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
a²=7b² ..1
a² is divisible by 7
a=7c
substituting values in 1
(7c)²=7b²
49c²=7b²
7c²=b²
b²=7c²
b² 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.
Answer:
Given √7
To prove: √7 is an irrational number.
Proof:
Let us assume that √7 is a rational number.
So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
√7 = p/q
Here p and q are coprime numbers and q ≠ 0
Solving
√7 = p/q
On squaring both the side we get,
=> 7 = (p/q)2
=> 7q2 = p2……………………………..(1)
p2/7 = q2
So 7 divides p and p and p and q are multiple of 7.
⇒ p = 7m
⇒ p² = 49m² ………………………………..(2)
From equations (1) and (2), we get,
7q² = 49m²
⇒ q² = 7m²
⇒ q² is a multiple of 7
⇒ q is a multiple of 7
Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√7 is an irrational number.
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
a²=7b² ..1
a² is divisible by 7
a=7c
substituting values in 1
(7c)²=7b²
49c²=7b²
7c²=b²
b²=7c²
b² 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