Proof that if $\forall a f(a) = g(a)$ then $f=g$ How do we prove formally that if:
$\forall a f(a) = g(a)$ 
$=>$
$f=g$
when
$f,g \in \mathbb F[x]$
 A: Let $f(x)$ and $g(x)$ be polynomials with coefficients in the field $\mathbb{F}$.
An element of $\mathbb{F}[x]$ is a formal expression. Such a formal expression $f(x)$ defines a function from $\mathbb{F}$ to $\mathbb{F}$. This kind of function is called a polynomial function. It is common to denote the polynomial $f(x)$ and the associated polynomial function by the same name. 
For this problem, that's not a good idea. So for any polynomial $P(x)$, denote its associated polynomial function by $P^\ast(x)$. In that notation, we want to prove that if $f^\ast(a)=g^\ast(a)$ for all $a$ in $\mathbb{F}$, then $f(x)=g(x)$.
We will need to assume that the field $\mathbb{F}$ is infinite. For if the field is finite, the result is no longer true.
Suppose $f^\ast(a)=g^\ast(a)$ for all $a$ in the field, but that $f(x)$ is not the same polynomial as $g(x)$. Then the polynomial $h(x)=f(x)-g(x)$ is not the zero polynomial.  
Since $f^\ast(a)=g^\ast(a)$ for all $a$ in $\mathbb{F}$, it follows that every $a$ in $\mathbb{F}$ satisfies the equation $h^\ast(a)=0$. If $\mathbb{F}$ is infinite, this is impossible. For a polynomial of degree $n\ge 1$ has at most $n$ roots.
A: The answer depends somewhat on what it means for $f=g$ to hold.  One approach is to consider functions as sets of ordered pairs, if $f(x)=x^2$ then $f=\{(1,1),(2,4),(3,9),(-3,9),\ldots\}$.  But not just any set of ordered pairs, we need each element of the domain to appear as the first element of an ordered pair exactly once.
If this is the definition of functions, then we prove this theorem via set inclusion.  For each $a$ in the domain, there is exactly one $f(a)$ such that $(a,f(a))\in f$.   But applying the hypothesis, $(a,(g(a))=(a,f(a))\in g$, so $f\subseteq g$.  Repeating in the other direction $g\subseteq f$ and hence $f=g$.
A: By definition two maps: $f\colon A\to B$ and $g\colon C\to D$ are equal if $A=C$ and $B=D$ and $\forall x\in A=C$ we have $f(x)=g(x)$.
A: As others have pointed out, this will only work in an infinite field. As a counterexample (for a finite field) you can take $\prod\limits_{a\in \mathbb F} (x-a)$, which is $0$ on $\mathbb F$, but its coefficients are different from the zero polynomial. 
For an infinite field, you can show that $f = 0$ iff $f(x) = 0 \forall x \in \mathbb F$ (a polynominal has at most $\operatorname{deg} (f) $ zeros), thus $f-g = 0$ holds iff $f(x) = g(x) \forall x \in \mathbb F$.
A: Using purely set theoretical definitions, recall that a function is a set of ordered pairs, $F$, with the following property: $$\forall x\forall y\forall z(\langle x,y\rangle\in F\land\langle x,z\rangle\in F\rightarrow y=z),$$
that is if $x$ is in the left coordinate of an ordered pair in $F$, then there is only one ordered pair that $x$ is its left coordinate.
This allows us to identify $y$ such that $\langle x,y\rangle\in F$ with $F(x)$.
Now to continue with the argument, recall that $\langle x,f(x)\rangle\in f$ and $\langle x,g(x)\rangle\in g$. By our assumption that $f(x)=g(x)$ we have that $f\ni\langle x,f(x)\rangle=\langle x,g(x)\rangle\in g$, therefore $f\subseteq g$ and $g\subseteq f$, so equality ensues.
A: While it is trivial that $f,g$ are equal as functions iff they agree on all arguments, it is less obvious that it implies equality as polynomials.
If coefficients are assumed to belong to an infinite field $\Bbb{F}$, then it is true that $f=g$ as polynomials iff $f(a)=g(a)$ for all $a \in \Bbb{F}$.
To prove this it suffices to consider polynomial $f-g$ has an infinite number of roots.  Therefore $f-g$ must be the zero polynomial, since otherwise the number of roots is bounded above by the degree of $f-g$.
In finite fields it is possible to give a nonzero polynomial whose roots include all elements of the field, so the above proof fails.
