Irrational Numbers

In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers. Informally, this means that an irrational number cannot be represented as a simple fraction. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals. As a consequence of Cantor's proof that the real numbers are uncountable(and the rationals countable) it follows that almost all real numbers are irrational.

Note:

• ∃ a (There is...)
• ∃ a₁, a₂, a₃ ... (There are...)

Theorem

∀a ∈ ℝ, a . 0 = 0

Proof

Because 0 is the additive identity, 0 + 0 = 0
Take any a ∈ ℝ

a . (0 + 0) = a . 0
a . 0 + a . 0 = a . 0 [Distributive Law]

∃ a . 0 ∈ ℝ s.t. -a . 0 + a . 0 = 0

Because a . 0 should have an additive inverse.

-a . 0 + (a . 0 + a . 0) = -a . 0 + a . 0 = 0
(-a . 0 + a . 0) + a . 0 = 0 [Associative Law]
0 + a . 0 = 0

a . 0 = 0 (Because 0 is the additive identity)

Examples

Prove or disprove the following claims.

1. a, b ∈ ℚ ⇒ 3a + 5b ∈ ℚ
2. a, b ∈ ℑ ⇒ a^b ∈ ℑ

i. True

write a = ^m₁ ⁄ _n1 & b = ^m₂ ⁄ _n2
m₁, n₁ ∈ ℤ & n₁ ≠ 0
m₂, n₂ ∈ ℤ & n₂ ≠ 0

3a + 5b = ^3m₁ ⁄ _n1 + ^5m₂ ⁄ _n2 = ^{3m₁n₂ + 5m₂n₁}⁄_n1n2

n₁, n₂, m₁, m₂ ∈ ℤ ⇒ 3m₁n₂, 5m₂n₁, n₁n₂ ∈ ℤ
and n₁ ≠ 0 & n₂ ≠ 0 ⇒ n₁n₂ ≠ 0

∴ 3a + 5b ∈ ℚ

ii. False.

Here is a counter example.
√2 ∈ ℑ & Consider √2^√2

Case 1 : √2^√2 ∈ ℚ
So we have a counter example.

Case 2 : √2^√2 ∈ ℑ
Let a = √2^√2 and b = √2
a, b ∈ ℑ
But a^b = (√2^√2)^√2 = √2^{√2 . √2} = √2² = 2 ∈ ℚ

∴ 2. is False