Rational Numbers Proof

Question

Apologies for the vagueness before, I'm new here. I hope this clears it up:

Show that, for all non zero $b\in \Bbb Z$, $${(0,b)}=((a',b')\in F:a'=0)$$ $$F=((a,b)\in \Bbb Z*\Bbb Z: b\ne 0))$$ where F defines $\Bbb Q$

The problem states that it is a special class and to show that $(0,b)$ is the identity for sum. Because of that, define $0\in \Bbb Q$ as $(0,b)$

I'm confused on how to solve it. I know that $(a,b) \sim (a',b') <=> ab'-ba'=0$ and that addition in $\Bbb Q$ is defined as $(a,b)+(a',b')=(ab'+a'b,bb')$

Answer 1

Without some context, it's hard to tell what theorems you're allowed to use here. But maybe you can multiply both sides of your equation by $b'$ to get $(0,b) = (a,1)$? Now the right side is an integer.

Answer 2

I assume you should have been taught that $(a,b) = (c,d)$ if and only if $a\times d = b\times c$. Further, you should have been taught that the second entry may never be zero.

As such, you will have $(0,b)=(a',b')$ if and only if $0\times b' = b\times a'$. In other words, you have $0=b\times a'$, but remembering that $b\neq 0$ that implies that $a'$ must be zero since $a$ and $b$ are elements of the integers which are well known to be an integral domain.

Similarly, $(b,b)=(a',b')$ if and only if $b\times b' = b\times a'$. By subtracting to one side and factoring, we have $0 = b\times (b'-a')$. By again noting the fact that $b$ is not allowed to be zero and the fact that the integers are an integral domain we get that it must be true that $(b'-a')$ must be zero. So we have $b'-a'=0$ which by adding implies that $b'=a'$