MATH 220 Homework 2 Solutions - Texas A&M University

MATH 220 Homework 2 Solutions

Mandatory Problems

Exercise 2.1.1

Let a, b, and c be integers. Prove that for all integers m and n, if a | b and a | c, then a | (bm + cn). Proof. Let m and n be integers. Assume a | b and a | c. Then there exist integers j and k such that b = aj and c = ak. (We must show that there exists an integer l such that bm + cn = al.) Observe that

bm + cn = ajm + akn = a(jm + kn). Put l = jm + kn. Then l is an integer such that bm + cn = al, and therefore a | (bm + cn).

Exercise 2.1.4

For every integer n, n2 + n + 5 is odd. Proof. Let n be an integer. We consider two cases.

Case 1: n is even. If n is even, there is some integer k such that n = 2k. Then we have

n2 + n + 5 = 4k2 + 2k + 4 + 1 = 2(2k2 + k + 2) + 1. Put l = 2k2 + k + 1. Then l is an integer such that n2 + n + 5 = 2l + 1, and therefore n2 + n + 5 is odd.

Case 2: n is odd. If n is odd, then there is some integer k such that n = 2k + 1. Then we have

n2 + n + 5 = (4k2 + 4k + 1) + (2k + 1) + 5 = 2(2k2 + 3k + 3) + 1. Put l = 2k2 + 3k + 3. Then l is an integer such that n2 + n + 5 = 2l + 1, and therefore n2 + n + 5 is odd.

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Exercise 2.1.11

Prove

that

for

all

real

numbers

x

and

y,

if

x

<

y,

then

x

<

x+y 2

<

y.

Proof.

Let

x

and

y

be

real

numbers

such that

x < y.

Then 2x = x + x < x + y, and thus

x<

x+y 2

.

Similarly,

x+y

<

y

+y

=

2y,

and

thus

x+y 2

<

y.

Combining

our

results,

we

obtain

x+y

x<

0. Observe that 0 (x - 1)2 = x2 - 2x + 1, so

2x x2 + 1 and thus

2 = 1 ? 2x 1 ? (x2 + 1) = x + 1

x

x

x

as desired.

Exercise 2.1.17

Prove

that

for

all

x, y

R+,

xy

x+y 2

,

with

equality

if

and

only

if

x

=

y.

Proof. Let x, y R+ be given. Observe that 0 (x - y)2 = x2 - 2xy + y2, and so 4xy

x2

+ 2xy + y2

= (x + y)2.

Thus

2xy

x + y,

and

therefore

xy

x+y 2

.

Assuming x = y, we obtain

xy

=

x2

=

x

=

2x

=

x

+

y .

2

2

Conversely,

assume

xy

=

x+y 2

.

Then

4xy

=

(x + y)2

=

x2

+ 2xy

+ y2,

and

rearranging

we

obtain

0 = x2 - 2xy + y2 = (x - y)2, so 0 = x - y and therefore x = y.

Exercise 2.1.21

Let a, b, c R with a = 0. Prove that the equation ax + b = c has a unique solution in R.

2

Proof.

Put

x0

=

c-b a

.

Then

x0

R

and

ax0 + b

= c,

so

the

equation

ax + b = c

has

a

solution.

If

now x1 is also a solution to the equation ax + b = c, then

0 = c - c = (ax0 + b) - (ax1 + b) = a(x0 - x1).

So x0 -x1 = 0, and thus x0 = x1. Therefore x0 is the unique solution to the equation ax+b = c.

Exercise 2.2.1

Let n be an integer. If n2 is even, then n is even.

Proof. Assume n is not even. Then n is odd, hence there is some k N such that n = 2k + 1. Thus n2 = (2k + 1)2 = 2(2k2 + 2k) + 1, and thus n2 is odd. Therefore, by contraposition, if n is even, then n2 is even.

Exercise 2.2.3

Prove that there are no integers m and n such that 8m + 26n = 1.

Proof. Assume, towards a contradiction, that there exist integers m and n such that 8m + 26n = 1. Then 2(4m + 13n) = 1, so 1 is even. But 1 is odd, a contradiction. Thus there are no integers m and n such that 8m + 26n = 1.

Exercise 2.2.10

Prove that for all integers m and n, if mn is even, then m is even or n is even. Proof. Assume that mn is even, and that m is not even. Then m is odd. Thus there exist integers j and k such that mn = 2j and m = 2k + 1. Then

2j = mn = 2kn + n, so n = 2(j - kn), where j - kn Z, and therefore n is even.

Exercise 2.2.13

Let x R. If for all > 0, |x| < , then x = 0.

Proof.

Assume, towards a contradiction, that x = 0.

Then

x 2

= 0, so

x 2

> 0.

Letting =

x 2

,

we obtain |x| < =

x 2

=

|x| 2

,

and

rearranging

yields

2

<

1.

This

is

a

contradiction,

and

therefore

x = 0.

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