Question

Prove that square root of 7 is an irrational number.​

Answer 1

Answer:

ANSWER

Let us assume that

7

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

7

=

b

a

⟹a=b

7

Squaring on both sides, we get

a

2

=7b

2

Therefore, a

2

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

so, we can write a=7p, for some integer p.

Substituting for a, we get 49p

2

=7b

2

⟹b

2

=7p

2

.

This means, b

2

is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

Hence,

7

is irrational.

Answer 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.