Chapter 1 · Relations and Functions

Example 25: $I_{\N}+I_{\N}$ is not onto

ExamplesOnto functions · Miscellaneous Examples

The question

Consider the identity function $I_{\N} : \N \to \N$ defined as $I_{\N}(x) = x\ \ \forall\, x \in \N$ ⁠. Show that although $I_{\N}$ is onto but $I_{\N} + I_{\N} : \N \to \N$ defined as $(I_{\N} + I_{\N})(x) = I_{\N}(x) + I_{\N}(x) = x + x = 2x$ is not onto.

Exam answer

$I_{\N}$ is onto: For every $y \in \N$ ⁠, $I_{\N}(y) = y$ ⁠, so $y$ is the image of $y \in \N$ ⁠. Thus $I_{\N}$ is onto.

$I_{\N} + I_{\N}$ is not onto: $(I_{\N} + I_{\N})(x) = 2x$ ⁠. Take $3 \in \N$ ⁠. If $2x = 3$ ⁠, then $x = \frac32 \notin \N$ ⁠. So 3 has no pre-image in $\N$ ⁠. Thus $I_{\N} + I_{\N}$ is not onto.

Hence $I_{\N}$ is onto but $I_{\N} + I_{\N}$ is not onto.

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

The one idea this problem is built on

Being onto is not preserved by adding functions. The identity hits every natural number, because each $y$ is its own preimage. But adding the identity to itself doubles every value, and doubling a natural number always gives an even number. So odd numbers such as 3 are never reached, and one unreached element of the codomain is enough to show a function is not onto.

Given, to show, and the plan

Given: $\N = \{1,2,3,\dots\}$ ⁠, the identity $I_{\N}(x) = x$ ⁠, and the sum $(I_{\N}+I_{\N})(x) = 2x$ ⁠, both from $\N$ to $\N$ ⁠.
To show: (a) $I_{\N}$ is onto. (b) $I_{\N}+I_{\N}$ is not onto.
Plan: (1) Prove (a) by producing a preimage for an arbitrary $y$ ⁠. (2) Understand how the sum of two functions is defined, and check $2x$ really lands in $\N$ ⁠. (3) Prove (b) with one counterexample: an element of the codomain with no preimage. (4) Describe exactly which elements are missed.

Basics you need

1Onto (surjective)

$f : A\to B$ is onto if every $y\in B$ equals $f(x)$ for at least one $x\in A$ ⁠. To prove it, take an arbitrary $y$ and construct $x$ ⁠. To disprove it, find one $y\in B$ that no $x\in A$ reaches. Basics: one-one and onto

2The identity function

$I_A : A\to A$ ⁠, $I_A(x) = x$ ⁠: every element is sent to itself. Basics: what a function is

3Sum of two functions

For $f, g : A\to\R$ ⁠, the sum $f+g$ is the function defined pointwise: $(f+g)(x) = f(x) + g(x)$ for each $x\in A$ ⁠. (Class XI, Relations and Functions.)

4Even and odd natural numbers

$n$ is even when $n = 2k$ for some integer $k$ ⁠, and odd otherwise. No number is both, so an odd number is never of the form $2x$ with $x$ a whole number. Basics: integers, even and odd

5One counterexample disproves "for all"

"Onto" is a claim about every element of the codomain. A single element that fails is enough to show the claim is false. Basics: proving and disproving

Solution

$I_{\N}$ is onto

Take any $y\in\N$ (the codomain). We need an $x\in\N$ (the domain) with $I_{\N}(x) = y$ ⁠. Choose $x = y$ ⁠. It is a natural number, and
$I_{\N}(x) = I_{\N}(y) = y.$
Since $y$ was arbitrary, every element of the codomain has a preimage, so $I_{\N}$ is onto.

The textbook says "clearly $I_{\N}$ is onto"; this step spells out why. To prove onto we must handle every $y$ ⁠, so we use a letter $y$ that could be any natural number and find its preimage explicitly. For the identity the preimage is $y$ itself.

Onto: for each $y$ in the codomain there is an $x$ in the domain with $f(x)=y$ (Definition 6, Basic 1).
What the function $I_{\N}+I_{\N}$ is

By the pointwise definition of a sum (Basic 3), for each $x\in\N$ ⁠,
$(I_{\N}+I_{\N})(x) = I_{\N}(x) + I_{\N}(x) = x + x = 2x.$
So $I_{\N}+I_{\N}$ is the "doubling" function. It does map $\N$ into $\N$ ⁠: if $x$ is a natural number then so is $2x$ ⁠. A few values:

$x$ 1 2 3 4 5 …

$(I_{\N}+I_{\N})(x) = 2x$ 2 4 6 8 10 …

Before testing "onto", we must know exactly what the function does and confirm that its outputs lie in the stated codomain $\N$ ⁠. The sum of functions is defined value by value, so adding the identity to itself just adds $x$ to $x$ at each point.

$(f+g)(x) = f(x)+g(x)$ (Basic 3). The sum of two natural numbers is a natural number, so $2x = x + x \in \N$ ⁠.
$I_{\N}+I_{\N}$ is not onto: the number 3 is never reached

Take $y = 3$ ⁠, which is in the codomain $\N$ ⁠. Is there an $x\in\N$ with $2x = 3$ ⁠? Solving the equation,
$2x = 3 \iff x = \frac{3}{2} = 1.5.$
This is the only number satisfying $2x = 3$ ⁠, and it is not a natural number. So no $x$ in the domain $\N$ has $(I_{\N}+I_{\N})(x) = 3$ ⁠. One element of the codomain has no preimage, hence $I_{\N}+I_{\N}$ is not onto.

To disprove onto we need just one element of the codomain that is missed (Basic 5). Solving $2x = 3$ finds the only possible candidate for a preimage, so once that candidate is ruled out, no preimage exists at all. The candidate fails because it is not in the domain $\N$ ⁠: the domain matters as much as the formula.

Dividing both sides of $2x = 3$ by 2 is reversible, so $x = \frac32$ is the unique solution. $\frac32$ lies strictly between 1 and 2, so it is not a natural number.
Exactly which numbers are missed

Every output $2x$ is even, so every odd natural number is missed: 1, 3, 5, 7, … . And every even number $y = 2k$ is reached, by $x = k$ ⁠. So the range of $I_{\N}+I_{\N}$ is the set of even natural numbers, a proper part of the codomain $\N$ ⁠.

Each input $x$ sends one arrow to $2x$ ⁠. The arrows only ever land on even numbers. Every odd number, including 3, is left without an arrow, so the function is not onto $\N$ ⁠.

The textbook needs only one missed element, but seeing all of them explains the failure: doubling can only produce even numbers. Note that 1 is missed too (the smallest output is $2\cdot1 = 2$ ⁠), so the number 1 would have served equally well as the counterexample.

An odd number is never $2\times$ (a natural number) (Basic 4). The range of a function is the set of outputs actually reached; "onto" means range = codomain.
Conclude

$I_{\N}$ is onto (Step 1), but $I_{\N}+I_{\N}$ is not onto (Step 3). So adding two onto functions can produce a function that is not onto.

This is the point of the example: properties like "onto" belong to a function as a whole and need not survive arithmetic on functions. The same doubling function is studied on its own in Example 8, where it is shown to be one-one but not onto on $\N$ ⁠. On $\R$ instead of $\N$ it would be onto, because $x = \frac32$ would then be allowed (Example 9).

$x$	1	2	3	4	5	…
$(I_{\N}+I_{\N})(x) = 2x$	2	4	6	8	10	…

Result

$I_{\N}$ is onto, since every $y\in\N$ is $I_{\N}(y)$ ⁠. But $I_{\N}+I_{\N}$ ⁠, which sends $x$ to $2x$ ⁠, is not onto: the element $3$ of the codomain would need $x = \frac32\notin\N$ ⁠. In fact it misses every odd number.

Check it with numbers

The values $2\cdot1, 2\cdot2, \dots, 2\cdot9999$ run $2, 4, 6, \dots, 19998$ ⁠. The site's verification script confirms that 3 is not among them, that no odd number below 1000 is among them, and that every even number below 1000 is (as $2\cdot\frac{y}{2}$ ⁠). Near 3 the outputs jump from $2\cdot1 = 2$ straight to $2\cdot2 = 4$ ⁠, skipping 3.

Common mistakes

Accepting $x = \frac32$ ⁠. The domain is $\N$ ⁠; a preimage must be a natural number.
Thinking the sum of onto functions must be onto. This example shows it fails, even when both summands are the same onto function.
Proving onto for $I_{\N}$ with examples. " $I_{\N}(5) = 5$ " for a few values is not a proof; use an arbitrary $y$ ⁠.
Confusing one-one with onto. $2x$ is one-one ( $2x_1 = 2x_2 \Rightarrow x_1 = x_2$ ⁠); it is the onto property that fails.

What to take away

To show a function is not onto, name one element of the codomain and show that the equation $f(x) = y$ has no solution in the domain.
Whether a function is onto depends on the domain and codomain, not just the formula: $2x$ is onto $\R$ but not onto $\N$ ⁠.
Sums of functions are defined pointwise, and they can lose properties the summands had. Example 26 shows the same for one-one.

Every fact used, and where it comes from

Fact	Used in	Source
Onto (surjective)	Steps 1, 3, 5	Definition 6, Section 1.3 (Lesson: types of functions)
Identity function $I_{\N}(x) = x$	Steps 1, 2	Given in the question; also used in Section 1.4
Sum of functions $(f+g)(x) = f(x)+g(x)$	Step 2	Class XI, Relations and Functions; stated in the question
$2x = 3 \iff x = \frac32\notin\N$	Step 3	Algebra; $\N = \{1,2,3,\dots\}$
Odd numbers are never $2x$ ⁠, $x\in\N$	Step 4	Definition of even and odd
Missed and reached values	Step 4, Check	Checked over large samples in the site's verification script

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Example 25: IN+INI_{\N}+I_{\N}IN​+IN​ is not onto

Basics you need

Solution

INI_{\N}IN​ is onto

What the function IN+INI_{\N}+I_{\N}IN​+IN​ is

IN+INI_{\N}+I_{\N}IN​+IN​ is not onto: the number 3 is never reached

Exactly which numbers are missed

Conclude

Every fact used, and where it comes from

Example 25: $I_{\N}+I_{\N}$ is not onto

$I_{\N}$ is onto

What the function $I_{\N}+I_{\N}$ is

$I_{\N}+I_{\N}$ is not onto: the number 3 is never reached