For any numbers $a, b,$ and $c,$ $a + b = a + c$ if and only if $b = c$

Question

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, it est, it's a function. — Michael Rozenberg, Sep 22 '20 at 15:57
@MichaelRozenberg so the $\Longleftarrow$ part it is a consequence of the fact that $+$ is an operation? — Air Mike, Sep 22 '20 at 16:01
$+$ 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. — Michael Rozenberg, Sep 22 '20 at 16:03

Michael Rozenberg · Accepted Answer · 2020-09-22T18:09:31.093

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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 $\mathbb R$.

edited Sep 22 '20 at 18:09

answered Sep 22 '20 at 16:16

Michael Rozenberg

194,933

I know it’s tacky to correct somebody’s Latin, but i.e. stands for “id est”. – Lubin Sep 22 '20 at 18:08
@ Lubin It was typo. I fixed. Thank you! – Michael Rozenberg Sep 22 '20 at 18:10
Please strive not to post further dupe answers to dupes of FAQs.For site organization is is better to instead link to prior answers (which hopefully have been polished over time due to feedback). Further it helps to localize information on a topic. – Bill Dubuque Sep 22 '20 at 19:03

score 0 · Answer 2 · answered Sep 22 '20 at 16:57

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Observe that $b = (b+a)+ (-a)$ and $c = (c + a) + (-a)\\$. So $$(b+a)+ (-a) = (c + a) + (-a) $$ Now using $\Rightarrow$, you can add $a$ both sides and observe that $a+b = a+c$ .

answered Sep 22 '20 at 16:57

Infinity_hunter

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For any numbers $a, b,$ and $c,$ $a + b = a + c$ if and only if $b = c$

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