Prove that is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

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

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let take that $3 + 2 \sqrt{5}$ is a rational number.

So we can write this number as

3 + 2 \sqrt{5}

\frac{a}{b}

Here a and b are two co prime number and b is not equal to 0

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}

irrational

number so it contradicts.

Hence,

3 + 2 \sqrt{5}

is a irrational number.