prove that √2 is irrational.
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Answer:
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
=ba where, a and b, are co-prime i.e. their HCF is 1
⇒(2
)2=(ba)2
⇒2=b2a2
⇒2b2=a2
⇒2∣a2[∵2∣2b2 and 2b2=a2]
⇒2∣a...(i)
⇒a=2c for some integer c
⇒a2=4c2
⇒2b2=4c2[∵2b2=a2]
⇒b2=2c2
⇒2∣b2[∵2∣2c2]
⇒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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