prove that √3 is irrational
Answers
Step-by-step explanation:
If possible , let
3
be a rational number and its simplest form be
b
a
then, a and b are integers having no common factor
other than 1 and b
=0.
Now,
3
=
b
a
⟹3=
b
2
a
2
(On squaring both sides )
or, 3b
2
=a
2
.......(i)
⟹3 divides a
2
(∵3 divides 3b
2
)
⟹3 divides a
Let a=3c for some integer c
Putting a=3c in (i), we get
or, 3b
2
=9c
2
⟹b
2
=3c
2
⟹3 divides b
2
(∵3 divides 3c
2
)
⟹3 divides a
Thus 3 is a common factor of a and b
This contradicts the fact that a and b have no common factor other than 1.
The contradiction arises by assuming
3
is a rational.
Hence,
3
is irrational.
Answer:root 3 is 1.72..
it is a non recurring and non terminating number which indicates its not a rational number hence its an irrational number .
Step-by-step explanation: