How do we prove that if $f(x)=g(x)$ then $f(x)+h(x)=g(x)+h(x)$.

Question

Im trying to prove that you can add an arbitrary expression to both sides of an equality and you retain equality.

In Real Analysis it is axiomatic that if you have an equality of real values, you can add the same real number to both sides, and you retain the equality. But generalizing that to all arbitrary expressions, even exiting the real numbers, is what Im trying to achieve.

Answer 1

First of all, let's observe that it is not relevant if the above property is true for every $x$ in the domain or for just one, because it can be studied "$x$ by $x$".

In a certain sense the property you asked about is always true, if it is well defined.

On one hand it is clearly necessary that in the codomain you have an additive structure (otherwise “+” does not make any sense), not necessarily commutative, because you put both $h(x)$ on the right side.

On the other hand, if you have an additive structure, then let $G$ be the codomain, and $+:G \times G: \to G$ the addition: then $R_{h(x)}:G \to G; R_{h(x)}(g):=h(x)+g$ is a well defined function.

Thus $f(x)+h(x)= R_{h(x)}(f(x))=R_{h(x)}(g(x))=g(x)+h(x).$ (This because $f(x)=g(x)$ and so the have the same image through $R_{h(x)}$)

Notice that it is not even requested to exist an additive inverse. It would be necessary if you asked the inverse property: "$f(x)+h(x)=g(x)+h(x) \implies f(x)=g(x)$?"

Answer 2

The actual meaning of equality of two things is that substitution of one for the other, is nullipotent - i.e. it has no effect.

More formally, for all predicates $P$

$x=y\iff (P(x)\iff P(y))$

So any true statement involving $x$ remains true when $x$ is exchanged for $y$ and vice versa.

Answer 3

From the viewpoint of formal logic, this is called the "substitution" property of equality. We begin with axioms of identity that say that, for any expression $E$ (which is well defined) we have $E = E$.

We then have axioms of substitution. If we have an expression $E = E$ and we already know that $A = B$, then we can replace any number of copies of $A$ within $E$ with $B$, as long as we preserve the proper structure.

So, for example, we would have $$ f(x) + h(x) = f(x) + h(x) $$ as a basic axiom. If we also know that $f(x) = g(x)$ then we can replace - substitute - one of the copies of $f(x)$ in the first equation with $g(x)$ to obtain $$ f(x) + h(x) = g(x) + h(x). $$ Of course, if these are functions, we may need to pay attention to make sure that $f(x)$, $g(x)$ and $h(x)$ are all defined.

Note that, when we work with these axioms, we do not "start with $f(x) = g(x)$ and add $h(x)$ to both sides". Instead we start with a longer equation that we already know is true, and we make a substitution of one piece by another piece that is equal to it.

In this way, the identity and substitution axioms allow us to manipulate equalities within formal logic in the way that we expect. This is exactly why we include these axioms: because we know how equality "should" work, and both of these axioms are motivated by that understanding.