MATH3030 Homework
Hong Yue,
Georgia College, Department of Mathematics,
Milledgeville, Georgia, 31061, USA, hong.yue@gcsu.edu
October 10, 2013
1
Week 5 Homework
Exercise 2.4.9, (3), 2.4.10, (1), (4), Exercise 2.5.3. (1), (3), (4), 2.5.4.
Exercise 2.4.9. Let x be a real number. Define the absolute value of x,
denoted |x|, by
x, if 0 ≤ x
|x| =
−x, if x < 0.
Let a and b be real numbers. Prove the following statements.
(3) |a − b| = |b − a|.
Proof: We prove it in for a ≥ b and a < b, respectively.
Case 1 a ≥ b: |a − b| = a − b and |b − a| = −(b − a) = a = b, therefore,
lhs=rhs.
Case 2 a < b: |a − b| = −(a − b) = b − a and |b − a| = b − a, therefore,
lhs=rhs.
Exercise 2.4.10. Let x and y be real numbers. Let x _ y and x ^ y be
defined by
x, if x ≥ y
y, if x ≥ y
x_y=
and x ^ y =
y, if x ≤ y,
x, if x ≤ y,
(Observe that x _ y is simply the maximum of x and y, and x ^ y is the
minimum, though our notation is more convenient for the present exercise
than writing max x, y and similarly for the minimum.)
1
Let a, b and c be real numbers. Prove the following statements. The
definition of absolute value is given in Exercise 2.4.9.
(1) (a _ b) + (a ^ b) = a + b.
(4) (a _ b) − (a ^ b) = |a − b|.
Proof: We prove (1) for a ≥ b and a < b, respectively.
Case 1 a ≥ b: lhs= (a _ b) + (a ^ b) = (a _ b) + (a ^ b) = a + b = rhs
Case 2 a ≤ b: lhs= (a _ b) + (a ^ b) = (a _ b) + (a ^ b) = b + a = rhs
The proof for (4) is similar:
Case 1 a ≥ b:
lhs= (a _ b) − (a ^ b) = (a _ b) − (a ^ b) = a − b = |a = b| = rhs
Case 2 a ≤ b:
lhs= (a _ b) − (a ^ b) = (a _ b) + (a ^ b) = b − a = |a − b| = rhs
Exercise 2.5.3. Prove or give a counterexample to each of the following
statements.
(1) For each non-negative number s, there exists a non-negative number
t such that s ≥ t.
Proof: Given s ≥ 0, let t = 2s , then t ≥ 0 and s ≥ t.
(3) For each non-negative number t, there exists a non-negative number
s such that s ≥ t.
Proof: ∀t ≥ 0, let s = 2t, then s ≥ 0 and s ≥ t.
(4) There exists a non-negative number s such that for all non-negative
numbers t, the inequality s ≥ t holds.
This statement is false. In order to show it is false we give a counterexample to show its negation is true:
For any non-negative number s, let t = s + 1 (a counterexample), then
t ≥ 0 and s < t.
Exercise 2.5.4. Prove or give a counterexample to each of the following
statements.
(1) For each integer a, there exists an integer b such that a|b.
Proof: ∀a, let b = a, then a|b.
(2) There exists an integer b such that for all integers a, the relation a|b
holds.
Proof: Let b = 0, then ∀a ∈ Z, a · 0 = 0, that is, a|0.
2
(3) For each integer b, there exists an integer a such that a|b.
Proof: ∀b, let a = b or a = 1, then a|b.
(4) There exists an integer a such that for all integers b, the relation a|b
holds.
Proof: Let a = 1, then a|b for all integers b.
3