# PROVING SQUARE ROOT OF 2 IS IRRATIONAL

To the contrary, let us assume that √2 is rational.

Since √2 is rational, it has to be in the form of 'a/b' where 'a' and 'b' are integers such that b ≠ 0.

or

we can assume that

√2 = a/b

Let 'a' and 'b' and have a common factor other than 1 which is 'c'.

Now, divide both numerator and denominator of the fraction a/b by the common factor 'c' and assume the result as x/y.

√2 = (a ÷ c)/(b ÷ c)

√2 = x/y

where 'x' and 'y' are relative prime. That is, 'x' and 'y' don't have a common factor other than 1.

So,

y√2 = x

Squaring both sides,

(y√2)2 = x2

2y2 = x2 ----(1)

Clearly x2 is a multiple of 2.

Since xis a multiple of 2, x also should be a multiple of 2.

For example, 64 is a perfect square and it is a multiple of 2.

64 = 2(32)

82 = 2(32)

When 64 is a multiple of 2, its square root value 8 is also is a multiple of 2.

8 = 2(4)

So, x = 2k, for some integer k.

Substitute x = 2k in (1).

2y2 = (2k)2

2y2 = 4k2

Divide both sides by 2.

y2 = 2k2

Clearly yis a multiple of 2.

Since yis a multiple of 2, y also should be a multiple of 2.

So, y = 2r, for some integer r.

Since both 'x' and 'y' are multiples of 2, it is clear that both 'x' and 'y' have the common factor 2.

But this contradicts the fact that 'x' and 'y' have no common factor other than 1.

This contradiction has arisen because of our incorrect assumption that √2 is rational.

So, we conclude that √2 is irrational.

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