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Homework Statement
Define the set Q[√2] to be the set {a + b√2 | a, b are rationals}, and define addition and multiplication as "usual" (so 2×4 = 8, 2 + 4 = 6, you know, the usual). Show that for any nonzero A in the set Q[√2], there exists an inverse element so that A×A-1 = 1Q[√2].
There is a hint saying that I'll need the fact that √2 is irrational. He asks "where" will I need this fact?
Homework Equations
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The Attempt at a Solution
So I proved that the number 1 itself is the multiplicative identity, and it is also definitely in this set Q[√2]. Now I need to show that there exists an inverse of A so that A×A-1 = 1. And of course A-1 must also be an element of Q[√2].
so let q = a + b√2, then find q-1 = c + d√2 so that
(a + b√2)(c + d√2) = 1.
So we have
ac + ad√2 + cb√2 + 2bd = 1
ac + 2bd + (ad + cb)√2 = 1.
Since √2 is irrational, α + β√2 = 1 implies α = 1, β = 0. This is where i need the fact that √2 is irrational (although i needed it to show 1 = 1Q[√2])
So ad + cb = 0 and ac + 2bd = 1. But how do show that solutions must always exist for these two equations, if I only have two equations but four unknowns? Am i even on the right track here?