the difference of two squares trick.

As established earlier, nothing non-zero multiplied by a real number yields zero. Thus, that system is not possible. The inverse is [a - b*sqrt(2)] / [a^{2} - 2b^{2}].