Download ASSIGNMENT 03: TRANSITION TO ADVANCED MATHEMATICS

ASSIGNMENT 03: TRANSITION TO ADVANCED
MATHEMATICS
Problem 1. (10 points each) Provide either a proof or a counterexample for
each of these statements.
(a) For all positive integers x, x2 + x + 41 is a prime.
FALSE.
If x = 41, then 412 + 41 + 41 = 41(41 + 1 + 1) = 41 ∗ 43, not prime.
(b) For integers a, b, c, if a divides bc, then a divides b or a divides c.
FALSE.
Let a = 6, b = 3, c = 4.
Then a = 6 divides bc = 3 · 4 = 12, but does not divide b or c.
(c) For all positive real numbers x, x2 − x ≥ 0.
FALSE.
Let x = 1/2, then (1/2)2 − 1/2 = −1/4 < 0.
(d) For all positive real numbers x, 2x ≥ x + 1.
FALSE.
√
Let x = 1/2, then 2 < 1/2 + 1 = 1.5.
(e) For any two irrational numbers x and y, xy is also irrational.
FALSE. √
√
√ √
√
Let x = 3 and y = 12, then xy = 3 · 12 = 36 = 6 ∈ Q.
Problem 2. (10 points each) Let x and y be a real numbers. Prove the
following:
(a) if x is irrational, then x − 5 is irrational.
Proof. We will show by contradiction. Let x be an irrational number,
and assume that x − 5 is rational. Since x − 5 ∈ Q, then x − 5 = pq ,
for some p, q ∈ Z, q 6= 0. Then
x=5+
p
5q + p
=
,
q
q
and 5q + p ∈ Z. Thus, x can be written as a fraction of two integers,
i.e. x is rational, a contradiction. Therefore, x−5 must be irrational.
(b) if x and y are rational, then xy is rational.
1
2
ASSIGNMENT 03: TRANSITION TO ADVANCED MATHEMATICS
Proof. Let x, y ∈ Q, then x = pq , y = rs , for some p, q, r, s ∈ Z with
q 6= 0 and s 6= 0. Then
p r
pr
xy = · = ,
q s
qs
where pr, qs ∈ Z with qs 6= 0 (as the product of two integers is an
integer). This shows that xy is rational, too.
Problem 3. (10 points) Prove that if n is an integer and 3n − 1 is odd, then
4n + 8 is divisible by 8.
Proof. Let n ∈ Z and 3n − 1 be odd. Then 3n − 1 = 2k + 1 for some k ∈ Z,
which implies 3n = 2k + 2 = 2(k + 1), which is even. Since 3 is odd, then n
must be even, therefore n = 2m, m ∈ Z. Hence,
4n + 8 = 4 · 2m + 8 = 8m + 8 = 8(m + 1).
As a result, if 3n − 1 is an odd integer, then 4n + 8 is divisible by 8.
Problem 4. (7 points) Exercise 2.1: Problem 4(a, b, d, f, g, h, j).
(a) N ⊆ Q − True: Every natural number is rational.
(b) Q ⊆ Z − False: example 23 is a rational, but not integer.
(d) ( 12 , 25 ) ⊆ Q − False: An interval contains irrationals, too (for exam√
√
ple, 3 is not rational and 12 < 3 < 52 ).
(f) R ⊆ Q − False: Real numbers include irrationals, too.
(g) [7, 10] ⊆ R − True.
(h) [2, 5] ⊆ {2, 3, 4, 5} − False: 2.5 is not an element of {2, 3, 4, 5}.
(j) (6, 9] ⊆ [6, 10) − True.
Problem 5. (8 points) Exercise 2.1: Problem 5(a, b, d, f, g, h, j, l).
(a) ∅ ∈ {∅, {∅}} − True.
(b) ∅ ⊆ {∅, {∅}} − True: The empty set is subset of every set.
(d) {∅} ⊆ {∅, {∅}} − True.
(f) {{∅}} ⊆ {∅, {∅}} − True: Since {∅} ∈ {∅, {∅}}.
(g) For every set A, ∅ ∈ A − False: ∅ ⊆ A.
(h) For every set A, {∅} ⊆ A − False.
(j) {1, 2} ∈ {{1, 2, 3}, {1, 3}, 1, 2} − False.
(l) {{4}} ⊆ {1, 2, 3, {4}} − True.
Problem 6. (5 points both) Exercise 2.1: Problems 11 and 12.
(2.1.11) We know that Y = {−3, −2, −1, 0, 1, 2, 3}. The set X is defined by
{x : x ∈ Z and |x| ≤ 3}. Being an integer and |x| ≤ 3, i.e. − 3 ≤ x ≤
3, implies that X = {−3, −2, −1, 0, 1, 2, 3}. Hence, X = Y .
(2.1.12) We know that Y = {1, 2, 3, 4, 5}. The set X is defined by {x√: x ∈
2
2
N and
√ x ≤ 30}. Being a natural number and x ≤ 30, i.e. − 30 ≤
x ≤ 30, implies that X = {1, 2, 3, 4, 5}. Hence, X = Y .