# Math Help - Closure and MIV

1. ## Closure and MIV

Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.

Q[Sqrt(2)] is define by a+bsqrt(2), a and b in Q.

MIV is existence of multiplicative inverse Ex: aa^-1=1

2. Originally Posted by JCIR
Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.
When you post a question why don't you define the terms?
I dare say that most of us have no idea what 'MIV' means!

3. Originally Posted by JCIR
Prove that Q[sqrt(2)] is closed under addition, and show that MIV Holds.

Q[Sqrt(2)] is define by a+bsqrt(2), a and b in Q.

MIV is existence of multiplicative inverse Ex: aa^-1=1
To construct MIV:

$\frac1{a+b\sqrt{2}} = \frac1{a+b\sqrt{2}} \times \frac{a-b\sqrt{2}}{a-b\sqrt{2}} = \frac{a}{a^2 - 2b^2}+\frac{-b\sqrt{2}}{a^2 - 2b^2}$

If $a,b \in \mathbb{Q}$, so will $\frac{-b}{a^2 - 2b^2}$ and $\frac{a}{a^2 - 2b^2}$

Thus $(a+b\sqrt{2})^{-1} \in \mathbb{Q}[\sqrt2]$

To prove its existence, start with the constructed form and multiply out and show that the product is 1

4. I suppose showing it is closed under addition is no problem?