More intro abstract algebra problems

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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?
 

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HallsofIvy
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This appears to be a fairly advanced class so surely you are already familiar with "rationalizing the denominator"! If [itex]x= 1/(a+b\sqrt{2})[/itex] then
[tex]x= \dfrac{1}{a+ b\sqrt{2}}\dfrac{a-b\sqrt{2}}{(a- b\sqrt{2}}= \dfrac{a- b\sqrt{2}}{a^2- 2b^2}[/tex]
Of course, both [itex]a/(a^2- 2b^2)[/itex] and [itex]-b/(a^2- 2b^2)[/itex] are rational as long as [itex]a^2- 2b^2[/itex] is not 0! That is the part that requires that [itex]\sqrt{2}[/itex] not be rational.
 
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Of course, the first thing i thought it was the reciprocal, but sometimes I get so worked up in these so-called "advanced" courses that i forget such simple things i learned in high school. I guess it's time i put it all together.

Thanks very much for your help, HallsofIvy!

Edit: while this may be an advanced class, the stuff we are doing is far from advanced!
 
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