# Determining groups not sure how to prove it.

1. Oct 23, 2011

### lostNfound

I'm going through a abstract algebra book I found and am trying to learn more about group theory by going through some of the proofs and practice sets, but am having trouble with the following:

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. Oct 23, 2011

### SteveL27

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.

3. Oct 24, 2011

### willem2

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.

Inverses are a little harder, altough a very common trick that you should definitely know can be used to get

$$\frac { 1 } { a + b \sqrt(2) }$$ in the form: c + d sqrt(2)

4. Oct 24, 2011

### HallsofIvy

Why did you delete this? I wouldn't say you should assume associativity but certainly you can just note that multiplication of real numbers is associative and this is just a subset of the real numbers. To show closure write the product of two such numbers as $(a+ b\sqrt{2})(c+ d\sqrt{2})$ and actually do the multiplication. What do you get? Show that it can be written as $u+ v\sqrt{2}$ by showing what u and v must be. The identity is $1= 1+ 0\sqrt{2}$, of course.

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

5. Oct 24, 2011

### lostNfound

Sorry I deleted that. I was trying to delete the post so people didn't feel like they needed to keep answering it. I was able to work out the answer on my own earlier. Thanks for the help though. I think we are on the same page.