# Is a function still a function if it doesn't have any rule?

From what I've read on the internet, I've concluded that function differs from relation in that function can only have one range per domain. So, if for example:

F={(1,3),(2,4),(3,6),(4,12)}


I don't think there's any rule in it ( f(x)=blabla ) cause I just randomly typed it out. As you can see, the F set has a perfect one-to-one correspondence. But since it doesn't have any rule, is it still considered a function? Thanks!

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Yes it is still a function! – Emin Mar 3 '14 at 19:21
@Emin Why? Can you cite any source? – Mouse Hello Mar 3 '14 at 19:22
What would you mean by a rule? – ploosu2 Mar 3 '14 at 19:23
@ploosu2 for example: f(x)=3+2x. I don't know whether or not it's called a rule though lol – Mouse Hello Mar 3 '14 at 19:25
Most functions cannot be explicitly written as $f(x)=\text{some formula}$. It's normal, they are still functions. – Dan Shved Mar 3 '14 at 19:26

The only property that needs to have a relation to be a function is that (as you've said) function can only have one range per domain. Roughly speaking it 'means' that in a part of time (domain) you can be just in one place (range). It doesn't need to have any formal rule for a given relation (as $y=f(x)$) to be a function, it needs just to fulfill the condition of the definition. So yes, the given relation is a function.
Indeed, your function does have a rule--infinitely-many possible rules, in fact. For example, if $g(x)$ is literally any function $\Bbb R\to\Bbb R,$ we could say that $F$ is the function from $\{1,2,3,4\}$ to $\Bbb R$ given by $$F(x)=\frac{3(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}+\frac{4(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}+\frac{6(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+\frac{12(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}+(x-1)(x-2)(x-3)(x-4)g(x).$$ Letting $g(x)=0$ makes $F$ the unique polynomial function of degree $3$ passing through the $4$ points you gave, restricted to the $x$-values $1,2,3,4.$ See here for more.
A function need not have a rule, though. More generally, given sets $A$ and $B,$ we say that $f$ is a function from $A$ to $B$ if $f$ is a relation whose domain is $A,$ having $B$ as a codomain, and such that for every $x\in A$ there is a unique $y\in B$ such that $\langle x,y\rangle\in f.$