X;y;z 2 R. Use the axioms of the real numbers to prove the following.

Math 242: Principles of Analysis Fall 2016

Homework 1 Part B solutions

1. Let x, y, z R. Use the axioms of the real numbers to prove the following.

(a) If x + y = x + z then y = z.

Solution. By Axiom 5(a), there is a number -x such that x + (-x) = 0. Adding -x to the equation x+y = x+z gives (-x)+(x+y) = (-x)+(x+z). By Axiom 2, this becomes ((-x)+x)+y = ((-x)+x)+z. By Axiom 1, (-x)+x = x+(-x) = 0, so we have 0 + y = 0 + z, which then becomes y + 0 = z + 0 by Axiom 1, and then y = z by Axiom 4(a).

(b) -(-x) = x.

Solution. By Axiom 5(a), the number -(-x) is the unique number such that -x + (-(-x)) = 0. But -x + x = x + (-x) b Axiom 1, and x + (-x) = 0 by Axiom 5(a), so -x + x = 0, so by uniqueness, x = -(-x).

(c) If x = 0, then (x-1)-1 = x. Solution. By Axiom 5(b), since x = 0, there is a unique number x-1 such that x ? x-1 = 1. If x-1 = 0, we would have x ? 0 = 1. But by Theorem 1.3.1(f) this would imply 0 = 1, which contradicts Axiom 4(b). Thus x-1 = 0, so by Axiom 5(b) there exists a number (x-1)-1 such that x-1 ? (x-1)-1 = 1. Multiplying both sides of this by x gives

x ? (x-1 ? (x-1)-1) = x ? 1 = x

by Axiom 4(b). On the other hand,

x ? (x-1 ? (x-1)-1) = (x ? (x-1)) ? (x-1)-1 = 1 ? (x-1)-1 = (x-1)-1 ? 1 = (x-1)-1

Axiom 2 Axiom 5(b) Axiom 1 Axiom 4(b).

Thus (x-1)-1 = x.

(d) -(x + y) = (-x) + (-y) Solution. By Axiom 5(a), -(x + y) is the unique number such that (x + y) + (-(x + y)) = 0. Thus, it suffices to show that (x + y) + ((-x) + (-y)) = 0. To

1

prove this, we write

(x + y) + ((-x) + (-y)) = x + (y + ((-x) + (-y))) = x + (y + ((-y) + (-x))) = x + ((y + (-y)) + (-x)) = x + (0 + (-x)) = x + ((-x) + 0) = x + (-x) =0

Axiom 2 Axiom 1 Axiom 2 Axiom 5(a) Axiom 1 Axiom 4(a) Axiom 5(a)

(e) -(x ? y) = (-x) ? y.

Solution. By Axiom 5(a), -(x?y) is the unique number such that x?y+(-(x?y)) = 0. Thus we just need to show that x ? y + (-x) ? y = 0. But

x ? y + (-x) ? y = y ? x + y ? (-x) = y ? (x + (-x)) =y?0 =0?y =0

Axiom 1 Axiom 3 Axiom 5(a) Axiom 1

since we proved in class that 0 ? y = 0 for any y R.

(f) If x ? y = 0 then either x = 0 or y = 0. Solution. Suppose x ? y = 0. If y = 0 then we are done. Otherwise y = 0, and Axiom 5(b) implies that there is a real number y-1 such that y ? y-1 = 1. Thus

x=x?1 = x ? (y ? y-1) = (x ? y) ? y-1 = 0 ? y-1

=0

Axiom 4(b)

Axiom 2 hypothesis

again since 0 ? x = 0 for any real number x.

(g) If x > 3 then x3 + 2x > 33.

Solution. First, since 2 > 0, Axioms 1 and 9 imply 2x = x ? 2 > 3 ? 2 = 6. Next, since 3 > 0, Axiom 7 implies x > 0. Thus, Axiom 9 implies x2 > 3 ? x. But Axiom 1 implies 3 ? x = x ? 3 and Axiom 9 implies x ? 3 > 3 ? 3 = 9. Hence Axiom 7 implies x2 > 9. Axioms 9 and 1 then imply x3 > 9 ? x = x ? 9 > 3 ? 9, so Axiom 7 implies x3 > 27. Finally, Axiom 8 implies x3 + 2x > 27 + 6 = 33.

(h) If 0 < x < 1, then x2 < x. Solution. Since x > 0, Axiom 9 implies x = 1 ? x > x ? x = x2.

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(i) |xy| = |x||y| Solution. If x 0 and y 0, then xy 0, so |xy| = xy = |x||y|. If x < 0 and y < 0, then 0 = x + (-x) < 0 + (-x) = -x, so -x > 0 and likewise -y > 0. Thus (-x)(-y) > 0 by Axiom 9. Thus x ? y = -((-x) ? y) = (-x) ? (-y) > 0, so |xy| = xy and |x||y| = (-x) ? (-y) = xy. If x 0 and y < 0, then |x| = x and |y| = -y. Axiom 9 implies x ? y < 0, so |xy| = -(xy) = x ? (-y) = |x||y|. The case x < 0 and y 0 is similar.

2.

Use

induction

to

show that

n k2

=

n(n + 1)(2n + 1)

for all

n N.

6

k=1

Solution.

Let

Sn

denote

the

statement:

n k2

=

n(n + 1)(2n + 1) .

6

k=1

In the base case n = 1, statement S1

1 takes the form k2 =

1(1 + 1)(2 + 1) .

6

Since

k=1

both sides of the equation equal 1, the statement S1 is true.

Now suppose that Sn is true for some integer n 1. We then need to prove that Sn+1

must

also

be

true.

Now

Sn+1

is

the

statement

n+1 k2

=

(n + 1)(n + 2)(2n + 3) .

6

If

we

k=1

consider the left hand side of this equation, and use the fact that Sn is true, we have

(

)

n+1

n

k2 =

k2 + (n + 1)2

k=1

k=1

= n(n + 1)(2n + 1) + (n + 1)2 6

n+1

=

(n(2n + 1) + 6(n + 1))

6

=

n

+

1

(2n2

+

7n

+

) 6

6

(n + 1)(n + 2)(2n + 3)

=

.

6

Therefore the statement Sn+1 is true. Hence the implication Sn = Sn+1 holds for

all n 1, so by the principle of induction Sn is true for all n 1.

3.

Use

induction

to

show that

n k3

( n(n + 1) )2 =

for all

n N.

2

k=1

Solution.

Let

Sn

denote

the

statement:

n k3

=

( n(n + 1) )2 .

2

k=1

In

the

base

case

n

=

1,

statement

S1

takes

the

form

1 k3

=

( 1(1 + 1) )2 .

2

Since

both

k=1

sides of the equation equal 1, the statement S1 is true.

3

Now Sn+1

suppose that Sn is true for must also be true. Now Sn+1

some is the

isnttaetgeemr ennt n+11.k3W=e(th(nen+w1i)s(hn 2

to +

2p)r)o2ve

.

that If we

k=1

consider the left hand side of this equation, and use the fact that Sn is true, we have

()

n+1

n

k3 =

k3 + (n + 1)3

k=1

=

( k=1 n(n

+

1) )2

+

(n

+

1)3

2

=

(n

+

1)2

(n2

+

4(n +

) 1)

4

= (n + 1)2 (n + 2)2

(

4 (n +

1)(n

+

2)

)2

=

.

2

Therefore the statement Sn+1 is true. Hence the implication Sn = Sn+1 holds for all n 1, so by the principle of induction Sn is true for all n 1.

n 1

n

4. Use induction to show that

=

for all n N.

k(k + 1) n + 1

k=1

1 1

1

Solution. The statement S1 reads

=

. This is clearly true since

k(k + 1) 1 + 1

k=1

both

sides

are

equal

to

1 2

.

Now suppose Sn is true for some n 1. Then

(

)

n+1 1

n 1

1

=

+

k(k + 1)

k(k + 1) (n + 1)(n + 2)

k=1

k=1

n

1

=

+

n + 1 (n + 1)(n + 2)

since Sn is true

n(n + 2)

1

=

+

(n + 1)(n + 2) (n + 1)(n + 2)

n2 + 2n + 1 =

(n + 1)(n + 2)

n+1

=

,

n+2

so Sn+1 is true. Therefore Sn = Sn+1 for all n 1, and the principle of induction implies Sn is true for all n 1.

5. Find an integer n0 such that n5 < 3n for all n n0. Use induction to prove your assertion.

4

Solution. Let Sn denote the statement n5 < 3n. By trial and error, we see that n5 3n if 1 n 10, but that 115 = 161051 < 177147 = 311. Thus Sn is false if n 10 but S11 is true. We claim that Sn is true for all n 11. The base case S11 is true, so now fix some n 11 and suppose Sn is true for this n. We want to show that Sn+1 is true, i.e. that (n + 1)5 < 3n+1. Since Sn is true we have

n5 < 3n,

so multiplying both sides by 3 gives

3n5 < 3 ? 3n = 3n+1.

If we can show that (n + 1)5 < 3n5, then by transitivity we would have (n + 1)5 < 3n+1

as desired. But

(n + 1)5 < 3n5

(n + 1)5 n5 < 3

( 1 )5 1 + < 3. n

Now

since

n

11

we

have

1+

1 n

1+

1 11

=

12 11

and

therefore

( 1 )5 ( 12 )5 248832

1+ <

=

< 3.

n

11

161051

Thus we have

(n + 1)5 < 3n5 < 3n+1

and therefore Sn+1 is true. Hence the implication Sn = Sn+1 is true for all n 11 so by the principle of induction Sn is true for all n 11.

6.

Let

x1 = 0

and

let xn+1

=6+

2 3

xn

for

n 1.

(a) Compute xn for n = 2, 3, 4, 5. Solution.

2

2

x2

=

6

+

3 x1

=

6

+

(0) 3

=

6

2

2

x3

=

6

+

3 x2

=

6

+

(6) 3

=

10

2

2

38

x4

=

6

+

3 x3

=

6

+

(10) 3

=

3

2

2

130

x5

=

6

+

3 x4

=

6

+

(38/3) 3

=

9

(b) Prove by induction that xn < 18 for all n N.

Solution. Let Sn be the statement xn < 18. Since x1 = 0 < 18, S1 is true. Now

suppose Sn is true for some n 1. Then xn < 18. Multiplying this inequality by

2 3

gives

2 3

xn

< 12.

Adding

6

to

both

sides

then

gives

6+

2 3

xn

<

18.

By

definition

xn+1

=

6+

2 3

xn,

so

xn+1

<

18,

and

thus

Sn+1

is

true.

Hence

Sn

=

Sn+1 for all

n 1, so by induction Sn is true for all n 1.

5

(c) Prove by induction that xn+1 > xn for all n N.

Solution. Let Sn be the statement xn+1 > xn. Since x2 = 6 > 0 = x1, S1 is

true. Now suppose Sn is true for some n 1. Then xn+1 > xn. Multiplying this

inequality

by

2 3

gives

2 3

xn+1

>

2 3

xn

.

Adding

6

to

both

sides

then

gives

6+

2 3

xn+1

>

6+

2 3

xn.

By

definition

xn+1

=

6+

2 3

xn

and

xn+2

=

6+

2 3

xn+1

,

so

xn+2

>

xn+1,

and

thus Sn+1 is true. Hence Sn = Sn+1 for all n 1, so by induction Sn is true

for all n 1.

7.

Let

A = {2 -

5 n+3

: n N}.

Find

lub(A) and glb(A), and prove

your

assertions.

Solution.

For

any

n

1,

we

have

n+3

4,

so

5 n+3

5 4

,

which

implies

-

5 n+3

-

5 4

,

and

thus

2-

5 n+3

3 4

.

Thus

3 4

is

a

lower

bound

for

A.

But

3 4

A

(take

n

=

1),

so

any

number

larger

than

3 4

is

not

a

lower

bound

for

A.

Thus

glb(A)

=

3 4

.

Next,

since

2-

5 n+3

<

2

for

all

n

N,

2

is

an

upper

bound

for

A.

Now

suppose

x

<

2.

Then 2 - x > 0. By the Archimedian Property, there exists some n N such that

n>

5 2-x

- 3.

Thus

n+3

>

5 2-x

,

and

since

2-x > 0

this

implies

(2 - x)(n + 3)

>

5.

Since

n+3

> 0,

this

then

implies

2-x

>

5 n+3

,

and

thus

2-

5 n+3

>

x.

But

2-

5 n+3

is

an element of A, so x is not an upper bound for A. Therefore lub(A) = 2.

8. Suppose A and B are nonempty subsets of R such that x y for every x A and every y B. Show that lub(A) glb(B).

Solution. There are a couple of ways to prove this.

First, given x A, since x y for every y B it follows that x is a lower bound for B and thus x glb B. Since this is true for every x A it follows that glb B is an upper bound for A and therefore glb B lub A.

Alternately, one could do a proof by contradiction. Suppose lub A > glb B. Then lub A is not a lower bound for B so there exists some y B such that y < lub A. Then this y cannot be an upper bound for A, so there must exist some x A such that x > y. This contradicts the initial assumption that x y for every x A and every y B.

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