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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:


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
up vote 1 down vote accepted

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.

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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 $\langle x,y\rangle\in f.$

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