1
Question
Prove that √2. is an irrational number by contradiction method
Prove that √2. is an irrational number by contradiction method
Open in App
Solution
Let √2 be a rational number then
√2 = p/q
squaring both the sides we get
2=p2/q2
(2p)2=q2 {equation 1}
this implies that q32 is divisible by 2 and then can also be said that q is divisible by 2
hence can be written as
q=2k where k is an integer
squaring both sides
q2 = (2k)2
from equation 1
(2k)2=(2p)2
and
p2 = 2k2
hence we can say 2 is the common factor in p and q and this is a contradiction to the fact that p and q are co prime numbers
hence √2 cannot be expressed as p/q
hence √2 is an irrational number.
√2 = p/q
squaring both the sides we get
2=p2/q2
(2p)2=q2 {equation 1}
this implies that q32 is divisible by 2 and then can also be said that q is divisible by 2
hence can be written as
q=2k where k is an integer
squaring both sides
q2 = (2k)2
from equation 1
(2k)2=(2p)2
and
p2 = 2k2
hence we can say 2 is the common factor in p and q and this is a contradiction to the fact that p and q are co prime numbers
hence √2 cannot be expressed as p/q
hence √2 is an irrational number.
Suggest Corrections
43
Similar questions