Examine whether underoot 2 is rational or irrational numbers
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I think root 2 Is a irrational no because it is non repeating and non terminating no.. in short we can express it in point .. but rational no requires p÷q form and q Should not be 0 ..
hope it helps you but I am not sure about this answer ...
hope it helps you but I am not sure about this answer ...
Answered by
24
heya,
if possible √2 be rational and simplest form is p/q
then ,a and b are integer it have no common factor other than 1 and bis not equal to 0.
then ,
√2=a/b =)2 =a^2/b^2( squaring on both side)
2b^2=a^2
2 divides a^2
and so ,(2 divides 2b^2 )
=) 2 divides a ,but 2 is prime and divides by b^2 =) 2 divides b .
let a=2c for some integer c .
putting a =2c in (1) we get .
2b^2=4c^2 =)b^2=2c^2
=)2 divides b^2 ( so,2 divides 2c^2 )
=)2 divides b
so, 2 is prime and 2 divides b^2 =)2 divides b
thus two is common factor of a and b .
but this is contradiction fact that a and b have no common. common factor other than 1 .
the contradiction arises by assuming that √2 is rational .
hence √2 is irrational..
hope it help you
@rajukumar.
if possible √2 be rational and simplest form is p/q
then ,a and b are integer it have no common factor other than 1 and bis not equal to 0.
then ,
√2=a/b =)2 =a^2/b^2( squaring on both side)
2b^2=a^2
2 divides a^2
and so ,(2 divides 2b^2 )
=) 2 divides a ,but 2 is prime and divides by b^2 =) 2 divides b .
let a=2c for some integer c .
putting a =2c in (1) we get .
2b^2=4c^2 =)b^2=2c^2
=)2 divides b^2 ( so,2 divides 2c^2 )
=)2 divides b
so, 2 is prime and 2 divides b^2 =)2 divides b
thus two is common factor of a and b .
but this is contradiction fact that a and b have no common. common factor other than 1 .
the contradiction arises by assuming that √2 is rational .
hence √2 is irrational..
hope it help you
@rajukumar.
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