Question

10. Prove that √7 is an irrational number,​

Answer 1

Answer:

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

Answer 2

Step-by-step explanation:

Exactly √7 is an irrational number.

We know that in a rational p and q are integers.

Now √7= √7÷1

So, we see that √7 is not an integer and 1 is the only integer.

Hence, √7 is an irrational number .