- #1

stripes

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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}= 1

_{Q[√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 = 1

_{Q[√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?