Geometry for elementary school/A proof of irrationality

In mathematics, a rational number is a real number that can be written as the ratio of two integers, i.e., it is of the form

a/b where a and b are integers and b is not zero. An irrational number is a number that cannot be written as a ratio of two integers, i.e., it is not of the form

a/b .

History of the theory of irrational numbers

The discovery of irrational numbers is usually attributed to Pythagoras, more specifically to the Pythagorean Hippasus of Metapontum, who produced a proof of the irrationality of the $\sqrt{2}$. The story goes that Hippasus discovered irrational numbers when trying to represent the square root of 2 as a fraction (proof below). However, Pythagoras believed in the absoluteness of numbers, and could not accept the existence of irrational numbers. He could not disprove their existence through logic, but his beliefs would not accept the existence of irrational numbers and so he sentenced Hippasus to death by drowning. As you see, mathematics might be dangerous.

Irrationality of the square root of 2

One proof of the irrationality of the square root of 2 is the following proof by contradiction. The proposition is proved by assuming the negation and showing that that leads to a contradiction, which means that the proposition must be true.

The term coprime is used in the proof. Two integers are coprimes of non of them divides the other.

1. Assume that $\sqrt{2}$ is a rational number. This would mean that there exist integers a and b such that a / b = $\sqrt{2}$.
2. Then $\sqrt{2}$ can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)² = 2.
3. It follows that a² / b² = 2 and a² = 2 b².
4. Therefore a² is even because it is equal to 2 b² which is obviously even.
5. It follows that a must be even. (Odd numbers have odd squares and even numbers have even squares.)
6. Because a is even, there exists a k that fulfills: a = 2k.
7. We insert the last equation of (3) in (6): 2b² = (2k)² is equivalent to 2b² = 4k² is equivalent to b² = 2k².
8. Because 2k² is even it follows that b² is also even which means that b is even because only even numbers have even squares.
9. By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).

Since we have found a contradiction the assumption (1) that $\sqrt{2}$ is a rational number must be false. The opposite is proven. $\sqrt{2}$ is irrational.