A friend of mine just asked me how to prove that if $a=b$ then $a+c=b+c$, where $a,b$ and $c$ are real numbers, I'm not sure what I should answer. I have a book called introduction to logic and to the theory of the deductive sciences by Alfred Tarski, which is about propositional logic, and I remember reading that two things $a$ and $b$ are equal if any proposition that is true about $a$ is also true about $b$ and vice-versa. However I think this isn't very formal.

I haven't taken any set theory course, I think that another way to justify it is to say that sum is a function and since the ordered pairs $(a,c)$ and $(b,c)$ are equal then $+(a,c)=+(b,c)$. But I'm not too convinced.

If we use the standard axioms how would we justify $a+c=b+c$ using the mainstream axioms of today. I think there is something Zermelo-Frankl with choice. Would these be enough, what properties of the real numbers do we need? Can we prove it using the usual construction of the real numbers and Zermelo-Frankl?

As you can probably see I am not very knowledgeable about these topics, so I would like a delicate explanation.

Many thanks and regards.


marked as duplicate by MJD, user26486, GEdgar, Community Apr 25 '15 at 1:47

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    $\begingroup$ This follows by the "well-defined" nature of $+$. $\endgroup$ – Thomas Andrews Apr 25 '15 at 0:39
  • $\begingroup$ If $a=b$, then $f(a)=f(b)$, where $f$ is any function with $a$ in its domain. $\endgroup$ – user26486 Apr 25 '15 at 0:56
  • $\begingroup$ I realize now that you did not specify what $a$, $b$, $c$ and $=$ are up until into the last third of the question, which points to reals. So we all projected our own ideas onto this, ranging from $\mathbb{N}$ to $\mathbb{R}$ to programming languages. $\endgroup$ – mvw Apr 25 '15 at 1:12
  • $\begingroup$ they are real numbers $\endgroup$ – Jorge Fernández Hidalgo Apr 25 '15 at 1:12

To elaborate on Thomas Andrews' comment:

Addition is well-defined: For all $a,b,c,d\in\mathbb{R}$, if $a=b$ and $c=d$, then $a+c=b+d$. The well-defined nature of addition in $\mathbb{R}$ is taken as an axiom. Hence, if $a=b$ and $c=d$, we have $$ a+c=b+d\Longleftrightarrow a+c=b+c, $$ but I think there is something more interesting you could show, namely cancellation of addition.

Cancellation of addition: For all $a,b,c\in\mathbb{R}$, if $a+c=b+c$, then $a=b$. [Of course, this is the converse of what is taken as an axiom.]

Proof. Pick $a,b,c\in\mathbb{R}$ and suppose $a+c=b+c$. By the existence of additive inverses (another axiom), there exists $-c\in\mathbb{R}$ such that $c+(-c)=0$. Since addition is well-defined, we have that $$ [a+c]+(-c)=[b+c]+(-c), $$ which by the associative property of addition (another axiom) we may write as $$ a+[c+(-c)]=b+[c+(-c)]. $$ This yields $a+0=b+0$, which becomes $a=b$. $\blacksquare$

I'm not sure if that description/explanation is really what you were looking for, but hopefully you found it helpful.


You start with $$ a \oplus c = x \quad (*) $$ where $x$ is the resulting value, then note that $a = b$, so you replace $a$ with $b$ in equation $(*)$. This gives $$ b \oplus c = x $$ Of course this means $$ a \oplus c = x = b \oplus c $$ and thus $$ a = b \Rightarrow a \oplus c = b \oplus c $$

Let us redo this for variables $a, b, c \in \mathbb{R}$ and the standard addition and equality there.

$\mathbb{R}$ is a field which includes that it is closed under addtion. So because $a, c \in \mathbb{R}$, we have $$ a + c = x \in \mathbb{R} $$ If we say $a = b$ this means we are talking the same unique element from $\mathbb{R}$ here, giving it two different names $a$ and $b$. Again changing the name $a$ to $b$ in the expression "a + c" does not change anything, as we add the same real elements like before, so "b + c" must give the same value $x$: $$ x = a + c = b + c $$ So again $a = b \Rightarrow a + c = b + c$.

  • $\begingroup$ Why can we replace $b$ with $a$ without altering the result? $\endgroup$ – Jorge Fernández Hidalgo Apr 25 '15 at 0:39
  • $\begingroup$ Because both $a$ and $b$ have the same value, which is what $a = b$ means. $\endgroup$ – mvw Apr 25 '15 at 0:40
  • $\begingroup$ so two things are equal if the have the same value? $\endgroup$ – Jorge Fernández Hidalgo Apr 25 '15 at 0:41
  • 1
    $\begingroup$ For the sake of the equality relationship, yes. We could now argue if $a$ and $b$ are the same object, thus just two different names for the same thing, or if they are two different things, just having the same value. $\endgroup$ – mvw Apr 25 '15 at 0:43
  • 1
    $\begingroup$ Could be helpful, possibly, to mention the properties of equality on $\mathbb{R}$ (i.e., that $=$ is reflexive, symmetric, and transitive) and that you use closure of addition on $\mathbb{R}$ to conclude that $a+c=x\in\mathbb{R}$. Honestly, I find it somewhat difficult to get a good grasp on what exactly OP is after in terms of a proof/explanation. $\endgroup$ – Daniel W. Farlow Apr 25 '15 at 1:02

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