Math, asked by chhemanthkumar2345, 7 months ago

show that √7 is irrational number by the method of contradiction​

Answers

Answered by Lovelyfriend
2

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that7=ba

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that7=ba⟹a=b7

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that7=ba⟹a=b7Squaring on both sides, we get

Let us assume that 7 is rational. Then, there exist co-prime positive integers a and b such that7=ba⟹a=b7Squaring on both sides, we geta2=7b2

Therefore, a2 is divisible by 7 and hence, a is also divisible by7

Therefore, a2 is divisible by 7 and hence, a is also divisible by7so, we can write a=7p, for some integer p.

Therefore, a2 is divisible by 7 and hence, a is also divisible by7so, we can write a=7p, for some integer p.Substituting for a, we get 49p2=7b2⟹b2=7p2.

Therefore, a2 is divisible by 7 and hence, a is also divisible by7so, we can write a=7p, for some integer p.Substituting for a, we get 49p2=7b2⟹b2=7p2.This means, b2 is also divisible by 7 and so, bis also divisible by 7.

Therefore, a2 is divisible by 7 and hence, a is also divisible by7so, we can write a=7p, for some integer p.Substituting for a, we get 49p2=7b2⟹b2=7p2.This means, b2 is also divisible by 7 and so, bis also divisible by 7.Therefore, a and b have at least one common factor, i.e., 7.

Answered by vishakhabothra003
2

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.

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