# HOW TO PROVE THE GIVEN NUMBER IS IRRATIONAL

A real number that is not rational is called an irrational number.

Theorem to Remember :

Let p be a prime number and a be a positive integer. If p divides a2, then p divides a.

Question 1 :

Prove that √2 is an irrational number.

Solution :

Let √2 be a rational number.

Then it may be in the form a/b

√2  =  a/b

Taking squares on both sides, we get

2  =  a2/b2

2b=  a2

a2 divides 2 (That is 2/a2)

Then a also divides 2.

Let a  =  2c

2b2  =  a2

By applying the value here, we get

2b2  =  (2c)2

2b2  =  4c2

b2  =  2c2

b2 divides 2 (That is 2/b2)

Then b also divides 2.

From this, we come to know that a and b have common divisor other than 1. It means our assumption is wrong. Hence √2 is irrational.

Question 2 :

Prove that √3 is an irrational number.

Solution :

Let √3 be a rational number.

Then it may be in the form a/b

√3  =  a/b

Taking squares on both sides, we get

3  =  a2/b2

3b2  =  a2

a2 divides 3 (That is 3/a2)

Then a also divides 3.

Let a  =  3c

3b2  =  a2

By applying the value here, we get

3b2  =  (3c)2

3b2  =  9c2

b2  =  3c2

b2 divides 3 (That is 3/b2)

Then b also divides 3.

From this, we come to know that a and b have common divisor other than 1. It means our assumption is wrong. Hence √3 is irrational.

Question 3 :

Prove that 3√2 is a irrational.

Solution :

Let us assume 3√2  as rational.

3√2  =  a/b

√2  =  a/3b

Since √2 is irrational

Since 3, a and b are integers a/3b be a irrational number. So it contradicts.

Hence 3√2 is irrational number.

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