Prove that √7 is an irrational number.
Answers
let us assume that √7 be rational.
then it must in the form of p / q.
As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.
√7 = p / q
√7 x q = p
squaring on both sides
7q² = p² ------1.
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p²= 49c²
subsitute p² in eqn(1) we get
7q² = 49 c²
q² = 7c²
q is divisble by 7
thus q and p have a common factor 7.
there is a contradiction to our assumption
as our assumsion p & q are co prime but it has a common factor.
so that √7 is an irrational.
Happy Learning..!!
Answer:
Let us assume that √7 be rational.
then it must in the form of p / q [q ≠ 0] [p and q are co-prime]
√7 = p / q
=> √7 x q = p
squaring on both sides
=> 7q²= p² ------ (1)
p² is divisible by 7
p is divisible by 7
p = 7c [c is a positive integer] [squaring on both sides ]
p² = 49 c² --------- (2)
Subsitute p2 in equ (1) we get
7q²= 49 c²
q² = 7c²
=> q is divisible by 7
The q and p have a common factor 7.
It is a contradiction as our assumsion p & q are co prime but it has a common factor.
So that √7 is an irrational.[tex][/tex]