Irrational number

An irrational number is any real number which is not rational. More systematically, it is the set of numbers which cannot be represented as the quotient of two integers $p$ and $q$, where $q \ne 0$, thus having a non-repeating, non-terminating decimal representation.

Properties

If $a$ is an irrational number, $b$ and $b'$ are rational numbers, $b'\neq 0$, $x$ is a real number and $n$ is a positive integer, then:

1. $a+b$, $a-b$, $b-a$ and $-a$ are irrational.
2. $a\times b'$, ${\frac {a}{b'}}$ and ${\frac {b'}{a}}$ are irrational.
3. $a \times b$ is rational if and only if $b=0$. The same thing also true for $\frac{b}{a}$.
4. $|a|$ is irrational.
5. $|x|$ is irrational if and only if $x$ is irrational.
6. $\sqrt[n]{a}$ is irrational.
7. If $b$ cannot be expressed as another rational number to the power of n (or in other word $\log _{n}(b)$ is irrational) then ${\sqrt[{n}]{b}}$ is irrational.

Examples

Common examples of irrational numbers are roots of numbers. Miscellaneous examples include numbers that are also transcendental such as pi and e.

Name	Representation	Value
Square root of 2	$\sqrt{2}$	1.41421356
Square root of 3	$\sqrt{3}$	1.73205081
Square root of 5	$\sqrt5$	2.23606798
Pi	$\pi$	3.14159265
Euler's number	e	2.71828183
The Golden ratio	$\varphi$	1.61803399

Proofs

$\sqrt{2}$ is an irrational number:

Theorem. Square root of 2 is irrational

Prerequisites: Rules of exponents Proof. Proof by contradiction: Assume $\sqrt{2}$ is rational. It can then be represented as an irreducible fraction of two integers, p and q. Therefore, $\frac{p}{q} = \sqrt{2}$ $p = q\sqrt{2}$ $p^2 = 2q^2$ Since q is an integer, then 2q2 is even, and so is p2. Since p2 is even, then p must be even. If p is even, there exists an integer a such that p = 2a. Substituting, $(2a)^2 = 2q^2$ $4a^2 = 2q^2$ $2a^2 = q^2$ Therefore, q2 must be even, and it follows that q must be even. $\frac{p}{q}$ can then be reduced (by 2) which contradicts the earlier statement (that it is irreducible). Therefore, $\sqrt{2}$ is irrational.

$\sqrt{2}+\sqrt{3}$ is an irrational number:

Theorem. The sum of square root of 2 and square root of 3 is irrational

Prerequisites: Property #1, #2, #6, and #7 of irrational numbers Proof. We have $({\sqrt {2}}+{\sqrt {3}})^{2}=5+2{\sqrt {6}}$. Since 5 and 2 are rational and $\sqrt{6}$ is irrational (by property #7), by property #1 and #2, we have $5+2{\sqrt {6}}=({\sqrt {2}}+{\sqrt {3}})^{2}$ is irrational. Therefore, by property #6, ${\sqrt {({\sqrt {2}}+{\sqrt {3}})^{2}}}=|{\sqrt {2}}+{\sqrt {3}}|={\sqrt {2}}+{\sqrt {3}}$ is irrational. Therefore, $\sqrt{2}+\sqrt{3}$ is irrational.