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Given (For all $a,b\in Q$, $a+b\in Q$ and $ab\in Q$)

This was a two part question.

Part a) is to prove that $Q$ is closed under addition and multiplication.

Part b) is prove that if $a\in Q$ and $a\ne0$, then $\frac1a\in Q$.

I proved part a but I'm stuck on part b... I know that I have to let $a=c+d\sqrt2$ and some how try and move things around to try and make it "look" like $p+q\sqrt2$ but I can't seem to get it.

Mario Carneiro
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Jared
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  • Is $q$ the set of rational numbers? That's usually denoted by $\Bbb Q$ (you write $\Bbb Q$ to get it), and your set $Q$ is then usually denoted as $\Bbb Q(\sqrt2)$ (written like this: $\Bbb Q(\sqrt{2})$). A hint for b: write $a\cdot\frac1a=(a_1+a_2\sqrt 2)(b_1+b_2\sqrt2)=1$, and work from there – Arthur Feb 04 '15 at 06:17
  • q is the set of rational numbers, sorry i think i should type in word to show the proper notations... but Ill try that way also, but I just thought that if you are trying to prove that 1/a exists in Q then you cant use 1/a in your proof? – Jared Feb 04 '15 at 06:32
  • It is unfortunate that both $Q$ and $\mathbb{Q}$ are used in the question. The notation $\mathbb{Q}$ is fairly standard for the rationals, but $Q$ for all $p+q\sqrt{2}$ where $p$ and $q$ range over the rationals is not. – André Nicolas Feb 04 '15 at 07:04
  • Similar question for $\sqrt7$: http://math.stackexchange.com/questions/324186/f-ab-sqrt7-mid-a-b-in-mathbbq-closed-under-addition-subtracti Another related question with $\mathbb Z$ instead of $\mathbb Q$: http://math.stackexchange.com/questions/142717/the-ring-ab-sqrt2-mid-a-b-in-mathbbz I was not able to find post about precisely this question, but I think that it is very probable that it has been posted on this site before. – Martin Sleziak Feb 04 '15 at 08:47
  • Hint $\ $ Rationalize the denominator. $\ $ – Bill Dubuque Feb 04 '15 at 19:33

1 Answers1

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$\frac{1}{a+b\sqrt{2}}=\frac{a-b\sqrt{2}}{(a+b\sqrt{2})(a-b\sqrt{2})}=\frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$

Wojowu
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  • I like your explanation. I began to wonder if it is necessary to show $a^2-2b^2\neq0$? I'm pretty certain it's impossible but can't think of a proof. – Karl Feb 04 '15 at 06:26
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    The irrationality of $\sqrt{2}$ gives that neither factor of $a^2-2b^2$ can be zero. – Pockets Feb 04 '15 at 06:38
  • I was driving to work when I realised! Sometimes I don't see the wood for the trees. – Karl Feb 04 '15 at 07:19
  • @Karl The original Pythagorean proof of the irrationality of $\sqrt2$ is often, in modern mathematics, reduced to "The equation $a^2 = 2b^2$ has no solution among the integers." It's not a big leap to conclude that there are no solutions among the rationals. – Arthur Feb 04 '15 at 11:10
  • @Karl his is known as rationalizing the denominator. See here for many examples. – Bill Dubuque Feb 04 '15 at 19:36
  • I understand rationalizing the denominator. The proof relies on $a\neq2b^2$ or there'd be a divide zero error. This morning I didn't realise this was just $(a/b)=\sqrt{2}$ and not a problem due to $\sqrt{2}$ irrational. I've been kicking myself for being so daft all day. – Karl Feb 04 '15 at 19:58