# Question about the validity of a group of rational numbers

I know that I need to prove the four criteria of a group (closure, associative, identity, and inverse), but I'm having a hard time with closure. When you input rational numbers, I don't see how you can get a rational number out. Any input would be greatly appreciated!

• That isn't a group of rational numbers! – Angina Seng Sep 14 '18 at 6:32
• So the question is invalid? I cannot prove that it is a group, either with addition or multiplication? – mbb254 Sep 14 '18 at 6:37
• The question is valid. – Angina Seng Sep 14 '18 at 6:49
• Hint: Can you find a larger group which this is a subset of? – Tobias Kildetoft Sep 14 '18 at 6:52

You can still prove this a group under addition, the question is perfectly valid since only a and b are required to be rational. Consider first the identity property. Taking $a=b=0$ we see that this is trivially satisfied, since 0 is the additive identity. For closure, let $a+\sqrt{2}b$ and $c+ \sqrt{2}d$. Then their sum is given by $(a+c)+(b+d) \cdot \sqrt{2}$ which is of the form of the other elements of the group since addition of two rational numbers gives a rational number. Thus closure also holds. Inverses are similar since for any $a+b \cdot \sqrt{2}$ we have that $-a-b \cdot \sqrt{2}$ is also an element of this group since $-a$ and $-b$ are rational. I leave associativity and part b to you.

$R$ is a group, so you need prove that $Q[\surd2]$ is a subgroup of $R$ (w.r.t. adition, resp., without $0$, w.r.t. multiplication). I.e. prove a closure w.r.t +, resp. *, and -, resp $^{-1}$

I.e.

a] $(a+b\surd2) + (c+d\surd2) =(a+c) +(c+d) \surd2$

$-(a+b\surd2 ) =-a-b\surd2$

b] $(a+b\surd2)(c+d\surd2)=(ac+2bd)+(ad+bc)\surd2$

$(a+b\surd2)^{-1} = a/(a^2-2b^2) +(-b\surd2)/(a^2-2b^2)$

and $a+b$, $c+d$ ... are rational

PS: $a-2b^2 \neq 0$ for rational $a, b$