Chapter 1 · Relations and Functions

Exercise 1.2 · Q12 (MCQ): $f(x)=3x$ on $\R$

Exercise 1.2Types of functions · Exercise 1.2

The question

Let $f : \R \to \R$ be defined as $f(x) = 3x$ ⁠. Choose the correct answer.

(A) $f$ is one-one onto
(B) $f$ is many-one onto
(C) $f$ is one-one but not onto
(D) $f$ is neither one-one nor onto.

Exam answer

$3x_1 = 3x_2 \Rightarrow x_1 = x_2$ ⁠, so $f$ is one-one. For $y \in \R$ ⁠, $f\big(\frac{y}{3}\big) = y$ ⁠, so $f$ is onto.

Answer: (A) $f$ is one-one onto

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Multiplying by 3 can be undone by dividing by 3, and dividing by 3 is always possible in $\R$ ⁠. Undoing in one direction means equal outputs come from equal inputs (one-one); undoing in the other means every real $y$ comes from the real number $\frac{y}{3}$ (onto). The domain matters: on $\R$ the number $\frac{y}{3}$ is always available, which would not be true on $\N$ or $\Z$ ⁠.

Given, to find, and the plan

Given: $f(x) = 3x$ ⁠, with domain $\R$ and codomain $\R$ ⁠.
To find: Which of (A)–(D) correctly describes $f$ ⁠.
Plan: (1) Prove one-one for arbitrary inputs. (2) Prove onto by solving $3x = y$ for an arbitrary real $y$ and checking. (3) Compare with each option and say why each wrong option fails.

Basics you need

1One-one, and "many-one"

$f$ is one-one if $f(x_1)=f(x_2) \Rightarrow x_1 = x_2$ for all $x_1, x_2$ ⁠; otherwise it is called many-one (Definition 5). To prove one-one, start from arbitrary $x_1, x_2$ ⁠. Basics: one-one and onto

2Onto

$f : X\to Y$ is onto if for every $y \in Y$ there is $x \in X$ with $f(x) = y$ (Definition 6). To prove it, take an arbitrary $y$ ⁠, construct $x$ ⁠, and check $x$ is in the domain.

3Dividing by a non-zero number

If $3a = 3b$ ⁠, dividing both sides by $3 \ne 0$ gives $a = b$ ⁠. And for any real $y$ ⁠, $\frac{y}{3}$ is again a real number.

4Reading a multiple-choice question

Each option makes two claims, one about one-one and one about onto. An option is correct only if both claims are true. Basics: proving and disproving

Solution

One-one: divide by 3

Let $x_1, x_2$ be any real numbers with $f(x_1) = f(x_2)$ ⁠. Then
$3x_1 = 3x_2 \ \Longrightarrow\ x_1 = x_2 \qquad(\text{divide both sides by } 3).$
So equal outputs always come from equal inputs: $f$ is one-one.

Because one-one is a claim about every pair of inputs, we use arbitrary $x_1, x_2$ rather than examples. Dividing by 3 is valid because $3 \ne 0$ ⁠. (If the function were $0\cdot x$ ⁠, this step would fail, and indeed that function sends every input to 0.)

One-one: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ (Definition 5). Division by a non-zero number (Basic 3).
Onto: every real $y$ is $3\cdot\frac{y}{3}$

Take an arbitrary $y \in \R$ ⁠. Solving $3x = y$ gives the candidate $x = \frac{y}{3}$ ⁠. This is a real number, so it lies in the domain $\R$ ⁠, and
$f\!\left(\frac{y}{3}\right) = 3\cdot\frac{y}{3} = y. \ ✓$
So every element of the codomain is an image: $f$ is onto.

The graph of $y = 3x$ (horizontal and vertical scales differ) is a straight line through the origin that extends without bound in both directions. Every horizontal line meets it exactly once: $y = 2$ meets it only at $x = \frac23$ ⁠. Exactly once means one-one and onto.

For onto we must reach every real $y$ ⁠, so we start from an arbitrary one, solve for the candidate input, and then check both that it is in the domain and that it maps to $y$ ⁠. The domain check is where $\R$ matters: if the domain were $\N$ ⁠, as in Example 8 with $2x$ ⁠, the candidate $\frac{y}{3}$ would often not be a natural number, and onto would fail.

Onto: for an arbitrary $y$ in the codomain, find $x$ in the domain with $f(x)=y$ (Definition 6). Basics: one-one and onto

Examine every option

We have shown: $f$ is one-one (Step 1) and onto (Step 2). Now test each option's two claims.

Option	Claims	Verdict
(A)	one-one, onto	Correct. Both claims are exactly what Steps 1 and 2 proved.
(B)	many-one, onto	Wrong. "Onto" is true, but "many-one" (not one-one) is false: Step 1 shows no two different inputs share an output.
(C)	one-one, not onto	Wrong. "One-one" is true, but "not onto" is false: every $y$ is the image of $\frac{y}{3}$ ⁠.
(D)	not one-one, not onto	Wrong. Both claims are false, by Steps 1 and 2.

An option is correct only if both of its claims hold. (B) and (C) each contain one true claim, so they must be rejected by pointing to the false one. Only (A) agrees with both findings.

Bijective = one-one and onto (Definition 7). Lesson: types of functions

Result

The correct answer is (A): $f(x) = 3x$ is one-one and onto, that is, bijective. One-one because $3x_1 = 3x_2$ gives $x_1 = x_2$ ⁠; onto because every real $y$ equals $f\!\left(\frac{y}{3}\right)$ ⁠. (Textbook answer: A.)

Check it with numbers

Preimages: $y = 2$ comes from $x = \frac23$ ⁠, since $3\cdot\frac23 = 2$ ✓; $y = -7$ comes from $x = -\frac73$ ⁠, since $3\cdot\left(-\frac73\right) = -7$ ✓. The inputs $-2, -1, 0, 1, 2$ give the five different outputs $-6, -3, 0, 3, 6$ ⁠, with no repeats ✓.

Common mistakes

Choosing (C) by thinking of $\N$ or $\Z$ ⁠, where $3x$ misses numbers like 1 and 2. Here the domain and codomain are $\R$ ⁠, where $\frac{y}{3}$ is always available.
Proving one-one with examples. Seeing that $f(1) \ne f(2)$ is not a proof; the argument must use arbitrary $x_1, x_2$ ⁠.
Forgetting to check the preimage lies in the domain. On $\R$ it does, but this is exactly the check that fails on $\N$ ⁠.

What to take away

$f(x) = ax$ (and more generally $ax + b$ ⁠) with $a \ne 0$ is a bijection from $\R$ to $\R$ ⁠.
Whether a function is onto depends on the domain and codomain: $3x$ is onto $\R$ but not onto $\Z$ ⁠.
Compare with Q11: $x^4$ fails both properties, $3x$ passes both.

Every fact used, and where it comes from

Fact	Used in	Source
Dividing by $3 \ne 0$ ⁠; $\frac{y}{3}\in\R$ for $y\in\R$	Steps 1, 2	Properties of real numbers
Definitions of one-one (many-one), onto, bijective	Steps 1–3	Definitions 5, 6, 7, Section 1.3 (Lesson: types of functions)
Same method for $2x$ on $\R$ ⁠, and its failure on $\N$	Steps 1, 2	Example 9 and Example 8, Section 1.3
All numerical values	Check	Recomputed exactly (fractions) in the site's verification script

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Exercise 1.2 · Q12 (MCQ): f(x)=3xf(x)=3xf(x)=3x on R\RR

Basics you need

Solution

One-one: divide by 3

Onto: every real yyy is 3⋅y33\cdot\frac{y}{3}3⋅3y​

Examine every option

Every fact used, and where it comes from

Exercise 1.2 · Q12 (MCQ): $f(x)=3x$ on $\R$

Onto: every real $y$ is $3\cdot\frac{y}{3}$