# How to determine whether the function is one-to-one? Differential Calculus

Determine whether the function f is one-to-one

f(t) is the number of people in line at a movie theater at time t.

• Don't you mean $f(t)$? Do you understand what a one-to-one function is? – Git Gud Sep 25 '13 at 16:39
• Not necessarily one to one. We could have $f(t_0)=17$, and at some later time $t_1$, $f(t_1)=17$. This could happen in a couple of ways: (i) The line has not moved or (ii) It has moved, $3$ people have entered the theatre, but $3$ have joined the end of the line. – André Nicolas Sep 25 '13 at 16:41
• One to one means that you can (in theory) uniquely figure out the input if you know the input. – copper.hat Sep 25 '13 at 17:01

Informally, we can think of a one-to-one function as one that maps distinct elements in the domain to distinct elements in the codomain. Or, in other words, if $f$ maps $a$ and $b$ map to the same thing, then $a=b$.
Formally, a function $f:A \rightarrow B$ is called one-to-one if $f(a)=f(b)$ implies $a=b$. Equivalently, if $a \neq b$, then $f(a) \neq f(b)$.
In this question, we have a function $f:T \rightarrow \mathbb{Z}^{\geq 0}$ defined by $f(t)$ is the number of people in line at a movie theater at time $t$, and $T$ is the set of times for which "time" is defined. The task is to find two distinct times $t_1 \in \mathbb{R}$ and $t_2 \in \mathbb{R}$ for which $f(t_1)=f(t_2)$.
• Suppose there are $N<\infty$ people in existence. We pick $N+1$ distinct points of time $t_1,t_2,\ldots,t_{N+1}$ (assuming the theater has been open longer than an instant, since time is continuous, such points of time exist). Then the pigeonhole principle implies there are two points in time $t_i$ and $t_j$ in which $f(t_i)=f(t_j)$; in other words, we have $N+1$ numbers $$f(t_1),f(t_2),\ldots,f(t_{N+1})$$ that all belong to $\{1,2,\ldots,N\},$ so they can't all be distinct.
• $0$ is also a possible value of $f$... – user103402 Nov 20 '13 at 3:06