17. Prove that √3 is an irrational number.
Answers
Answered by
0
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∣a
2
[∵3∣3b
2
]
⇒3∣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∣b
2
[∵3∣3c
2
]
⇒3∣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.
Attachments:
Similar questions