Prove that √3+√3 is
is an irrational
number
Answers
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∣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.
Answer:
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.
Step-by-step explanation:
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