Question

show that 6-2 √3 is an irrational number​

Answer 1

Step-by-step explanation:

let us assume that 6 - 2 √3 is is rational number

6 - 2 √3 = p/q ( where pand q are coprime integers and q are not equal to 0)

√3 = p/ 3q-5

now we prove that √ 3 is an irrational number

let us again assume that √3 is irrational number

√3 = p/q ( where pand q are coprime integers and q are not equal to 0)

√3q = p

squaring both sides

(√3q)*2 = p*2

= p*2

q*2 = p*2/3

3 divides p*2

3 is divide p*2

3 is factor of p*2 ___(1)

and now substituting the value of p = 3r in q*2 = (3r*2)/3

q*2 = 9r/3

q*2 = 3r

q*2/3 =r

3 divides q*2

3 is divide q*2

3 is factor of q*2 ___(2)

equation (1)and(2) ,3 is common factor of p and q

This contradicts our assumption.so , √3 is irrational no.

6 - 2 √3 is rational no so, √3 is rational number but we can prove that√3 is irrational no so, 6 - 2 √3 is irrational no. This contradicts our assumption 6 - 2 √3 is irrational no