SET & INDUCTIVE PROOFS ANYONE?

What we need to prove is C ⊆ B, i.e. that every element x ∈ C implies x ∈ B.

Proof by contradiction - assume there is some element z in C which is not in B, i.e. z ∈ C and z ∉ B

Assume

a) (A ∩ B) = (A ∩ C)

b) (A ∪ C) ⊆ (A ∪ B).

c) z ∈ C

d) z ∉ B.

i) If z ∈ C (c), z ∈ (A ∪ C) and therefore also z ∈ (A ∪ B) (b).

ii) But if z ∉ B (d) and z ∈ (A ∪ B) (i), then z ∈ A and z ∉ B, i.e. z ∈ A\B.

iii) If z ∈ A\B (ii) then z ∈ A, and then if z ∈ C (c) and A then we can say that z ∈ (A ∩ C) and hence z ∈ (A ∩ B) (a).

iv) But if z ∈ A\B then z ∉ (A ∩ B). Contradiction with (iii).

For the 2nd problem. First show that it is true for 2

2(2) - 1 = 3

2² - 1 = 3

Now assume that it is true for x. Is it true for x + 1?

i.e is 3 + 5 + 7 + ... (2(x+1)-1) = (x+1)² - 1 ?

So we have 3 + 5 + 7 + ... (2x-1) + (2(x+1) - 1)

we know, by our assumption that the first part

3 + 5 + 7 + ...(2x-1) is equal to x² - 1

so the expression becomes

x² - 1 + (2(x+1) - 1 which expands into

x² - 1 + 2x + 2 - 1

or, selective combining terms

x² +2x + 1 - 1

since x² +2x + 1 = (x+1)²

we can change

x² +2x + 1 - 1

into

(x+1)² - 1

and we are done. Since assuming it is true for x allows us to

show that it is true for x + 1, showing that it was true for 2, implies

that it is true for 3. Then if it is true for 3, it must be true for 4, etc.