My question is as following : Let $A$ and $B$ be some languages, there exist a language $C$ such that $A\le C$ and $B\le C$, where "$\le$" means "reducible to", so $A\le C$ means there is a mapping $f$ from the strings over $A$s alphabet to strings over $C$s alphabet such that $x\in A$ iff $f(x)\in C$.
I need to either prove it or find an example to contradict the statement
I'm not really sure on how to solve such thing. May anyone help?
Edit :
Well, After thinking for some while I managed to conclude we need to define $C$ such that it will include both $A$ and $B$,
So I defined $C$ as follows :
$$C = \{0w\mid w \in A\} \cup \{1w \mid w \in B\}$$
Now I defined the reduction function : $$ f(x) =\begin{cases} 0x & \text{if $x \in A$}\\ 1x & \text{if $x \in B$} \end{cases} $$
the problem now is that I want to show either $x \in A$ iff $f(x) \in C$ or $x \in B$ iff $f(x) \in C$, but the fact $x$ isn't in $A$ does not conclude $f(x)$ not in $C$ so I tried $f(x) \in C$ iff $x \in A$ or $x \in B$. but I'm not exactly sure it's alright, may anyone help?
Your $C=0A\cup 1B$ will work, but to have $A\le C$ and $B\le C$ you'll need to have two mapping reductions: $f_1:\Sigma_A\rightarrow\Sigma_C$ and $f_2:\Sigma_B\rightarrow\Sigma_C$ such that $x\in A\Longleftrightarrow f_1(x)\in C$ and $x\in B\Longleftrightarrow f_2(x)\in C$.
Define $$ f_1(x)=0x\qquad f_2(x)=1x $$ Now it's easy to see that $f_1$ gives a mapping reduction $A\le C$: if $x\in A$ then $f(x)=0x\in C$ and if $x\notin A$ then $0x\notin C$ (since any string in $C$ starting with $0$ must be followed by a string in $A$). The same reasoning gives a reduction $B\le C$ using $f_2$.
By the way, welcome to the site!