Prove that 3 + 2 \sqrt{} 5 is irrational.

Foundation

Mathematics Foundation

Surds

Properties of Real Number

Irrational Number

Question

Prove that

3 + 2 \sqrt 5

is irrational.

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Let us assume

3 + 2 \sqrt{5}

+ is rational.

So we can write this number as

3 + 2 \sqrt{5} = \frac{a}{b}

---- (1)

Here a and b are two co-prime number and b is not equal to zero.
Simplify the equation (1) subtract 3 both sides, we get

2 \sqrt{5} = \frac{a}{b} - 3

2 \sqrt{5} = \frac{a - 3 b}{b}

Now divide by 2 we get

\sqrt{5} = \frac{a - 3 b}{2 b}

Here a and b are integer so

\frac{a - 3 b}{2 b}

is a rational number, so

\sqrt{5}

should be a rational number.

But

\sqrt{5}

is a irrational number, so it is contradict.

Therefore,

3 + 2 \sqrt{5}

is irrational number.