Is there any unique function $c=f(a, b)$ such that if I know only $c$, I can deduce unique $a$ and $b$ from it? I want to know a relation $c = f(a, b)$ (where $a, b, c$ are real numbers) such that if I only know $c$ I can deduce $a, b$ inputs from it.
As an example consider $f(a, b) = a+b$, then $f(2,3) = 5$ but if I know $5$ as an output only,  I cannot get $2, 3$ uniquely because $f(1,4), f(5,0)$ all of them give the same result $5$. 
Does there exist such function which can make me guess input $a, b$ as unique?
also a, b belongs to every real number R. ie. I should be able to pick any a, b from R. 
 A: Sure, for example $f$ might be the function that intertwines the decimal digits of its arguments so that for example
$$ \begin{align}f(\sqrt{2000000},\pi)&=f(1414.213562373095\ldots, 3.141592653589\ldots)\\&=10401043.211431556922367533059859\ldots\end{align}$$
where reconstructing the digits of the arguments is readily possible.
A: What you are asking for is an injective (or even more, bijective) function from $\mathbb{R}^2$ to $\mathbb{R}.$
In general, there are many ones because it can be shown that they have the same cardinality (see Hagen von Eitzen's brilliant answer).
But it can also be shown that there are not continuous ones.
Indeed, consider the unit circle $S^1$ centered in the origin, and the restriction of $f$ to the circle: it is a continuous function on a compact set, so it has a point of maximum, say $M$, such that $f(M)\geq f(x) \ \forall x \in S^1$ and a point of minimum, say $m$. Clearly they're not the same, because otherwise we obtain the whole circle with the same value, against the hypotheses on $f$. 
So, because they're different you can joint them with two different arcs of the unit circle. On these arcs, by the intermediate value property, we can surely find two different points $c$ and $c'$ such that $f(c)-f(c')=\frac{f(M)+f(m)}2.$
This goes against the hypothesis of the injective function $f$.
If you're interested in this kind of theorems, you can read the Borsuk-Ulam theorem: it's not really easy to prove but it's great.
