A function \(f\) from a set \(A\) to a set \(B\) is a relation \(AfB\) such that for every \(a \in A,\) there is exactly one pair \((a,b) \in f.\)
Instead of writing \(AfB,\) we write \(f:A \rightarrow B.\) Instead of writing \((a,b) \in f,\) we write \(f(a) = b.\)
Let \(A = \{1,2,3\}\) and \(B = \{w,x,y,z\}.\) The relation \(f = \{(1,w), (2,y), (3,w)\}\) is a function from \(A\) to \(B.\) Using the notation above, we would define the function as follows: \[f:A \rightarrow B\] \(f(1) = w, f(2) = y, f(3) = w.\)
The relation \(f = \{(1,y),(2,z),(3,w),(1,x)\}\) is not a function. By definition, a function from \(A\) to \(B\) must have exactly one pair for each element of \(A\) but \(f\) has two pairs for the number \(1.\)
The relation \(f = \{(1,z),(2,x)\}\) is not a function. By definition, a function from \(A\) to \(B\) must have exactly one pair for each element of \(A\) but \(f\) has no pairs for the number \(3.\)
The relation \(f\) from \(\mathbb{R}\) to itself defined by \(\{(x,x^2) : x \in \mathbb{R}\}\) is a function. In particular, \[f: \mathbb{R} \rightarrow \mathbb{R}\] is defined by \(f(x) = x^2.\) You can compute \(f\) when given a particular number: \[f(2) = 4, f(-1) = 1, f(5) = 25\] Every input has exactly one output.
On the other hand, \(f(x) = \sqrt{x}\) is not a function on \(\mathbb{R}\) because there is no number \(y\) in \(\mathbb{R}\) such that \(f(-1) = y.\) By definition, a function on \(\mathbb{R}\) must have exactly one pair for element of \(\mathbb{R}.\)