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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.

Asinomás
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2 Answers2

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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.

Daniel W. Farlow
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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 $a = b \Rightarrow a + c = b + c$.

mvw
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  • Why can we replace $b$ with $a$ without altering the result? – Asinomás Apr 25 '15 at 00:39
  • Because both $a$ and $b$ have the same value, which is what $a = b$ means. – mvw Apr 25 '15 at 00:40
  • so two things are equal if the have the same value? – Asinomás Apr 25 '15 at 00:41
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    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. – mvw Apr 25 '15 at 00:43
  • Note: This made me pull out Stoy's Denotational Semantics. – mvw Apr 25 '15 at 00:59
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    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. – Daniel W. Farlow Apr 25 '15 at 01:02
  • Yea, I answered first and then noticed, that the OP pulled out the axiomatic mace and realized that this is not as simple as I thought first. In fact the argumentation I have in mind does not care about what binary operation we are talking here and what set we are in. It is about the notion of equality by value of an expression. – mvw Apr 25 '15 at 01:09