prove that root 3 is irrtional
Answers
Step-by-step explanation:
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.
I hope it help you
Here is your Answer:
Let us assume √3 to be a rational number.
Then, there exist positive integers a and b such that √3=a/b
where, a and b, are co-prime i.e. their HCF is 1
Now,
√3=a/b
=> (√3)² = a²/b²
=> 3 = a²/b²
=> 3b² = a²
⇒3 divides a²
[∵3 divides 3b²]
⇒3 divides a...(i)
⇒Let a=3c for some integer c
⇒a = 3c
=> a² = 9c²
=> 3b² = 9c²
[∵a² =3b² ]
⇒b²=3c²
⇒3 divides b²
[∵3 divides 3c²]
⇒3 divides b...(ii)
From (i) and (ii), it is observed 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 was wrong.
Hence,
√3 is an irrational number.
Hope It Helps!!