√7 isa)an integerb)an irrational numberc)a rational numberd)none...
Lets assume that √7 is rational number. ie √7=p/q.
suppose p/q have common factor then
we divide by the common factor to get √7 = a/b were a and b are co-prime number.
that is a and b have no common factor.
√7 =a/b co- prime number
√7= a/b
a=√7b
squaring
a^2=7b^2 ....1
a� is divisible by 7
a=7c
substituting values in 1
(7c)^2=7b^2
49c^2=7b^2
7c^2=b^2
b^2=7c^2
b^2 is divisible by 7
that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.
√7 is irrational.
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√7 isa)an integerb)an irrational numberc)a rational numberd)none...
√7 isa)an integerb)an irrational numberc)a rational numberd)none...
Let us assume that root 7 is a rational number,
or,√7=p/q--(1).( where p and q are coprime integers and q is not equal to zero),
squaring both side in equation 1.
7=p²/q²,
7q²=p²--(2),
from( 2 ).7 is a factor of p square it means it is also divisible by p,
put p=7min (2).
7q²=49m²,
q²=7m²--(3).
from (3) 7 is a factor of q square it means it is also divisible by q,
and from two and three we can say that P and Q have a common factor 7 hence it contradict our assumption that p and q are coprime integers,
hence √7 is irrational