Hey all,

This is my first attempt at any sort of a proof in mathematics. I'm reading through the book 'understanding Analysis' by Abbott. Can you take a look and tell me what you think? Have I left out anything major as required by induction or contradiction?

Cheers.

**Chapter 1**

**Exercise 1.2.9. **

Show that the sequence (x1, x2, x3, . . . ) defined in Example 1.2.7 is bounded above by 2; that is, prove that Xn ≤ 2 for every n ∈ N.

**Example 1.2.7**

Let X1 = 1, and for each n ∈ N define

Xn+1 = (1/ 2)Xn + 1.

**Proof by Induction**

where n ∈ N

X1 = 1 <= 2

Xn+1 = (1/ 2)Xn + 1 <= 2

Xn/2 <= 1

Xn <= 2

**Proof by Contradiction**

where n ∈ N

Xn !<= 2

Xn > 2

Xn/2 > 1

Xn/2 + 1 > 2 = Xn+1

if n = 0

Xn+1 = 1 !> 2

Therefore Xn+1 <= 2

**Reasoning via analysis**

Let X1 = 1, and for each n ∈ N define

Xn+1 = (1/ 2)Xn + 1.

Thus Xn/2 is always (denominator -1) / denominator for all Xn

example n= 1 Xn/2 = ½. N = 2, Xn/2 = ¾, n = 3, Xn/2 = 7/8 etc...

Therefore (1/ 2)Xn + 1 is always something less then unity + unity which will always be less then 2.

**Exercise 1.2.10. **

Let y1 = 1, and for each n ∈ N define yn+1 = (3yn + 4)/4.

(a) Use induction to prove that the sequence satisfies yn < 4 for all n ∈ N.

**(a) Proof by induction**

y1 = 1 < 4, for each n ∈ N

yn < 4

3yn < 12

(3yn +4)/4 < 4 = yn+1

Therefore

yn = (4yn+1 -4 )/ 3 < 4

yn+1 < 4

or

yn < yn+1

yn < (3yn +4)/4

yn < 4

**Exercise 1.2.11**

If a set A contains n elements, prove that the number of

different subsets of A is equal to 2^n. (Keep in mind that the empty set ∅ is

considered to be a subset of every set.)

**Proof via induction**

for each n ∈ N

If n >=0, A>=1

A = 2^n

2^n >= 1

using log rules

n >= 0

n +1 >= 1

using logs

2^(n+1) > 2

Therefore

2^(n+1) = An+1

An+1 > An > A