Exercises

Determine if the function $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ is one-to-one, onto or neither.

This function is not one-to-one. For example $f(1) = f(-1) = 1$.

The function is also not onto since, for example, there is no x such that $f(x) = -1$ but -1 is in the codomain.

Determine if the function $f:[0, \pi] \to [-1, 1]$ defined by $f(x) = \cos(\theta)$ is one-to-one, onto or neither.

This function is one-to-one. It starts at $(0, 1)$ and decreases continuously to $(\pi, -1)$. The y-values cover every point in that range so the function is also onto.

Determine if the function $f:[0, \pi] \to [-1, 1]$ defined by $f(\theta) = \sin(\theta)$ is one-to-one, onto or neither.

This function isn't one-to-one, e.g. $f(0) = f(\pi) = 0$ and it isn't onto since it never takes on any negative values.

Determine if the function $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = \lceil x \rceil $ is one-to-one, onto or neither.

This function isn't one-to-one. For example, $.5\in\mathbb{R}$ but there isn't a real number such that $\lceil x \rceil = .5$.

This function also isn't one-to-one. For example, $f([0.1, 0.9]) = {1}$.

Determine if the function $f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is one-to-one, onto or neither.

This function is one-to-one. Suppose $x_1, x_2\in\mathbb{R}$ then if

taking the cube root of both sides gives us $x_1 = x_2$. Another way of looking at this would be to observe that the cube root of a number is unique.

This is also an onto function because the square root is defined for every value of x, i.e. if we pick any real number y and take $x=y^{1/3}$ then

Determine if the function $g:\mathbb{Z} \to [0,1]$ defined by $ g(x) = \begin{cases} 1, & x \text{ is even} \\ 0, & x \text{ is odd} \end{cases}$ is one-to-one, onto or neither.

This function isn't one-to-one, e.g. $f(2) = f(4) = 1$. It is, however, onto since its co-domain is limited to 0 and 1. Any odd number maps to 0 and any even number mapes to 1.

Determine if the function $\pi:\mathbb{N} \to \mathbb{N}$ where $\pi(n)$ is the number of prime numbers less than or equal to $n$ us one-to-one, onto or neither.

This function isn't one-to-one. For example, $\pi(6) = \pi(3 \cdot 2) = 2$ and $\pi(15) = \pi(3 \cdot 5) = 2$

This function is onto. For any natural number, $n$, let $x$ be the product of the first $n$ prime numbers. Then we would have $\pi(x) = n$. You can do this for an arbitrarily large number because, as we saw earlier in our discussion of number theory, there are infinitely many primes to chose from.

$f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$

$f:[0, \pi] \to [-1, 1]$ defined by $f(x) = \cos(\theta)$

This function can't be made onto without restricting the domain which would eliminate some of the original function's values.

$f:[0, \pi] \to [-1, 1]$ defined by $f(x) = \sin(\theta)$

$f:\mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$

In the previous section, notice that I didn't ask for a way you could restrict the function that made it one-to-one while still being equal to the original function. Why wouldn't it be possible to do that?

The problem that prevents a function from being one-to-one is that there are two different points with the same y-coordinate, e.g. (1, 3) and (2, 3). There's no way to fix this without changing or removing one of these points and that will leave you with a function that's no longer equal to the original.

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of a function $f:A \to C$ which is not onto if possible and explain why it isn't possible if not.

Does the definition of a function require every member of the codmain to be paired with an element of the domain?

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of a function $f:A \to C$ which is not onto if possible and explain why it isn't possible if not.

This is possible since the definition of a function doesn't require every element of the codomain to have a pairing. There are many possibilities but the simplest just takes every element of $A$ and maps it to a single element of $B$, e.g. $f(x) = a, \forall x\in A$.

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of a function $f:A \to C$ which is onto if possible and explain why it isn't possible if not.

This is possible although you'll have to double up on one of the domain values, e.g.

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of functions $f:A \to B$ and $g:B \to C$ where $f$ is onto but the composition $f \circ g$ isn't or explain why you can't.

This is possible. Define $f$ so that it maps each number to the letter with that position in the alphabet, e.g. $f(1) = a$ and $f(2) = b$. Then define $g$ so that it maps every element to $m$. $f$ is onto but the composition will take every element of $A$ and map it to the single element of $C$.

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of functions $f:A \to B$ and $g:B \to C$ where $f$ is not one-to-one but the composition $f \circ g$ is or explain why you can't.

Think about the domains and codomains and how they match up (or don't).

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of functions $f:A \to B$ and $g:B \to C$ where $f$ is not one-to-one but the composition $f \circ g$ is or explain why you can't.

This isn't possible because the sets don't match up properly. $g$ will be passing values from $C$ into $f$ but $f$ is defined for values in $A$.

For the sets $A=\{1, 2, 3, 4\}$, $B = \{a, b, c, d\}$ and $C = \{m, n, p\}$, give an example of functions $f:A \to B$ and $g:B \to C$ where $f$ is not one-to-one but the composition $g \circ f$ is or explain why you can't.

This isn't possible. The composition will be a function from $A$ to $C$ which, ultimately, will have to map a set with four elements to a map with three elements. There's no way to do that without assigning one of the domain values to two of the co-domain values.

If $f:A \to B$ and $g:A \to B$ are both one-to-one, does $(f + g):A \to B$ have to be one-to-one?

This doesn't have to be the case. Consider, for example, $f, g:\mathbb{R}\to\mathbb{R}$ defined by $f(x) = x$ and $g(x) = -x$. These two functions are both clearly one-to-one but $(f + g)(x) = x + (-x) = 0$ which is also clearly not one-to-one.

If $f:A \to B$ and $g:A \to B$ are both onto, does $(f + g):A \to B$ have to be onto?

This doesn't have to be the case and we can use the same functions from the previous question here as well: $f, g:\mathbb{R}\to\mathbb{R}$ defined by $f(x) = x$ and $g(x) = -x$. These two functions are both clearly onto but $(f + g)(x) = x + (-x) = 0$ which is also clearly not onto.

As we saw in the previous questions, we could make this onto by restricing the codommain but that's something you would have to do on a case by base basis.