Question

1
(1) 2
$1 \div \sqrt{2}$
prove as irrational
​

Answer 1

Answer:

To prove 1/√2 is irrational

To prove 1/√2 is irrationalLet us assume that √2 is irrational 

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2p

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sides

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem 

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer)

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 1

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² 

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² by theorem p is also divisible by 2

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² by theorem p is also divisible by 2But p and q are coprime

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² by theorem p is also divisible by 2But p and q are coprimeThis is a contradiction which has arisen due to our wrong assumption

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² by theorem p is also divisible by 2But p and q are coprimeThis is a contradiction which has arisen due to our wrong assumption∴1/√2 is irrational

To prove 1/√2 is irrationalLet us assume that √2 is irrational 1/√2 = p/q (where p and q are co prime)q/p = √2q     = √2psquaring both sidesq²   = 2p²                                                  .....................(1)By theorem q is divisible by 2∴ q = 2c ( where c is an integer) putting the value of q in equitation 12p² = q² = 2c² =4c²p² =4c² /2 = 2c²p²/2 = c² by theorem p is also divisible by 2But p and q are coprimeThis is a contradiction which has arisen due to our wrong assumption∴1/√2 is irrational