Prove that √2 is irrational also prove of √3, √5, √7

1.	Prove that √2 is irrational also prove of √3, √5, √7
Answer» Ans: √2 Let us assume that √2 is a rational number Take a and b are co prime numbers where b is not equal to 0 Such that √2 =a/b √2b= a take it as eqn 1 Squaring both side (√2b) 2 = a2 (2b) 2 = a2 take it as eqn two b2= a2/2 2 divides a2 Also 2 divides a ( if a prime number p divides a2 then divides a also) Let a = 2c Substitute in 2 (2b)2 =( 2c) 2 2b2 = 4c2 2b2 / 4= c2 b2 /2 = c2 2 divides b2 Also 2 divides b Therefore, 2 divides both a and b also a and b are co prime number. Our assumption that √2 is a rational number is wrong. √2 is a irrational number. 1.\xa0Let\xa0√2 be a rational number\xa0Therefore,\xa0√2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q\xa0≠ 0On squaring both sides, we get\xa0 p²= 2q² ...(1)Clearly, 2 is a factor of 2q²⇒ 2 is a factor of p² [since, 2q²=p²]⇒ 2 is a factor of p\xa0Let p =2 m for all m ( where m is a positive integer)Squaring both sides, we get\xa0 p²= 4 m² ...(2)From (1) and (2), we get\xa0 2q² = 4m² ⇒ q²= 2m²Clearly, 2 is a factor of 2m²⇒ 2 is a factor of q² [since, q² = 2m²]⇒ 2 is a factor of q\xa0Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1 Therefore, Our supposition is wrongHence\xa0√2 is not a rational number i.e., irrational number.

Discussion