I was reading about the field of real numbers $\mathbb{R},$ and a basic question arose in my mind.
How one should prove that, for any numbers $a, b,$ and $c,$ $a + b = a + c$ if and only if $b = c?$
In order to prove this statement, we have to prove two things. I will start with the part usually known as the cancellation rule.
$\implies.$ Let $a, b,$ and $c$ be any numbers. Suppose that $a + b = a + c.$ Since $a$ is any number, we know from the filed axioms, that exists a number $-a$ such that $a + (-a) = (-a) + a= 0.$ Hence, adding $(-a)$ to both sides, we have $(-a) + (a + b) = (-a) + (a + c).$ By the Associative Law, we deduce that $((-a) + a) + b = ((-a) + a) + c.$ Since $(-a) + a = 0,$ we get $0 + b = 0 + c.$ Since $0$ is the additive identity, it follows that $b = c.$ $\square$
This part is really straightforward. Although, I have no idea what to do about the $\Longleftarrow$ part.
I think of this last part as the statement that $+$ is well-defined in $\mathbb{R},$ which means that for all $x, y, z, w \in \mathbb{R},$ if $x = y$ and $z = w,$ then $x + z = y + w,$ but I’m not sure if these two things are related to each other.
Another thing that “bores” me is the fact that I use this fact to prove the $\implies$ part when I added $(-a)$ to both sides and stated that they still equal.
In short:
How can I prove that if $b = c,$ then $a + b = a + c?$ Is this related to the fact that $+$ is well-defined? Do we assume that $+$ is well-defined (i.e., is it an axiom), or can it be proved?
Yes. Because $+$ it's an operation, id est, it's a function:
$$+:\mathbb R^2\rightarrow\mathbb R.$$ If $+((a,b))\neq+((a,c))$, so we obtain a contradiction with the definition of a function.
$a+b$ and $a+c$ must be the same element of $\mathbb R$.