This is a question I have regard the proof of the cancellation law for addition in Apostol's Calculus. We are told that the sum of two real numbers x and y is x+y and that it is uniquely determined. We are also given a few axioms. The ones that matter are the commutative law, associative law and the existence of negatives.

Cancellation law for addition: If $a + b = a + c$, then $b = c$.

We assume that $a + b = a + c$. By the Existence of Negatives Axiom, we know that there is a number $y$ such that $y + a = 0$.

My question regards this step:

Since sums are uniquely determined we have $y + (a + b) = y + (a + c)$. Why is it allowed to do this step? What is the reasoning behind it?

  • 1
    $\begingroup$ This follows from the fact that the addition map $+\colon A\times A\rightarrow A$ is a well-defined map (where all $a,b,c,y\in A$ in your example). Since $a+b = a+c$, we have $(y,a+b) = (y,a+c)$ in $A\times A$ and hence $y+(a+b) = +(y,a+b) = +(y,a+c) = y+(a+c)$. $\endgroup$
    – Claudius
    Mar 29, 2016 at 11:00

1 Answer 1


Cancellation law for addition:- If $x,y,z\in\mathbb{R}$ and $x+y=x+z$, then $y=z$

Proof: $$x+y= x+z \qquad \qquad (1)$$

Since, $x\in\mathbb{R}$ (existence of additive inverse). Now adding $-x$ to both sides of $(1)$, we get $(-x)+(x+y) = (-x)+(x+z)$

$\Rightarrow (-x+x)+y =(-x+x)+z$ (by associativity of addition)

$\Rightarrow 0+y = 0+z$ ( by property of additive inverse)

$\Rightarrow y = z$ ( by property of additive identity)


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