### Real Analysis Proofs

### 1 ♦ 0 = 0.a

0.a = 0.a

0.a = (0 + 0).a [

0.a = 0.a + 0.a [

-0.a + 0.a = -0.a + 0.a + 0.a [

0 = (-0.a + 0.a) + 0.a [

0 = 0 + 0.a

0.a = (0 + 0).a [

*Since 0 is the additive identity*]0.a = 0.a + 0.a [

*Distributive Axiom*]-0.a + 0.a = -0.a + 0.a + 0.a [

*Adding (-0.a) to both sides, the additive inverse of 0.a*]0 = (-0.a + 0.a) + 0.a [

*Associative Axiom*]0 = 0 + 0.a

**0**=**0.a**[*Since 0 is the additive identity*] ♠### 2 ♦ -a = -1(a)

(-a) = (-a)

(-a) = (-a) + 0 [

(-a) = (-a) + 0.a [

(-a) = (-a) + (1 + (-1))a [

(-a) = (-a) + (1.a + (-1)a) [

(-a) = (-a) + (a + (-1)a) [

(-a) = ((-a) + a) + (-1)a [

(-a) = 0 + (-1)a [

(-a) = (-a) + 0 [

*Additive Identity*](-a) = (-a) + 0.a [

*Since 0 = 0.a (By above theorem (1))*](-a) = (-a) + (1 + (-1))a [

*Additive Inverse of 1*](-a) = (-a) + (1.a + (-1)a) [

*Distributive Axiom*](-a) = (-a) + (a + (-1)a) [

*Multiplicative Identity*](-a) = ((-a) + a) + (-1)a [

*Associative Axiom*](-a) = 0 + (-1)a [

*Additive Identity*]**(-a)**=**(-1)a**[*Additive Identity*] ♠### 3 ♦ -(a + b) = (-a) + (-b)

-(a + b) = -(a + b)

–(a + b) = (-1)(a + b) [

(-1)(a + b) = (-1)(a) + (-1)(b) [

–(a + b) = (-1)(a + b) [

*By above theorem (2)*](-1)(a + b) = (-1)(a) + (-1)(b) [

*Distributive Axiom*]**-(a + b)**=**(-a) + (-b)**[*By above theorem (2)*]### ♦ Sup(A + B) = Sup(A) + Sup(B)

∀ a ∈ A and ∀ b ∈ B;

a ≤

∴ a + b ≤

∴

By definition of

Let ε > 0;

∃a ∈ A and ∃b ∈ B such that a + ε/2 ≥

a + b + ε ≥

a + b ≥

Since a + b ≤

(1) and (2) ⇒

a ≤

**Sup(A)**and b ≤**Sup(B)**∴ a + b ≤

**Sup(A) + Sup(B)**∴

**Sup(A) + Sup(B)**is an upper bound of [A + B]By definition of

**Sup(A + B)**,*. ∴***Sup(A + B)**is the least upper bound of [A + B]**Sup(A + B)**≤**Sup(A) + Sup(B)**-- (1)Let ε > 0;

∃a ∈ A and ∃b ∈ B such that a + ε/2 ≥

**Sup(A)**and b + ε/2 ≥**Sup(B)**a + b + ε ≥

**Sup(A) + Sup(B)**a + b ≥

**Sup(A) + Sup(B)**- εSince a + b ≤

**Sup(A + B)**;**Sup(A + B)**≥ a + b ≥**Sup(A) + Sup(B)**- ε ∴**Sup(A + B)**≥**Sup(A) + Sup(B)**- ε ⇒**Sup(A + B)**≥**Sup(A) + Sup(B)**-- (2)(1) and (2) ⇒

**Sup(A + B) = Sup(A) + Sup(B)**♠
## Comments

## Post a Comment