Exercise 2.2 answers: Relations and Functions

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1. 3x – y = 0 means y = 3x.
2. Check every x from 1 to 14: y = 3x must also lie in A, that is, y ≤ 14.
3. x = 1,2,3,4 give y = 3,6,9,12, all in A. x = 5 gives y = 15, which is not in A, so no larger x works.
4. So R = {(1,3), (2,6), (3,9), (4,12)}.
5. Domain is the set of first entries: {1,2,3,4}. Range is the set of second entries: {3,6,9,12}. Codomain is the whole set A, as stated in the definition.

AnswerDomain = {1,2,3,4}, Codomain = {1,2,...,14} = A, Range = {3,6,9,12}

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1. x must be a natural number less than 4, so x can be 1, 2 or 3.
2. For each x, y = x + 5: x=1 gives y=6, x=2 gives y=7, x=3 gives y=8.
3. R in roster form = {(1,6), (2,7), (3,8)}.
4. Domain = {1,2,3} (the x-values). Range = {6,7,8} (the y-values).

AnswerR = {(1,6), (2,7), (3,8)}; Domain = {1,2,3}; Range = {6,7,8}

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1. Go through every pair (x,y) with x ∈ A, y ∈ B and check whether x − y is odd.
2. x=1: 1−4=−3 (odd), 1−6=−5 (odd), 1−9=−8 (even). Keep (1,4),(1,6).
3. x=2: 2−4=−2 (even), 2−6=−4 (even), 2−9=−7 (odd). Keep (2,9).
4. x=3: 3−4=−1 (odd), 3−6=−3 (odd), 3−9=−6 (even). Keep (3,4),(3,6).
5. x=5: 5−4=1 (odd), 5−6=−1 (odd), 5−9=−4 (even). Keep (5,4),(5,6).
6. R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}.

AnswerR = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}

Watch this explained “One relation, two sentences”, 5:49 into A relation is nothing more than a chosen part of the product

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(i) in set-builder form

1. The figure shows each element of P sending an arrow to the element of Q that is 2 less than it: 5→3, 6→4, 7→5.
2. So the rule is: the second entry is 2 less than the first entry.

AnswerR = {(x, y) : y = x − 2, x ∈ P, y ∈ Q}

(ii) roster form

1. Read off the arrows as ordered pairs: (5,3), (6,4), (7,5).
2. So R = {(5,3), (6,4), (7,5)}.
3. Domain (first entries) = {5,6,7}. Range (second entries) = {3,4,5}.

AnswerR = {(5,3), (6,4), (7,5)}; Domain = {5,6,7}; Range = {3,4,5}

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(i) Write R in roster form

1. For each a in A, find every b in A that a divides exactly (b ÷ a leaves no remainder).
2. a=1 divides every element: pairs (1,1),(1,2),(1,3),(1,4),(1,6).
3. a=2 divides 2,4,6: pairs (2,2),(2,4),(2,6).
4. a=3 divides 3,6: pairs (3,3),(3,6).
5. a=4 divides only 4: pair (4,4). a=6 divides only 6: pair (6,6).

AnswerR = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}

(ii) Find the domain of R

1. The domain is the set of all first entries used in R.
2. Every element of A appears as some a, so the domain is all of A.

AnswerDomain of R = {1, 2, 3, 4, 6}

(iii) Find the range of R

1. The range is the set of all second entries used in R.
2. Every element of A also appears as some b, so the range is all of A.

AnswerRange of R = {1, 2, 3, 4, 6}

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1. The domain is every x used: {0,1,2,3,4,5}.
2. The range is x+5 for each of those x: 5,6,7,8,9,10.

AnswerDomain = {0,1,2,3,4,5}; Range = {5,6,7,8,9,10}

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1. Prime numbers less than 10 are 2, 3, 5, 7.
2. Cube each one: 2³=8, 3³=27, 5³=125, 7³=343.
3. R = {(2,8), (3,27), (5,125), (7,343)}.

AnswerR = {(2,8), (3,27), (5,125), (7,343)}

Watch this explained “A sentence, turned into pairs”, 3:49 into A relation is nothing more than a chosen part of the product

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1. n(A) = 3, n(B) = 2, so n(A × B) = 3 × 2 = 6.
2. A relation from A to B is any subset of A × B, and a set with 6 elements has 2⁶ subsets.
3. So the number of relations = 2⁶ = 64.

Answer64

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1. Pick any two integers a and b.
2. The difference of two integers is always another integer — that never fails, whatever a and b are.
3. So the condition ‘a − b is an integer’ is automatically true for every pair of integers.
4. That means every pair (a, b) with a, b ∈ Z belongs to R, so R is all of Z × Z.
5. The domain (the first entries used) is every integer, and the range (the second entries used) is every integer too.

AnswerDomain of R = Z (all integers), Range of R = Z (all integers).

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