prove that root 7 isirrational by the method of contradiction
Answers
Answered by
3
Hey dear
Here is ur answer
Let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q2= p2 ------> (1)
p2 is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p2 = 49 c2 --------- > (2)
subsitute p2 in equ (1) we get
7q2 = 49 c2
q2 = 7c2
=> q is divisble by 7
thus q and p have a common factor 7.
there is a contradiction
as our assumsion p & q are co prime but it has a common factor.
so that √7 is an irrational.
Hope it helps u
Have a great day
Here is ur answer
Let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q2= p2 ------> (1)
p2 is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p2 = 49 c2 --------- > (2)
subsitute p2 in equ (1) we get
7q2 = 49 c2
q2 = 7c2
=> q is divisble by 7
thus q and p have a common factor 7.
there is a contradiction
as our assumsion p & q are co prime but it has a common factor.
so that √7 is an irrational.
Hope it helps u
Have a great day
krusheed:
re send the answer please
Similar questions