PROVING IRRATIONAL NUMBERS BY CONTRADICTION

Solved Examples

Example 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.

Example 2 :

5 -  √3 is irrational.

Solution :

Let 5 - √3 be a rational number.

Then it may be in the form a/b

5 - √3  =  a/b

Taking squares on both sides, we get

5 - (a/b)  =  √3

(5b - a)/b  =  √3

a, b and 5 are rational numbers. Then the simplified value of (5b - a)/b must be rational. But it is clear that √3 is irrational.

So, it contradicts our assumption. Hence 5 - √3 is irrational.

Example 3 :

3 + 2√5 is irrational.

Solution :

Let 3 + 2√5 be a rational number.

Then it may be in the form a/b

3 + 2√5  =  a/b

Taking squares on both sides, we get

3 - (a/b)  =  2√5

(3b - a)/b  =  2√5

(3b - a)/2b  =  √5

a, b, 3 and 2 are rational numbers. Then the simplified value of (3b - a)/2b must be rational. But it is clear that √5 is irrational.

So, it contradicts our assumption. Hence 3 + 2√5 is irrational.

Example 4 :

√2 + √5 is irrational.

Solution :

Let √2 + √5 be a rational number.

Then it may be in the form a/b

√2 + √5  =  a/b

Taking squares on both sides, we get

2 + 5 + 2√10  =  a²/b²

7 + 2√10  =  a²/b²

2√10  =  (a²/b²) - 7

2√10  =  (a²- 7b²)/b²

√10  =  (a²- 7b²)/2b²

a, b, 7 and 2 are rational numbers. Then the simplified value of (a²- 7b²)/2b² must be rational. But it is clear that √10 is irrational.

So, it contradicts our assumption. Hence √2 + √5 is irrational.

Example 5 :

3√2

Solution :

Let 3√2 be a rational number.

Then it may be in the form a/b

3√2  =  a/b

√2  =  a/3b

a, b and 3 are rational numbers. Then the simplified value of a/3b must be rational. But it is clear that √2 is irrational.

So, it contradicts our assumption. Hence 3√2 is irrational.

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