Show that if $x \in \mathbb{Q}$, then there is exactly one $y \in \mathbb{Q}$ such that $x+y=0$

Question

Show that if $x \in \mathbb{Q}$, then there is exactly one $y \in \mathbb{Q}$ such that $x+y=0$. (Here 0 means $\langle 0,1 \rangle$)

Answer 1

Let $x \in \mathbb{Q}$. By definition, there exist $a, b \in \mathbb{Z}$ where $b \neq 0$ such that $x = \frac{a}{b}$. Clearly, $y = \frac{-a}{b} \in \mathbb{Q}$ since this is similarly a ratio of integers (and the integers are closed under multiplication). We have $$x + y = \frac{a}{b} + \frac{-a}{b} = \frac{a + (-a)}{b} = \frac{0}{b} = 0.$$ This establishes the existence of such a $y$, so we may write $y = -x$.

To show that there is exactly one $y$, we must show uniqueness. Suppose $y_1$ and $y_2$ both have the property that $x + y_1 = 0$ and $x + y_2 = 0$. Then: $$y_1 = y_1 + 0 = y_1 + (x + y_2) = (y_1 + x) + y_2 = 0 + y_2 = y_2.$$ Hence, this $y$ we found is unique.

Answer 2

You shouldn't assume that everyone is taking the same class as you are and defining the mathematical objects in the same way you are.

I'm assuming you have defined the rationals as the set of all ordered pairs of integers $(a,b); a,b \in \mathbb Z; b\ne 0$ but with the equivalence that $(a,b) = (c,d)$ if and only if $ab=bd$. And with addition and multiplication defined as $(a,b)\times (c,d) = (ac,bd)$ and $(a,b)+(c,d) = (ad+bc,bd)$.

In which case, for any $q=(a,b)$ then $(a,b) + (-a,b) = (ab+b(-a),b^2) = (ab-ab,b^2)= (0,b^2)$ and we have $(0,b^2) = (0,1)$ because $0*1 = b^2*0 = 0$.

And we know that if $(a,b)+(c,d) = (ad + bc,bd)=(0,1)$ then $ad+bc=(ad+bc)*1 = (bd)*0 = 0$. So $bc=(-a)b$ and therefore $(-a,b)=(c,d)$ and $(-a,b)$ is the only such rational number.