prove that √2 is irrational number
Answers
ANSWER
Let us assume on the contrary that 2is 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=4
a=2
Answer:
Let √2 be a rational number
Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0
On squaring both sides, we get
p²= 2q² ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p² [since, 2q²=p²]
⇒ 2 is a factor of p
Let p =2 m for all m ( where m is a positive integer)
Squaring both sides, we get
p²= 4 m² ...(2)
From (1) and (2), we get
2q² = 4m² ⇒ q²= 2m²
Clearly, 2 is a factor of 2m²
⇒ 2 is a factor of q² [since, q² = 2m²]
⇒ 2 is a factor of q
Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1
Therefore, Our supposition is wrong
Hence √2 is not a rational number i.e., irrational number.