Download Homework #3 - Han

MATH 2001 Homework
Han-Bom Moon
Homework 3 Solution
Section 1.5.
• Do not abbreviate your answer. Write everything in full sentences.
• Write your answer neatly. If I couldn’t understand it, you’ll get 0 point.
• You may discuss with your classmates. But do not copy directly.
1. Prove that the sum of any two rational numbers is rational.
a
c
Let x, y be two rational numbers. Then x = and y = for some a, b, c, d ∈ Z
b
d
and b, d 6= 0. Then
a c
ad + bc
x+y = + =
b d
bd
and bd 6= 0. Because ad + bc, bd are integers, x + y is also a rational number.
2. Let a, b, and c be three integers such that a|b and a|c. Prove that for any integers
x and y, a|(xb + yc).
Let a, b, c ∈ Z. Suppose that a|b and a|c. Because a|b, b = ak1 for some k1 ∈ Z.
Similarly, from a|c, c = ak2 for some k2 ∈ Z. Then for every x, y ∈ Z,
xb + yc = xak1 + yak2 = a(xk1 + yk2 ).
Therefore a|(xb + yc).
3. Let a, b be two positive real numbers. Show that
√
2ab
ab ≥
.
a+b
(The left side is called the geometric mean and the right side is called the harmonic
mean of a and b.)
Let a, b be two positive real numbers. Then (a − b)2 ≥ 0.
a2 − 2ab + b2 ≥ 0
(a + b)2 = a2 + 2ab + b2 ≥ 4ab
Because a + b is a positive number,
p
√
√
a + b = (a + b)2 ≥ 4ab = 2 ab.
√
√
√
ab(a + b) ≥ 2 ab · ab = 2ab
√
2ab
ab ≥
a+b
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MATH 2001 Homework
Han-Bom Moon
√
2ab
if and only if a = b.
a+b
Let a, b be two positive real numbers. Suppose that a = b. Then
4. Prove that for two positive real numbers a and b,
√
√
ab =
Conversely, suppose that
and
√
a2 = a =
ab =
2a2
2ab
=
.
2a
a+b
2ab
. Then
a+b
√
(a + b) ab = 2ab,
ab =
√
ab
(a + b) = 2 √ = 2 ab.
ab
√
(a + b)2 = (2 ab)2 = 4ab
a2 + 2ab + b2 = 4ab
(a − b)2 = a2 − 2ab + b2 = 0
Therefore a − b = 0, so a = b.
5. By using contraposition, show that if n2 is odd, then n is odd.
Let n ∈ Z. Suppose that n is even. Then n = 2k for some k ∈ Z. n2 = (2k)2 =
4k 2 = 2 · 2k 2 . Therefore n2 is even.
6. Let x and y be two real numbers. Prove that if x is rational and y is irrational,
then x + y is irrational.
Let x be a rational number and y be an irrational number. Suppose, on the cona
trary, that x + y is a rational number. Then x + y = for some a, b ∈ Z. Because
b
c
x is rational, x = for some c, d ∈ Z. Thus
d
y=
a c
ad − bc
a
−x= − =
.
b
b d
bd
So y is rational. It contradicts to the fact that y is irrational. Therefore x + y is an
irrational number.
7. Let’s assume the following statement:
Any integer n can be written exactly one of following three forms: 3k, 3k + 1,
and 3k + 2 for some integer k.
(We will discuss its generalization later.)
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MATH 2001 Homework
Han-Bom Moon
(a) By using above statement, show that for a integer n, if n2 is a multiple of 3,
then n is a multiple of 3.
Suppose, on the contrary, that n is not a multiple of 3. Then n = 3k + 1 or
n = 3k + 2 for some k ∈ Z. If n = 3k + 1, then
n2 = (3k + 1)2 = 9k 2 + 6k + 1 = 3(3k 2 + 2k) + 1.
If n = 3k 2 + 2, then
n2 = (3k + 2)2 = 9k 2 + 12k + 4 = 3(3k 2 + 4k + 1) + 1.
Therefore in any case, n2 is not a multiple of 3.
(b) Show that the following statement is not true: If n2 is a multiple of 4, then n
is a multiple of 4.
Consider n = 2. Then n2 = 4 = 4 · 1, so n2 is a multiple of 4. However, n is
not a multiple of 4.
8. Let a, b be two integers. Consider the following statement:
If a|b and b|a, then a = b.
Is it true? Prove it if it is true, and give a counterexample if it is false.
Let a = 3 and b = −3. Then a = 3 = (−1) · (−3) = (−1) · b, so b|a. Similarly,
because b = −3 = (−1) · 3 = (−1) · a, a|b. But 3 6= −3. Therefore the statement is
false.
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