HOW TO PROVE THE GIVEN NUMBER IS IRRATIONAL

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

2b²  =  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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