I was doing an exercise below that appears to be simple, but my textbook says that algebraic system defined below is not a field for $a,b \in \mathbb R$. I checked it with the definition of a field, but I don't see where $a$ and $b$ being real would make this system not be a field. Perhaps there is a mistake in the textbook as it is edition 1972. Here is the question:
Let $K$ be the set of all pairs $(a,b)$ such that:
a) $a,b \in \mathbb Q$
b) $a,b \in \mathbb R$
For every $(a,b),(c,d)\in K^2$:
$$(a,b)+(c,d)=(a+c,b+d)$$
$$(a,b)*(c,d)=(ac+2bd,ad+bc)$$
Additionally $(0,0)$ is an identity for the first operation and $(1,0)$ is an identity for the $2$nd one.
Prove that set $K$ with operations defined in this way is a field in case a), but not in case b).
What's the inverse element for $*$ ?
Suppose $(a,b)\ne (0,0)$ then is there an inverse for $(a,b)$ always?
If $(a,b)*(x,y)=(ax +2by,ay+bx)= (1,0)$
Then $x=-y\frac ab$ (assuming $b\ne 0$) and $-\frac{a^2}by+2by=1$ or $y=\frac 1{2b-\frac {a^2}b}$ (assuming $2b-\frac {a^2}b \ne 0$).
So that's the inverse. $(a,b)^{-1}= (-y\frac ab,\frac 1{2b-\frac {a^2}b})$
That is, if we assume that either $b\ne 0$ and $2b -\frac {a^2}b \ne 0$.
If $b = 0$ (and $(a,b)\ne (0,0)$ so $a\ne 0$) then we need $ay +bx =ay=0$ and $y$ must be $0$ Then we need $ax=1 $ so $x =\frac 12$.
So $(a,0)^{-1} = (\frac 1a, 0)$.
If $b \ne 0$ but $2b - \frac {a^b}b =0$ then $2b^2 = a^2$ and $\frac ab=\sqrt 2$. Well, that never happens if $a,b \in \mathbb Q$ and we dont have to worry about it. And so the inverse will always occur if $(a,b)\ne (0,0)$ and $a,b\in \mathbb Q$.
But if $a,b \in \mathbb R$ and $\frac ab = \sqrt 2$.... well then we are stapled six ways to Sunday. Aren't we?
If we try to solve $(\sqrt 2b, b)*(x,y)=(1,0)$ where $b \ne 0$ we get $\sqrt 2bx+2by, \sqrt 2by + bx)=(1,0)$ so $(\sqrt 2x + 2y)=\frac 1b$ and $\sqrt 2y+x = 0$. So $x=-\sqrt 2y$ and so $\sqrt 2x + 2y = \sqrt 2(-\sqrt 2)y + 2y=-2y+2y=0\ne \frac 1b$ so this is impossible.
So $(\sqrt 2b, b); b\ne 0$ will not have an inverse.