- #1

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Prove that G={a+b*sqrt(2) | a,b E R; a,b not both 0} is a group under ordinary multiplication.

Any help would be awesome!

- Thread starter lostNfound
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- #1

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Prove that G={a+b*sqrt(2) | a,b E R; a,b not both 0} is a group under ordinary multiplication.

Any help would be awesome!

- #2

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Where are you stuck? You know that you need to verify each of the group axioms. Is your set closed under the given binary operation? Is there an identity? etc.

Prove that G={a+b*sqrt(2) | a,b E R; a,b not both 0} is a group under ordinary multiplication.

Any help would be awesome!

- #3

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Closure is easy, you just take two arbirtrary elements a+b sqrt(2) and c + d sqrt(2), compute the product and show that it also is a member of G.So I know that in order to prove it is a group, there are several things that have to be confirmed, including associativity, closure, inverses, and identity. Associativity seems like it can be assumed, but the others till have to be proved. I'm having most of my trouble with the closure and inverse proofs, especially that for closure.

Inverses are a little harder, altough a

[tex] \frac { 1 } { a + b \sqrt(2) } [/tex] in the form: c + d sqrt(2)

- #4

HallsofIvy

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Why did you delete this? I wouldn't say you shouldSo I know that in order to prove it is a group, there are several things that have to be confirmed, including associativity, closure, inverses, and identity. Associativity seems like it can be assumed, but the others till have to be proved. I'm having most of my trouble with the closure and inverse proofs, especially that for closure.

And the multiplicative inverse of [itex]a+ b\sqrt{2}[/itex] is [itex]1/(a+ b\sqrt{2})[/itex]. Rationalize the denominator to show how that can be written in the form [itex]u+ v\sqrt{2}[/itex].

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