CBSE Class 10 Answered
is ?6 a irrational number
Asked by himanish gothwal | 16 Apr, 2013, 10:29: PM
Expert Answer
Yes, we prove the statement by contradiction i.e. Assume that sqrt(6) is a rational number.
Then there exist positive integers p and q such that the hcf(p,q) = 1 and p/q = sqrt(6).
Square both sides: p^2 / q^2 = 6,
p^2 = 6q^2.
Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).
This is only possible if 2 is a factor of p.
Let p =2k. Then k is a positive integer as well.
Thus, 4k^2 = 6q^2,
2k^2 = 3q^2.
As 2 is a factor of the LHS, 2 is also a factor of the RHS.
But this is only possible if 2 is a factor of q.
=> hcf(p,q) >= 2. Contradiction!
Thus sqrt(6) is irrational.
Then there exist positive integers p and q such that the hcf(p,q) = 1 and p/q = sqrt(6).
Square both sides: p^2 / q^2 = 6,
p^2 = 6q^2.
Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).
This is only possible if 2 is a factor of p.
Let p =2k. Then k is a positive integer as well.
Thus, 4k^2 = 6q^2,
2k^2 = 3q^2.
As 2 is a factor of the LHS, 2 is also a factor of the RHS.
But this is only possible if 2 is a factor of q.
=> hcf(p,q) >= 2. Contradiction!
Thus sqrt(6) is irrational.
Answered by | 17 Apr, 2013, 06:02: AM
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