Question 143565
A rational number is a number that can be expressed as the quotient or ratio of two integers.  The term rational has nothing whatever to do with sanity; its root is the word 'ratio.'  Irrational numbers cannot be expressed as the quotient of two integers.  For example, there are no two integers p and q such that {{{sqrt(2)=p/q}}}.  See following (borrowed from Wikipedia) for a proof:


One proof of the number's irrationality is the following proof by <i>infinite descent</i>. It is also a <i>reductio ad absurdum|proof by contradiction</i>, which means the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, which means that the proposition must be true.

1. Assume that {{{sqrt(2)}}} is a rational number, meaning that there exists an integer a and an integer b such that a / b = {{{sqrt(2)}}} .
2. Then {{{sqrt(2)}}}  can be written as an <i>irreducible fraction</i> a / b such that a and b are <i>coprime</i> integers and (a / b)<sup>2</sup> = 2.
3. It follows that a<sup>2</sup> / b<sup>2</sup> = 2 and a<sup>2</sup> = 2 b<sup>2</sup>. (<i>(a / b)<sup>n</sup> = a<sup>n</sup> / b<sup>n</sup></i>)
4. Therefore a<sup>2</sup> is even because it is equal to 2 b<sup>2</sup>. (2 b<sup>2</sup> is necessarily even because its divisible by 2—that is, (2 b<sup>2</sup>)/2 = b<sup>2</sup> — and numbers divisible by two are even by definition.)
5. It follows that a must be even as (squares of odd integers are also odd, referring to b) or (only even numbers have even squares, referring to a).
6. Because a is even, there exists an integer k that fulfills: a = 2k.
7. Substituting 2k from (6) for a in the second equation of (3): 2b<sup>2</sup> = (2k)<sup>2</sup> is equivalent to 2b<sup>2</sup> = 4k<sup>2</sup> is equivalent to b<sup>2</sup> = 2k<sup>2</sup>.
8. Because 2k<sup>2</sup> is divisible by two and therefore even, and because 2k<sup>2</sup> = b<sup>2</sup>, it follows that b<sup>2</sup> is also even which means that b is even.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).
::<i>Q.E.D.</i>


{{{pi}}}, {{{e}}} (the base of the natural logarithms), and anything like {{{sqrt(x)}}} where {{{x}}} is <i> not </i> a perfect square, are all irrational, but this is by no means an exhaustive list.  There are an infinite number of irrational numbers, just as there are an infinite number of rational numbers.  In fact, there are an infinite number of irrational numbers between any two rational numbers, and an infinite number of rational numbers between any two irrational numbers.