MATH327 – HOMEWORK SOLUTIONS HOMEWORK #4

Section 1.5: Problems 1, 2, 3, 5 Section 2.1: Problems 1,4,7,9,10,14

Krzysztof Galicki Problem 1.5.1.

a) Yes, it is reflexive, antisymmetric, and transitive and therefore it is a partial order on ℝ. It is also a total order.

b) No, as it is not reflexive. For example, it is not true that 2 < 2.

c) (ℝ, ⪯), where a ⪯ b means a^{2} ≤ b^{2} is not a partial order as it is not antisymmetric. While a ⪯ b and b ⪯ a imply that a^{2} = b^{2}, the latter does not mean that a = b.

d) This relation is not antisymmetric. Consider (2, 3) and (2, 1). Clearly (2, 3) ⪯ (2, 1) and (2, 1) ⪯ (2, 3) but (2, 3) ≠ (2, 1).

e) This is a partial order but not a total order. Consider, for example, (2, 3) and (3, 4). Neither (2, 3) ⪯ (3, 4), nor (3, 4) ⪯ (2, 3).

f) This relation is not antisymmetric. Consider two words w_{1} = cat and w_{2} = pit (in this example the alphabet is the standard English alphabet). They have the same length so w_{1} ⪯ w_{2} and w_{2} ⪯ w_{1}. But they are not the same words.

Problem 1.5.2.

a) {1, 10, 100, 1000, 1001, 101, 1010, 11, 110, 111}

b) {1, 11, 111, 110, 10, 101, 1010, 100, 1001, 1000} Problem 1.5.3.

a) a ≺ b ≺ c ≺ d (other follow by transitivity).

b) a ≺ b and c ≺ d.

c) a ≺ b, a ≺ d, and c ≺ d.

d) a ≺ b, a ≺ c, and a ≺ d.

e) a ≺ d ≺ e, b ≺ e, and b ≺ c ≺ f (other follow by transitivity).

f) a ≺ f, g ≺ f, d ≺ b ≺ c, d ≺ h ≺ i, ≺ c, e ≺ i (all the other relations follow by transitivity).

Problem 1.5.5.

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b)

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∘ ∘  ∘  ∘  ∘ ∘  ∘  ∘  ∘  ∘  ∘ ∘  ∘  ∘  ∘ ∘ ∅ {a}  {b}  {c}  {d} {a, b} {a, c} {a, d} {b, c} {b, d} {c, d} {a, b, c} {a, c, d} {a, b, d} {b, c, d} {a, b, c, d} (b’) The Hesse diagram for (P({a, b, c, d}), ⊆)

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Problem 2.1.1

a) It is not a function as f(3) = 1 and f(3) = 3.

b) It is not a function with domain {1, 2, 3, 4} as f(3) is not defined.

c) Yes, it is a function.

d) It is not a function as f(1) = 1 and f(1) = 2.

e) Yes, it is a function. Problem 2.1.4

a) f(n) = 2n.

b) f(n) = 1+ integer part of n/100.

c) f(n) = nmod3 + 1.

d) f(n) = n. Problem 2.1.7

{(1, −3), (3, 1), (5, 5), (2, 5), (4, 17)}

The function is not 1-1 as f(5) = f(2).

Problem 2.1.9

a) Write x^{2} = |x|^{2} and the rest follows from the definition \sqrt{x^{2}} = \sqrt{|x|^{2}} = |x| (|x| is always non-negative).

b) As both sides of this inequality are non-negative integers we can square it to get

|x + y| ≤ |x| + |y| ⇔ (x + y)^{2} ≤ (|x| + |y|)^{2} ⇔ x^{2} + y^{2} + 2xy ≤ x^{2} + y^{2} + 2|x||y| ⇔

⇔ xy ≤ |xy|.

The last inequality is obvious (you can verify by checking four cases of different sign of x and y).

Problem 2.1.10

a) It is not 1-1 as g(1) = g(−1). Neither it is onto as g(x) ≥ 1.

b) It is not 1-1 as g(1) = g(−1) but now it is clearly onto as any natural number can be written as |x| + 1. Simply take all positive integers and the zero for x.

Problem 2.1.14 Write f(x) = (x + 7)^{2} − 100.

a) 1-1 but not onto. f(n) is strictly increasing for n ≥ −7, hence it must be 1-1. Also, there is no natural number n such that f(n) = −98. This would mean that f(n) = (n + 7)^{2} − 100 = −98 or (n + 7)^{2} = 2 which has no solutions in natural numbers.

b) not 1-1 and not onto. To see this take f(−17) = f(−3) = 0, and observe that there is no integer such that f(m) = −98. This would mean that f(m) = (m+7)^{2}−100 = −98 or (m + 7)^{2} = 2 which has no integer solutions.

c) not 1-1 but onto. Just draw this parabola in the XY -plane.