Mathematical-Reasoning · Unit 21 · Video 2 · Interactive Practice

Equal Functions: Why x⁵ mod 5 Is the Identity on R₅

IKey Formulas

Formula	Meaning
$R_n = \{0, 1, 2, \ldots, n - 1\}$	The set of remainders on division by $n$ ; every integer is congruent modulo $n$ to exactly one element of $R_n$
$a \bmod n = r$ , where $a \equiv r \pmod n$ and $0 \le r \lt n$	$\bmod$ is an operation that returns a number, $19 \bmod 6 = 1$ ; a congruence such as $19 \equiv 7 \pmod 6$ is a statement, true or false
$f = g$ provided $\text{dom}(f) = \text{dom}(g)$ , $\text{codom}(f) = \text{codom}(g)$ , and $f(x) = g(x)$ for every $x$ in the domain	Equality of functions: the definition never mentions formulas
$I_A : A \to A$ , $I_A(x) = x$ for every $x \in A$	The identity function on a nonempty set $A$ : every arrow goes straight across

Key Insight: A formula is only one way to describe a function's outputs. On $R_5$ the formulas $x^5 \bmod 5$ and $x$ look nothing alike yet describe the same function, while the diagonal count $d$ and $f(x) = \frac{x(x - 3)}{2}$ share a formula and are still different functions, because their domains differ.

IIBuilding a Function out of Remainders

Reducing mod $n$ gives each input in $R_n$ exactly one output in $R_n$ , though some remainders may never be outputs.

Rule n 6

IIISame Formula, Different Functions

The diagonal count $d$ and $f(x) = \frac{x(x - 3)}{2}$ share a formula, but only $f$ accepts every real input.

x 0

IVDifferent Formulas, Same Function?

Equality checks three things only: the domain, the codomain, and the output at every input.

n 5 f(x) codomain of f g(x) codomain of g

VQuiz Questions

Problem 1 · The Range of a Remainder Function

Given: $f : R_8 \to R_8$ , $f(x) = x^2 \bmod 8$ . Find the range of $f$ .

✅ Correct! The eight outputs are

0, 1, 4, 1, 0, 1, 4, 1

, so only three remainders are ever reached.

❌ Not quite. These are the squares

x^2

, before the remainder is taken. The operation

\bmod 8

always returns an element of

R_8

; for example

f(3) = 9 \bmod 8 = 1

❌ Not quite.

R_8

is the codomain, the set the outputs are allowed to lie in. The range contains only the outputs that actually occur, and

2

is never one of them.

❌ Close, but…

0

is an output too:

f(0) = 0

, and

f(4) = 16 \bmod 8 = 0

because

16 = 8 \cdot 2 + 0

Show solution

Compute $x^2$ for each $x \in R_8$ and take its remainder on division by $8$ :

$x$	$x^2$	$x^2 = 8q + r$	$f(x)$
$0$	$0$	$8 \cdot 0 + 0$	$0$
$1$	$1$	$8 \cdot 0 + 1$	$1$
$2$	$4$	$8 \cdot 0 + 4$	$4$
$3$	$9$	$8 \cdot 1 + 1$	$1$
$4$	$16$	$8 \cdot 2 + 0$	$0$
$5$	$25$	$8 \cdot 3 + 1$	$1$
$6$	$36$	$8 \cdot 4 + 4$	$4$
$7$	$49$	$8 \cdot 6 + 1$	$1$

The outputs are $0, 1, 4, 1, 0, 1, 4, 1$ , so $\text{range}(f) = \{0, 1, 4\}$ , a proper subset of the codomain $R_8$ . The output $1$ has four preimages, $1$ , $3$ , $5$ and $7$ , which a function allows.

Problem 2 · Same Rule, Different Codomain

Given: $f : R_4 \to R_4$ and $g : R_4 \to \mathbb{Z}$ , both defined by the rule $(x^2 + x) \bmod 4$ . Which statement is true?

✅ Correct! Conditions 1 and 3 hold, but condition 2 fails, and all three are required.

❌ Not quite. Equal functions need equal codomains as well. Here

\text{codom}(f) = R_4

and

\text{codom}(g) = \mathbb{Z}

, so condition 2 fails even though the outputs all match.

❌ Not quite. The rule takes a remainder, so every output of

g

lies in

R_4

; in fact

g

outputs only

0

and

2

. The functions differ, but for another reason.

❌ Not quite.

g

uses the same rule,

\bmod 4

included:

g(3) = 12 \bmod 4 = 0 = f(3)

. The outputs agree everywhere; look at the codomains.

Show solution

Check the three conditions of the definition.

1. Domains: $\text{dom}(f) = R_4 = \text{dom}(g)$ . ✓

2. Codomains: $\text{codom}(f) = R_4$ but $\text{codom}(g) = \mathbb{Z}$ , and $R_4 \ne \mathbb{Z}$ (for instance $5 \in \mathbb{Z}$ but $5 \notin R_4$ ). ✗

3. Outputs: $x^2 + x$ takes the values $0, 2, 6, 12$ at $x = 0, 1, 2, 3$ , with remainders $0, 2, 2, 0$ on division by $4$ . Both functions use this rule, so $f(x) = g(x)$ for every $x \in R_4$ . ✓

Condition 2 fails, so $f \ne g$ . Matching outputs are not enough on their own; the definition also compares the codomains.

Problem 3 · Two Formulas on $R_7$

Given: $f, g : R_7 \to R_7$ with $f(x) = 2x \bmod 7$ and $g(x) = 9x \bmod 7$ .

Is $f = g$ ?

What is the range of $f$ ?

✅ Correct!

f

and

g

are one function, and its outputs

0, 2, 4, 6, 1, 3, 5

fill all of

R_7

❌ Not quite.

27

is not an output:

g(3) = 27 \bmod 7 = 6

, since

27 = 7 \cdot 3 + 6

. That equals

f(3) = 6

❌ Not quite. The definition of equal functions never mentions formulas. Compare the outputs:

g(1) = 9 \bmod 7 = 2 = f(1)

, and so on at every input.

❌ Not quite. They agree everywhere, for example

f(1) = 2

and

g(1) = 9 \bmod 7 = 2

. Remember to reduce

9x

before comparing.

❌ Not quite. Compare

f(x)

and

g(x)

at each of the seven inputs.

❌ Not quite.

2x

is even, but its remainder need not be:

f(4) = 8 \bmod 7 = 1

❌ Not quite. These are the values of

2x

, before reduction. Every output of

f

lies in

R_7

; for example

f(6) = 12 \bmod 7 = 5

❌ Close, but…

f(0) = 0

, so

0

is in the range too.

❌ Not quite. List

f(x)

for

x = 0, 1, \ldots, 6

Show solution

Equality. Both functions have domain $R_7$ and codomain $R_7$ , so conditions 1 and 2 hold. For condition 3, $9x - 2x = 7x$ is a multiple of $7$ , so $9x \equiv 2x \pmod 7$ for every integer $x$ . Congruent integers have the same remainder on division by $7$ , so $9x \bmod 7 = 2x \bmod 7$ , that is, $g(x) = f(x)$ for every $x \in R_7$ . The table confirms it:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$2x$	$0$	$2$	$4$	$6$	$8$	$10$	$12$
$f(x)$	$0$	$2$	$4$	$6$	$1$	$3$	$5$
$9x$	$0$	$9$	$18$	$27$	$36$	$45$	$54$
$g(x)$	$0$	$2$	$4$	$6$	$1$	$3$	$5$

So $f = g$ .

Range. The outputs of $f$ are $0, 2, 4, 6, 1, 3, 5$ : every element of $R_7$ appears, so $\text{range}(f) = R_7$ , the whole codomain.

Problem 4 · Cubes and the Identity Function

Given: $f : R_3 \to R_3$ , $f(x) = x^3 \bmod 3$ , and $g : R_4 \to R_4$ , $g(x) = x^3 \bmod 4$ . Which statement is true?

✅ Correct!

f

sends

0, 1, 2

0, 1, 2

, while

g

sends

0, 1, 2, 3

0, 1, 0, 3

❌ Not quite. Check every input of

g

2^3 = 8 = 4 \cdot 2 + 0

, so

g(2) = 0

, while

I_{R_4}(2) = 2

. One mismatch is enough.

❌ Not quite.

8

is not an output of

f

, whose codomain is

R_3

f(2) = 8 \bmod 3 = 2

, since

8 = 3 \cdot 2 + 2

❌ Not quite. A similar-looking rule is not enough:

\text{dom}(f) = R_3

and

\text{dom}(g) = R_4

, so condition 1 fails (and the codomains differ too).

Show solution

$f$ versus $I_{R_3}$ . Both have domain and codomain $R_3$ . The outputs: $f(0) = 0$ , $f(1) = 1$ , and $f(2) = 8 \bmod 3 = 2$ because $8 = 3 \cdot 2 + 2$ . So $f(x) = x = I_{R_3}(x)$ for every $x \in R_3$ , and $f = I_{R_3}$ .

$g$ versus $I_{R_4}$ . Both have domain and codomain $R_4$ . The outputs: $g(0) = 0$ , $g(1) = 1$ , $g(2) = 8 \bmod 4 = 0$ because $8 = 4 \cdot 2 + 0$ , and $g(3) = 27 \bmod 4 = 3$ because $27 = 4 \cdot 6 + 3$ . At $x = 2$ , $g(2) = 0 \ne 2 = I_{R_4}(2)$ , so $g \ne I_{R_4}$ .

$f$ versus $g$ . Their domains $R_3$ and $R_4$ differ, so $f \ne g$ no matter how alike the rules look.

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