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This question introduces the surjective property of a function. You will need to be able to confidently identify the domain, codomain and range of a function before you can understand the surjective property.

There are two proof styles.

To prove that a function $f$ is surjective, start with a generic element $b$ from the codomain and, by solving $f(a) = b$ for $a$, show that $a$ is in the domain.
Always prove that a function $f$ is not surjective by contradiction. Start with a specific element $b \\in B$ and, by solving $f(a) = b$ for $a$, show that $a$ is not in the domain. This contradiction proves it is not possible for the function to be surjective.

The codomain of $f_1$ is $\\mathbb Z$. Suppose that $f_1$ is surjective, which means that the whole codomain gets mapped to, and in particular that there is some integer $a$ such that $f(a) = 1 \\in \\mathbb Z$. Then

$f_1(a) = 2a = 1$

and so $a = \\frac{1}{2}$. But $\\frac{1}{2} \\not\\in \\operatorname{Domain}(f_1) = \\mathbb Z$, so our assumption that $f_1$ is surjective must be false.

The codomain of $f_2$ is $\\mathbb Z$. For any $b \\in \\mathbb Z$ suppose that $f(a) = b$. Then

$f_2(a) = \\lceil a \\rceil = b$.

There are many values of $a$ which map to $b$, and one such value is just $a=b \\in \\operatorname{Domain}(f_2) = \\mathbb R$. So $f_2$ is surjective.

The codomain of $f_3$ is the positive rationals $\\mathbb Q^+$. For any $\\frac{p}{q} \\in \\mathbb Q^+$ suppose that $f(a) = \\frac{p}{q}$. Then

$f_3(a) = \\frac{1}{a} = \\frac{p}{q}$

and so $a = \\frac{q}{p} \\in \\operatorname{Domain}(f_3) = \\mathbb Q^+$. So $f_3$ is surjective.

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Consider the example

$f: \\mathbb Z \\mapsto \\mathbb R, \\quad f(x) = \\left|x\\right|$.

The set of possible inputs $\\mathbb Z$ is called the domain.
The set of possible outputs $\\mathbb R$ is called the codomain.

A function must be defined for every element of its domain, but the codomain may contain additional elements that are unused.

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An introduction to terminology about the surjective property of a function.

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$\\mathbb N$

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$\\mathbb Z^+$

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$\\mathbb Z$

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$\\mathbb R$

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The set of actual outputs is called the range and may be different to the codomain. In other words, the range is the set of all elements that are actually mapped to, or in the language of sets

$\\operatorname{Range}(f) = \\left\\{ b | \\, f(a) = b, a \\in \\mathbb Z\\right\\}$.

What is the range of $f$?

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$\\mathbb Z$

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$\\mathbb N$

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$\\mathbb R$

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If the codomain and range are equal the function is said to be surjective or onto. The function $f$ is not surjective because the codomain is different to the range. The codomain is

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$f_1 : \\mathbb Z \\mapsto \\mathbb Z, \\quad f_1(x) = 2x$

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$f_2 : \\mathbb R \\mapsto \\mathbb Z, \\quad f_2(x) = \\lceil x\\rceil$

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$f_3 : \\mathbb Q^+ \\mapsto \\mathbb Q^+, \\quad f_3(x) = \\frac{1}{x}$

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Select all functions that are surjective. For this question the set of positive rational numbers is denoted by $\\mathbb Q^+$.

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