Σ = {a, b}. We write #a(x) for the number of occurrences of the letter a in the word x and similarly for #b.
note: $\Sigma^*$ refers to the set of all finite strings (words) over Σ.
∀x∈$\Sigma^*$,∃y,z∈$\Sigma^*$ such that x=yz∧ [#a(y)=#b(z)].
In words this statement is saying that for all elements x in the language there a exists a y and a z that is also in the language such that x = yz AND the number of as in y is the same as the number of bs in z.
What I'm having a lot of trouble is the x = yz, does this imply two words concatenated?
eg. x = aabb, y = aa, z = bb and yz = aabb?
A push in the right direction for what type of proof would also be appreciated. I think the statement is false. Because $\epsilon$, the empty word, is part of the language. Using the example of a which can be also be expressed as the concatenated a$\epsilon$ there are no bs at all in the word! So the requirement of the number of as being the same the as the number of bs is impossible to meet.
Note: I changed my mind and have decided it is most likely that this statement was false. I did this when I remembered the existence of the empty word as well as the fact that we are proving this for ALL finite strings in the alphabet with no requirement that both letters be present in words for them to be valid.
First a couple of small points: yes, $x=yz$ means concatenation. Second, yes it is obvious that any string can be split into two strings, but the statement you are asked to prove says more than that: it says that the number of $a$s in the first part is the number of $b$s in the second part. I personally don't feel that this is obvious, because you won't get it by splitting $x$ just any old how. For example, if $$x=abababa$$ then $$x=(ab)(ababa)$$ doesn't work and you need to do something like $$x=(aba)(baba)\ .$$
Hint for the proof. There are probably various ways to do it but here is an idea. If the word $x$ has $m$ letters, say $x=x_1\cdots x_m$, define two sequences $$\eqalign{ u_n&=\langle\hbox{number of $a$s in $x_1\cdots x_n$}\rangle\cr v_n&=\langle\hbox{number of $b$s in $x_{n+1}\cdots x_m$}\rangle\cr}$$ for $n=0,\ldots,m$. Can you say anything about the sequences $u_n,v_n$? Can you explain why there is a value of $k$ such that $u_k=v_k$?