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 ⇒ ab ∈ ℑ

i. True

write a = m1n1 & b = m2n2
m1, n1 ∈ ℤ & n1 ≠ 0
m2, n2 ∈ ℤ & n2 ≠ 0

3a + 5b = 3m1n1 + 5m2n2 = 3m1n2 + 5m2n1n1n2

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 22

Case 1 : 22 ∈ ℚ
So we have a counter example.

Case 2 : 22 ∈ ℑ
Let a = 22 and b = 2
a, b ∈ ℑ
But ab = (22)2 = 22 . 2 = 22 = 2 ∈ ℚ

∴ 2. is False

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