prove that √7 is irrational
Answers
Step-by-step explanation:
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.
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Answer:
How do I prove by contradiction that √7 is irrational?
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.