Is there a law that you can add or multiply to both sides of an equation?

Question

It seems that given a statement $a = b$, that $a + c = b + c$ is assumed also to be true.

Why isn't this an axiom of arithmetic, like the commutative law or associative law?

Or is it a consequence of some other axiom of arithmetic?

Thanks!

Edit: I understand the intuitive meaning of equality. Answers that stated that $a = b$ means they are the same number or object make sense but what I'm asking is if there is an explicit law of replacement that allows us to make this intuitive truth a valid mathematical deduction. For example is there an axiom of Peano's Axioms or some other axiomatic system that allows for adding or multiplying both sides of an equation by the same number?

In all the texts I've come across I've never seen an axiom that states if $a = b$ then $a + c = b + c$. I have however seen if $a < b$ then $a + c < b + c$. In my view $<$ and $=$ are similar so the absence of a definition for equality is strange.

Answer 1

This is a basic property of equality. An equation like $$a=b$$ means that $a$ and $b$ are different names for the same number. If you do something to $a$, and you do the same thing to $b$, you must get the same result because $a$ and $b$ were the same to begin with.

For example, how do we know that Samuel Clemens and Mark Twain were equal in height? Simple: Because they were the same person.

How do we know that $a+c$ and $b+c$ are equal numbers? Because $a$ and $b$ are the same number.

Answer 2

This is an axiom of predicate logic. For example, check out this list of the axioms in predicate calulus, intended to be an ambient logic for ZFC set theory. Note axioms 13 and 14:

$$\vdash x=y\to (x\in z\to y\in z)$$ $$\vdash x=y\to (z\in x\to z\in y)$$

In set theory, the only basic atomic formulas are of the form $x=y$ or $x\in y$, so this, together with transitivity of equality (axiom 8), which will allow you to prove

$$\vdash x=y\to (x=z\to y=z)$$ $$\vdash x=y\to (z=x\to z=y),$$

is sufficient to prove by induction that for any predicate of the language $\varphi(x)$, it is a theorem that

$$\vdash x=y\to(\varphi(x)\leftrightarrow \varphi(y)).$$

And once you define class terms via $x\in\{x\mid\varphi\}\leftrightarrow\varphi$, you can prove

$$\vdash x=y\to A(x)=A(y)$$

for any class term $A(x)$ by converting it to the statement $$\vdash x=y\to (z\in A(x)\leftrightarrow z\in A(y))$$

using extensionality and applying the theorem above for predicates. This demonstrates how the rule $x=y\to f(x)=f(y)$ gets translated into a rigorous proof in a formal system.

Answer 3

If you are given that $$a = b$$ then you can always infer that $$f(a) = f(b)$$ for a function $f$. That's what it means to be a function. However, if you are given $$f(a) = f(b)$$ then you can't infer $$a = b$$ unless the function is injective (invertible) over a domain containing $a$ and $b$.

For your problem, $f(x) = x + c$.

Answer 4

This is a consequence of the substitution property. We know that $a + c = a + c$, right? Certainly $a + c$ is the same thing as $a + c$. But since $b$ is the same thing as $a$, it follows that $a + c = b + c$.

Answer 5

Equal means same. If two things are equal then they're the same, so whatever you do to one of them you can do to the other one and expect to get the same result, because they're the same -- there is no difference between them to allow you to get a different result.

Answer 6

You're perhaps missing the point of people saying 'That's what equality means.' This is the natural result of equality. The basic property of equality is that one equal thing can be substituted in for another. So if A = B, then you can substitute A in for B, or B in for A, anywhere it is found. If you want the longer proof:

A = B  (initial proposition)
A = A  (equality substitution)
A + C = A + C  (all things are equal to themselves)  
A + C = B + C  (equality substitution)

Answer 7

As others have pointed out, if a=b, then (a+c)=(b+c) follows from the replacement property of equality as follows:

everything is equal to itself       1 (a+c)=(a+c).
hypothesis                          2  a=b.
replacing the second a with b in 1  3 (a+c)=(b+c).

You can't infer that "if a < b then a + c < b + c" in this way, since a < a is false. Thus, < and = are not similar in this way.

Answer 8

I just want to throw in my two cents and go a bit deeper. First, the idea $a=b\rightarrow a+c=b+c$ is not an axiom of "arithmetic" per se. It's in fact a theorem derived from the following axiom:

Axiom of substitution. If $\phi(x)$ is a statement and: i) $\phi(a)$ holds; ii) $a=b$; then $\phi(b)$ holds. In logical symbols (which you probably can ignore): $$\forall a\forall b(\phi(a)\land a=b\rightarrow \phi(b))$$

What this really is is something of the meaning of equality. $a=b$ means that it is essentially redundant to write different letters - it's just notation. Thus, if $\phi(x)$ is the statement "$x$ is red", and we find a red object $a$, it's only natural that we can deduce that any object equal to $x$ is also red.

Next, (technically we need a lot of set theory to develop this purely foundationally, but we'll accept the idea of a function for now), we then have that if $a=b$, then $f(a)=f(b)$.

Proof. Let $\phi(x)$ be the statement $f(x)=f(a)$. Then $\phi(a)$ holds. Since $a=b$, $\phi(b)$ holds by the axiom and therefore $f(b)=f(a)\,$.$\square$

Now, as other answers have stated, we can simply let $f(x)=x+c$ and note that the above theorem implies what you are talking about. You can do similarly for multiplication by $c$.

One can think about the converse - in fact, it is not necessarily true that if $f(a)=f(b)$, then $a=b$. This is in fact a non-trivial condition known as one-to-one-ness or injectivity. In fact, the function $f(x)=x+c$ is injective since you can now subtract $c$ from each side, but for instance the function $g(x)=0x=0$ is not injective (why?)

Answer 9

As all the others have said, from $a=b$ you can conclude that $f(a)=f(b)$ for any function defined on a domain containing $\{a,b\}$.

There is a subtle point, however, that has not been addressed so far: An equation $$\Phi(x)=\Psi(x)\tag{1}$$ implicitly defines a solution set $S$ in the agreed domain of discourse. "Solving" such an equation means producing an "explicit" description of $S$, e.g., in the form of a list or of a parametric representation of $S$.

Applying some function $f$ to both sides of $(1)$ (say, squaring both sides) in order to obtain a simpler equation may very well alter (meaning: enlarge) the solution set $S$.