Real Analysis Proofs

1 ♦ 0 = 0.a


0.a = 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 [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) [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;
aSup(A) and bSup(B)

a + bSup(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 + bSup(A) + Sup(B) - ε

Since a + bSup(A + B);
Sup(A + B)a + bSup(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)

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