Prove that √2 is an irrational number.
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Step-by-step explanation:
Let us assume on the contrary that
2
is a rational number. Then, there exist positive integers a and b such that
2
=
b
a
where, a and b, are co-prime i.e. their HCF is 1
⇒(
2
)
2
=(
b
a
)
2
⇒2=
b
2
a
2
⇒2b
2
=a
2
⇒2∣a
2
[∵2∣2b
2
and 2b
2
=a
2
]
⇒2∣a...(i)
⇒a=2c for some integer c
⇒a
2
=4c
2
⇒2b
2
=4c
2
[∵2b
2
=a
2
]
⇒b
2
=2c
2
⇒2∣b
2
[∵2∣2c
2
]
⇒2∣b...(ii)
From (i) and (ii), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means that our supposition is wrong.
Hence,
2
is an irrational number.
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