# Square root of 2

Missing image
Square_root_of_2_triangle.png
The square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1.

The square root of 2, √2, also known as Pythagoras's constant, is the positive real number which, when multiplied by itself, gives the product 2. Its numerical value approximated to 65 decimal places Template:OEIS is:

1.41421 35623 73095 04880 16887 24209 69807 85696 71875 37694 80731 76679 73799

√2 was the first known irrational number. Geometrically, √2 is the length of a diagonal across a square with sides of one unit of length; this follows from Pythagoras' theorem.

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## History

The first approximation of this number was given in ancient Indian mathematical texts, the Sulbasutras (800 B.C. to 200 B.C.) as follows: Increase a unit length by its third and this third by its own fourth less the thirty-fourth part of that fourth. That is,

[itex]1 + \frac{1}{3} + \frac{1}{3 \cdot 4} - \frac{1}{3 \cdot4 \cdot 34} = \frac{577}{408} \approx 1.414215686[itex].

The discovery of the irrational numbers is usually attributed to Pythagoras or one of his followers, who produced a (most likely geometrical) proof of the irrationality of √2.

## Proof of irrationality

One proof of the number's irrationality is the following proof by contradiction. The proposition is proved by proving that the opposite of the proposition is true and showing that the proposition is false, which means that the proposition must be true.

1. Assume that √2 is a rational number, meaning that there exists an integer a and b so that a / b = √2.
2. Then √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 √2 is a rational number must be false. The opposite is proven. √2 is irrational.

This proof can be generalized to show that any root of any natural number is either a natural number or irrational.

## A different proof

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Irrationality_of_sqrt2.png

Another reductio ad absurdum showing that √2 is irrational is less well-known. It is an example of proof by infinite descent. It makes use of a classic ruler-and-compass construction, proving the theorem by a method similar to that employed by ancient Greek geometers.

Let ABC be a right isosceles triangle with hypotenuse length m and legs n. By the Pythagorean theorem, m/n = √2. Suppose m and n are integers. Let m:n be a ratio given in its lowest terms.

Draw the arcs BD and CE with centre A. Join DE. It follows that AB = AD, AC = AE and the ∠BAC and ∠DAE coincide. Therefore the triangles ABC and ADE are congruent.

Since ∠EBF is a right angle and ∠BEF is half a right angle, BEF is also a right isosceles triangle. Hence BF = m − n. By symmetry, DF = m − n, and FDC is also a right isosceles triangle. It also follows that FC = 2n − m.

Hence we have an even smaller right isosceles triangle, with hypotenuse length 2n − m and legs m − n. These values are integers even smaller than m and n and in the same ratio, contradicting the hypothesis that m:n is in lowest terms. Therefore m and n cannot be both integers, hence √2 is irrational.

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