Question

Proof that 3 - \sqrt{8} is irrational.

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Answer 1

Step-by-step explanation:

Given :-

3-√8

To find :-

Prove that 3-√8 is an irrational number ?

Solution :-

Given that :

3-√8

It can be written as 3-√(2×2×2)

=> 3 - 2√2

Let us assume that

3 - 2√2 is a rational number

It must be in the form of p/q , where p and q are integers and q ≠ 0

Let 3-2√2 = a/b

Where a and b are co-primes

=> -2√2 = (a/b) -3

=> 2√2 = 3-(a/b)

=> 2√2 = (3b-a)/b

=> √2 = (3b-a)/(2×b)

=> √2 = (3b-a)/2b

=>√2 is in the form of p/q

=> √2 is a rational number

But We know that

=> √2 is not a rational number.

=> √2 is an irrational number

This contradicts to our assumption that (3-√8) is a rational number.

3-√8 is not a rational number.

3-√8 is an irrational number.

Hence, Proved.

Answer:-

3-√8 is an irrational number.

Note :-

3 is a rational number

√8 is an irrational number

The difference between the rational and irrational is also an irrational number.

3-√8 is an irrational number.

Used Method :-

• Method of Contradiction or Indirect method

Points to know :-

• The numbers are in the form of p/q ,where p and q are integers and q≠0 called rational numbers. They are denoted by Q.

• The numbers are not in the form of p/q ,where p and q are integers and q≠0 called irrational numbers. They are denoted by Q' or S.

• If q is the rational number and s is the irrational number then

• q+s is an irrational number.

• q-s is an irrational number.

• q×s is an irrational number.

• q/s is an irrational number.