Chapter 25 · Defined Operations

Use Number Functions, Congruence, and Selection Operators

Some symbols combine two inputs. Some inspect one number. Some make a true-or-false comparison, and some simply choose the smaller or larger input. Read the local definition, identify the symbol family, and then use the right kind of reasoning.

Binary operationUnary functionCongruence relationMinimum & maximumInteger & fractional parts

Four different jobs

Ask what the symbol produces

(a,b)↦ a∧ b · two inputs produce one number

n↦φ(n) · one input produces one number

a ≡ b (mod m) · a statement is true or false

a△ b=min(a,b) · one input is selected

Symbol family

Decide what kind of mathematical object you are reading

Not complete

These descriptions can overlap. A minimum or maximum selector uses two inputs and returns one number, so it is also a binary operation. “Selection operator” is its more specific job. In the matching question, choose the most specific description. Thus7▽4 is a selection operator, although it is also a binary operation.

2→1

Binary operation

Uses two inputs and produces one output.

a∧ b

1→1

Unary function

Inspects one input and returns one value.

φ(n), ψ(n)

T/F

Relation

Makes a statement that is true or false.

a ≡ b (mod m)

pick

Selection operator

Returns one of the inputs according to a rule.

min(a,b), max(a,b)

Do not mix the jobs. A congruence statement does not “calculate an answer number.” A unary function does not need a second input. A selection operator never invents a third value—it returns one of the two values already present.

Classify each expression.

6∧15

φ(18)

17 ≡ 5 (mod 6)

7▽4

Need a hint?

Count inputs and identify the result. A selector is a special kind of binary operation.

Example 2

Combine the GCD and LCM in one defined operation

Not complete

In this problem only, [a,b] means the least common multiple and (a,b) means the greatest common divisor.

a∧ b=[a,b]+(a,b)=lcm(a,b)+gcd(a,b)

Left input a

Right input b

inputs14 and 4

→

two number factsGCD 2 · LCM 28

→

defined output30

Divisors and common divisors

Common divisors:

First common multiples

GCD2

LCM28

sum30

check56 = 56

The identity gcd(a,b)×lcm(a,b)=a× b provides a useful check.

Complete the worked example calculation for 14∧4.

GCD

LCM

14∧4

Need a hint?

Find the largest common divisor and first common multiple, then add them.

Guided Practice 2

Work backward through a GCD–LCM operation

Not complete

Why does the search cover every answer?

For positive x, lcm(6,x) is at least x and gcd(6,x) is at least 1. If their sum is 33, then x + 1 ≤ 33, so x ≤ 32. Searching through 40 therefore covers every possible worked solution.

A short hand solution

Call the GCD g. It divides 6 and the LCM, so it also divides their sum 33. Thus g is 1 or 3. If g=1, the LCM would be 32, which is not a multiple of 6. So g=3 and the LCM is 30. Using GCD × LCM = 6 × x gives 3 × 30 = 6x, hence x=15. Check: gcd(6,15)=3 and lcm(6,15)=30.

6∧ x=lcm(6,x)+gcd(6,x)=33

Fixed left input

Required output

Search positive inputs through

candidates checked40

matches1

matching inputs15

statusunique

x	GCD	LCM	operation value	comparison

The table tests every positive integer up to the displayed limit. Its status describes only that finite search. The bound x≤32 proves that the original 6∧x=33 search through40 is complete; changed settings need their own bound for a conclusion about all positive integers.

Finish the guided practice.

Value of x

gcd(6,x)

lcm(6,x)

Need a hint?

Use the complete GCD–LCM definition and the bound x≤32 for the practice problem.

Example 3

Compose the divisor-count function with itself

Not complete

Count exponent choices

18 = 2¹ × 3². A divisor may use zero or one copy of 2 (two choices) and zero, one, or two copies of 3 (three choices). Each combination gives exactly one divisor, so 2 × 3 = 6 divisors. In general, a prime with exponent e contributes e+1 choices, including zero copies.

φ(n)=number of positive divisors of n

Local meaning. The worked example defines φ this way for this problem.

Starting positive integer

Number of times to apply φ

Starting number

Positive divisors:

Exponent shortcut

If n=p₁^e₁p₂^e₂⋯, then the number of divisors is (e₁+1)(e₂+1)⋯.

starting value18

first output6

final output4

applications2

Complete the worked example composition.

φ(18)

φ(6)

φ(φ(18))

Need a hint?

The first divisor count becomes the new input; it is not the final answer yet.

Guided Practice 3

Use a divisor-sum function and compare function compositions

Not complete

ψ(n)=sum of all positive divisors of n

Starting positive integer

First function

Optional second function

Divisors of the current input

Two functions, two questions

φ(n)6

ψ(n)42

selected result42

number of divisors6

The same divisor list answers two different questions: “How many?” and “What is their sum?”

Complete the worked example practice for ψ(20).

List all positive divisors of 20 once each, separated by commas

ψ(20)

φ(20)

Need a hint?

List factor pairs to avoid missing a divisor; then count or add according to the chosen function.

Guided Practice 4

Read congruence as a same-remainder relation

Not complete

a ≡ b (mod m) ⇔ amod m=bmod m

The symbol ≡ makes a statement. It is true when the two remainders match and false when they do not. It does not mean ordinary equality.

Reference number a

Modulus m

Search from

Search through

reference remainder1

matching values5

first match1

last match9

Values congruent to 7 modulo 2:

Test one value b

7 ≡ 9 (mod 2) is true

Both remainders are 1.

Complete both original searches.

1≤b≤10 with b ≡ 7 (mod 2)

1≤ b≤20 with b ≡ 3 (mod 5)

What does congruence compare?

Need a hint?

Compute the reference remainder, then move by jumps equal to the modulus.

Exercise 5 · Test 25 Question 9

Let selection operators choose the smaller or larger input

Not complete

a△ b=min(a,b), a▽ b=max(a,b)

Reset the symbols. Earlier chapter problems used triangle-like symbols for different formulas. In this exercise only, the symbols select the smaller and larger inputs.

Explore the worked example expression with x

startx = 7

x△44

4▽(x△4)4

5▽[⋯]5

possible outputsonly 5

Because x△4≤4, the next maximum is exactly 4. The outside maximum then chooses 5. The nested expression has one possible value, no matter how x changes.

Apply the Test 25 selectors to exact fractions

In the test expression, ○ selects the larger input and △ selects the smaller input.

Make the four selections first

This is an exact-fraction adaptation of the test expression. Here ○ chooses the larger number and △ chooses the smaller.

[(2/3 ○ 17/26) + (5/8 △ 23/33)] ÷ [(1/3 △ 34/99) + (237/106 ○ 9/4)]

The first two selections belong to the numerator; the last two belong to the denominator.

Selection 1: 2/3 ○ 17/26Selection 2: 5/8 △ 23/33Selection 3: 1/3 △ 34/99Selection 4: 237/106 ○ 9/4

Compare your selections and calculation

selected numerator terms2/3 and 5/8

numerator31/24

denominator31/12

quotient1/2

Complete both selection investigations.

Value of 5▽(4▽(x△4))

Number of different possible values

Value of the Test 25 fraction

Need a hint?

Make each of the four selections first. Only then add the numerator and denominator terms.

Exercise 13

Split a nonnegative number into integer and fractional parts

Not complete

Read a nonnegative decimal as written: the digits before the decimal point form its integer part, and the digits after it form its fractional part. For example, 12.075 = 12 + 0.075. Keep the leading zeros in the fractional part. Do not round a value across an integer boundary.

[x]=integer part of x, {x}=fractional part of x

This laboratory follows the worked example’s nonnegative-decimal examples. For such values, x=[x]+{x} and 0≤{x}<1.

Number to split

integer part12

fractional part0.75

Solve the worked example system

a+[b]=15.3, {a}+b=7.8

Reason by parts

The first equation shows that the fractional part of a is 0.3. The second equation then gives b=7.8-0.3=7.5. Since [b]=7, the first equation gives a=15.3-7=8.3.

Verify a candidate pair

Both equations hold

Complete the practice problem.

[7.5]

{8.3}

Need a hint?

The fractional part is at least zero and below one. Read the first equation’s decimal part first.

Independent practice

Number-function and relation workshop

Not complete

Correct all eight questions to complete the workshop.

1. What kind of object is a ≡ b (mod m)?

2. If a∧ b=lcm(a,b)+gcd(a,b), find 18∧12.

3. If φ(n) counts positive divisors, find φ(36).

4. If ψ(n) sums positive divisors, find ψ(12).

5. Find φ(ψ(6)).

6. How many integers from 1 through 25 are congruent to 4 modulo 7?

7. If △ selects the smaller and ▽ the larger, find (9△14)+(9▽14).

8. Let ○ select the larger input and △ select the smaller. Evaluate [(2/3 ○ 17/26)+(5/8 △ 23/33)] ÷ [(1/3 △ 34/99)+(237/106 ○ 9/4)].

Need a hint?

Answers:relation;42;9;28;6;4;23;1/2. For18∧12 addGCD6 andLCM36. The divisors of36 have3×3 exponent choices. ψ(12)=1+2+3+4+6+12=28. Forφ(ψ(6)), firstψ(6)=12, thenφ(12)=6. The congruent values are4,11,18,25.

Six-response check

Exit ticket and certificate

Not complete

Here φ counts positive divisors and ψ sums them. For nonnegative t, [t] is the integer part and {t}=t−[t] is the fractional part.

1. What kind of mathematical object is ψ(n)?

2. Find 6∧15 under a∧b=gcd(a,b)+lcm(a,b).

3. Find φ(φ(18)).

4. List every b from 1 through 20 satisfying b ≡ 3 (mod 5).

5. In a+[b]=15.3 and {a}+b=7.8, find a.

6. In the same system, find b.

Need a hint?

Answers:unary function;33;4;3,8,13,18;8.3;7.5. UseGCD3+LCM30; divisor-count chain18→6→4; jumps of5 from3; and the decimal-part equations.

Optional reflection — not automatically gradedThis response is saved but does not affect mission completion.

Chapter 25 Achievement

Number Function, Congruence & Selector Navigator

This certifies that the learner can use GCD–LCM operations, divisor-count and divisor-sum functions, same-remainder congruence, minimum and maximum selectors, and integer and fractional parts with accurate local notation.

Lesson 25.5 complete