Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication

Question

We are given $G=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$ and asked to show that $G-\{0+0\sqrt{2}\}$ is a group under multiplication.

Let us first specify that multiplication takes place as expected:

$$(a+b\sqrt{2})(c+d\sqrt{2})=a(c+d\sqrt{2})+b\sqrt{2}(c+d\sqrt{2})=ac+2bd+(ad+bc)\sqrt{2}$$

It is clear that $1+0\sqrt{2}$ is the multiplicative identity.

But I have hit a blocker on finding the inverse. Dummit and Foote's hint is to 'rationalize the denominator' but I'm not sure what this means in this case. If I try straight up equating coefficients in

$$(a+b\sqrt{2})(c+d\sqrt{2})=1+0\sqrt{2}$$ then that's just $$ac+2bd=1$$ and $$ad+bc=0$$. That's not a very nice set up at all. Any hints?

Answer 1

Well, for any $a+b\sqrt{2}$ you have to find an inverse $c+d\sqrt{2}$, so your setup is actually pretty nice: it consists of a linear system

$$ \begin{pmatrix} a & 2b\\ b & a \end{pmatrix} \cdot \begin{pmatrix} c\\ d\\ \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix}. $$

You can compute the inverse of the matrix explicitly in terms of the determinant, multiply on the left and you will get a formula for $c$ and $d$ depending on $a$ and $b$.

Edit: also, note that if $a+b \sqrt{2} \neq 0$, then we cannot have $a^2 = 2b^2$ by a factorization argument. Hence the determinant is always nonzero for nonzero elements.

Answer 2

Hint:

Multiply top and bottom of $\;\dfrac1{a+b\sqrt2}\;$ by $\;{a-b\sqrt2}$

Answer 3

It suffices to show that $G\le (\Bbb R^*, \times)$, since any subgroup of a group is itself a group.

I will use the one-step subgroup test.

Clearly $1\in G$, so $G\neq \varnothing$.

Let $x=a+b\sqrt{2}, y=c+d\sqrt{2}\in G$. Then $a,b,c,d\in \Bbb Q$, so, in particular, $x\in \Bbb R^*$ as $x\neq 0$. Hence $G\subseteq \Bbb R^*$. Furthermore, we have

$$\begin{align} xy^{-1}&=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\\ &=\frac{a+b\sqrt{2}}{c+d\sqrt{2}}\times \frac{c-d\sqrt{2}}{c-d\sqrt{2}}\tag{1}\\ &=\frac{(a+b\sqrt{2})(c-d\sqrt{2})}{c^2-2d^2}\\ &=\frac{ac-2bd}{c^2-2d^2}+\frac{bc-ad}{c^2-2d^2}\sqrt{2}, \end{align}$$

which is in $G$ as neither $x$ nor $y$ is zero. Hence $xy^{-1}\in G$.

Hence $G\le (\Bbb R^*, \times)$.

Hence $G$ is a group.

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$(1)$: Here is what is meant by "rationalise the denominator".

Answer 4

Bear in mind the multiplication in $G$ is the same as the multiplication of the reals so whatever is true about multiplication of $\mathbb R$ will still be true of the multiplication of $G$.

Multiplication is associative. We don't have to prove that.

We do have to prove that mulitplicaition is closed on $G$. That is: if $w \in G$ and $u \in G$ then $wu \in G$. (which you did).

The identity in $\mathbb R$ is $1$ so the identity for $G$ will be $1$. We don't have to prove that $1*w = w$ for all $w\in G$ and we don't have to solve that if $e*w = w$ then $e=1$. We know that must be true. The only thing we have to prove is that $1$ is actually a member of $G$. (Which as $1 = 1 + 0\sqrt 2$ it obviously is.

And as the invers of $x$ is $\frac 1x$, we do know that the inverse of $w \in G$ will HAVE to be $\frac 1w$. The issue though is we have to prove that if $w \in G$ then $\frac 1w \in G$ also.

SO let's prove that:

If $w = a+ b\sqrt 2\ne 0$ then $\frac 1w = \frac 1{a+b\sqrt 2}$. And that is in $G$???? We can write $\frac 1{a+b\sqrt 2}$ as $c +d \sqrt 2$ where $c$ and $d$ is rational? Is that actually true? How would we show that? How would we figure out what $c,d $ are?

Well, the hint "rationalize the denominator" is very apt.

$\frac 1{a+b\sqrt 2} = \frac 1{a+b\sqrt 2}\cdot \frac {a-b\sqrt 2}{a-b\sqrt 2}=$

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

And as $a,b \in \mathbb Q$ we have $\frac a{a^2 -2b^2},-\frac b{a^2 -2b^2}\in \mathbb Q$. So $\frac 1{a+b\sqrt 2} = \frac a{a^2 -2 b^2}- \frac b{a^2 -2b^2}\sqrt 2\in G$.

And that's that.

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Actually $ac + 2bd = 1$ and $ad + bc = 0$ is a perfectly fine set up.

$d = -\frac {bc}a$ (assuming $a \ne 0$)

$ac + 2b(-\frac {bc}a) = ac- \frac {2b^2c}a =2$

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

$c = \frac 1{a-\frac{2b^2}a}=\frac {a}{a^2 -2b^2}$ (assuming $a-\frac{2b^2}a\ne 0$.... which it cant because $(a-\frac{2b^2}a)c = 1 \ne 0$.)

So $d =\frac {b \frac {1a}{a^2 -2b^2}}a= \frac {b}{a^2 - 2b^2}$

....

And if $a$ does equal $0$ we have

$ac + 2bd = 2bd =1$ and so $b\ne 0$ and $ad +bc = bc = 0$.

Now $b \ne 0$ as 1) $2bd =1 \ne 0$ and also because we are told $a + b\sqrt 2 \ne 0$ and if $a =0$ then....

SO $b\ne 0$ and $bc = 0$ so $c = 0$.

And $2bd= 1$ so $b = \frac 1{2b}$.

... in other words $b\sqrt 2\cdot x = 1 \implies x = \frac 1{2b}\sqrt 2$

Answer 5

We have $$ac+2bd=1,ad+bc=0$$ and that is nice because it is a system of linear equations in two variables. Multiplying first equation by $b$ and second one by $a$ and subtracting the two we get $$(2b^2-a^2)d=b$$ ie $$d=\frac{b} {2b^2-a^2}$$ and $$c=-ad/b=\frac{a} {a^2-2b^2}$$ thus $$c+d\sqrt {2}=\frac{a}{a^2-2b^2}+\frac{b}{2b^2-a^2}\sqrt{2}$$ is the inverse of $(a+b\sqrt{2})$.

I wonder why you stopped at those equations. Not every question is difficult and try to have more confidence in yourself.