prove that √3 is irrational
Answers
Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N. We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3. Hence the root of 3 is an irrational number.
ANSWER
Let us assume on the contrary that
3
is a rational number.
Then, there exist positive integers a and b such that
3
=
b
a
where, a and b, are co-prime i.e. their HCF is 1
Now,
3
=
b
a
⇒3=
b
2
a
2
⇒3b
2
=a
2
⇒3 divides a
2
[∵3 divides 3b
2
]
⇒3 divides a...(i)
⇒a=3c for some integer c
⇒a
2
=9c
2
⇒3b
2
=9c
2
[∵a
2
=3b
2
]
⇒b
2
=3c
2
⇒3 divides b
2
[∵3 divides 3c
2
]
⇒3 divides b...(ii)
From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.
Hence,
3
is an irrational number.