Recently I come across this definition of function. "A function F is a set of ordered pairs $( x , y)$, no two of which have the same first member. That is, if $(x, y) \in F$ and $(x, z) \in F$, then $y = z$."
It made me confused because as far as i understood that it suggests we can't have one to many function like $\sqrt{x}$. But why this is so?
On
The notion of function has been so successful in mathematics that it should not be tampered with.
Now it could very well be that in the realm of a certain problem you want to bring in some sort of "many-valued function". You can do this any time by making your "function" set-valued, or even multiset-valued. Then it is again a function in the official sense; but you would have to give explanations what "continuity" and the like would mean in your context.
An example: A polynomial $$p(z):=z^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0$$ with complex coefficients $a_k$ has $n$ complex roots, counted with multiplicity. These roots depend on the coefficient vector ${\bf a}=(a_0,a_1,\ldots,a_{n-1})$, but there is no function $$f:\quad{\mathbb C}^n\to{\mathbb C},\qquad {\bf a}\to f({\bf a})$$ producing one or several of these roots upon inputting the coefficient vector ${\bf a}$.
On the other hand you can define the set $M:={\mathbb N}^{\mathbb C}$ of multisets of complex numbers. Each such multiset $S$ is a function $S:\ {\mathbb C}\to{\mathbb N}$ which assigns to each complex number $z$ a certain integer multiplicity. Maybe $M$ is to large for our purpose; therefore we restrict to the set $M_n\subset M$ of multisets of cardinality $n$. Then the fundamental theorem of algebra tells us that there is a bona fide function $$f:\quad {\mathbb C}^n\to M_n,\qquad {\bf a}\to f({\bf a})$$ producing a well-defined multiset of complex numbers upon inputting a coefficient vector ${\bf a}$. After defining a suitable topology on $M_n$ one can even say that the multiset $f({\bf a})$ depends continuously on ${\bf a}$.
A function $f\colon X \to Y$ can be thought of as a subset $F \subseteq X \times Y$ such that
For every $x \in X$ there is a $y \in Y$ such that $(x, y) \in F$.
If $(x, y), (x, y') \in F$ then $y = y'$.
If you have a function $f\colon X \to Y$ then it corresponds to the ordered pairs $F = \{(x, f(x)) \mid x \in X\}$. Conversely if you have a set of pairs $F \subseteq X \times Y$ with the above properties then the associated function $f\colon X \to Y$ is defined by letting $f(x)$ be the unique element of $Y$ such that $(x, f(x)) \in F$.
It is indeed true that this means functions cannot be one to many. In set theory this is part of the definition of a function. A function $f\colon X \to Y$ is an assignment, to every $x \in X$, of a unique element $f(x) \in Y$.