HW1 Problem 4

Return to Homework 1, Glossary, Theorems

### Problem 4

Determine if the following relations define a function. If it is a function, determine whether it is one-to-one and onto. Justify your answers.

(a) $f: \mathbb{R} \rightarrow \mathbb{Z}, f(x) = int(x), \textrm{ where } int(x) \textrm{ is the floor of } x$

(b) $h: \mathbb{R} \rightarrow \mathbb{R}, h(t) = 2t+6$

**Solution**

(a)

- This is a function since $f$ will map any $x \in \mathbb{R}$ to some $y \in \mathbb{Z}$.
- This function is not one-to-one because multiple elements of the domain map to the same element in the codomain: $f(1.1) = f(1.3)$ but $1.1 \neq 1.3$.
- This function is onto. Let $x = y + r$ where $r$ is some real value on the interval $[0,1]$. Then, $f(x) = f(y + r) = int(y + r) = y$ according to the definition of the floor function.

(b)

- This is a function since $h$ will map any $x \in \mathbb{R}$ to some $y \in \mathbb{R}$.
- This function is one-to-one. Suppose $h(x_1) = h(x_2)$. Then, $2x_1 + 6 = 2x_2 + 6 \implies 2x_1 = 2x_2 \implies x_1 = x_2$.
- This function is onto. Let $x = \frac{1}{2} y - 3$. Then, $h(x) = h(\frac{1}{2} y - 3) = 2(\frac{1}{2} y - 3) + 6 = y - 6 + 6 = y$.