### 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...)
- ∃ a
_{1}, a_{2}, a_{3}... (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)

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.

- a, b ∈ ℚ ⇒ 3a + 5b ∈ ℚ
- a, b ∈ ℑ ⇒ a
^{b}∈ ℑ

i. True

write a =

m

m

3a + 5b =

n

and n

∴ 3a + 5b ∈ ℚ

write a =

^{m1}⁄_{n1}& b =^{m2}⁄_{n2}m

_{1}, n_{1}∈ ℤ & n_{1}≠ 0m

_{2}, n_{2}∈ ℤ & n_{2}≠ 03a + 5b =

^{3m1}⁄_{n1}+^{5m2}⁄_{n2}=^{3m1n2 + 5m2n1}⁄_{n1n2}n

_{1}, n_{2}, m_{1}, m_{2}∈ ℤ ⇒ 3m_{1}n_{2}, 5m_{2}n_{1}, n_{1}n_{2}∈ ℤand n

_{1}≠ 0 & n_{2}≠ 0 ⇒ n_{1}n_{2}≠ 0∴ 3a + 5b ∈ ℚ

ii. False.

Here is a counter example.

√2 ∈ ℑ & Consider √2

Case 1 : √2

So we have a counter example.

Case 2 : √2

Let a = √2

a, b ∈ ℑ

But a

∴ 2. is False

Here is a counter example.

√2 ∈ ℑ & Consider √2

^{√2}Case 1 : √2

^{√2}∈ ℚSo we have a counter example.

Case 2 : √2

^{√2}∈ ℑLet a = √2

^{√2}and b = √2a, b ∈ ℑ

But a

^{b}= (√2^{√2})^{√2}= √2^{√2}^{ . }^{√2}= √2^{2}= 2 ∈ ℚ∴ 2. is False

super

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