PROVING SQUARE ROOT OF 3 IS IRRATIONAL

Let us assume to the contrary that √3 is rational.

Since √3 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

√3 = 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.

√3 = (a ÷ c)/(b ÷ c)

√3 = 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√3 = x

Squaring both sides,

(y√3)² = x²

3y² = x² ----(1)

Clearly x² is a multiple of 3.

Since x² is a multiple of 3, x also should be a multiple of 3.

So, x = 3k, for some integer k.

Substitute x = 3k in (1).

3y² = (3k)²

3y² = 9k²

Divide both sides by 3.

y² = 3k²

Clearly y² is a multiple of 3.

Since y² is a multiple of 3, y also should be a multiple of 3.

So, y = 3r, for some integer r.

Since both 'x' and 'y' are multiples of 3, it is clear that both 'x' and 'y' have the common factor 3.

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 √3 is rational.

So, we conclude that √3 is irrational.

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