prove that root2 is an irrational number
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Let assume that √2 is rational in form of a/b where a&b are coprime we get,
√2=a/b
squaring both side we get,
2= a^2/b^2
a^2 =2b^2 ------(i)
2 divides a^2
2 divides a (from theorem 2 is prime no. which divides a^2 so it divides a also)
a=2c where c is some integer
put value of a in (i)
(2c)^2 = 2b^2
4c^2 =2b^2
b^2 =2c^2
2 divides b^2
2 divides b also ( from theorem )
from i and ii 2 is factor of both so a and b is not a coprime no.
the above contradiction rises because of our wrong assumption hence √2 is irrational.
hope it helps u ....
√2=a/b
squaring both side we get,
2= a^2/b^2
a^2 =2b^2 ------(i)
2 divides a^2
2 divides a (from theorem 2 is prime no. which divides a^2 so it divides a also)
a=2c where c is some integer
put value of a in (i)
(2c)^2 = 2b^2
4c^2 =2b^2
b^2 =2c^2
2 divides b^2
2 divides b also ( from theorem )
from i and ii 2 is factor of both so a and b is not a coprime no.
the above contradiction rises because of our wrong assumption hence √2 is irrational.
hope it helps u ....
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