Question

write the negation of statement √7 is rational​

Answer 1

Answer:

root7 is not an irrational number : False . Negation of the statement is equal to the opposite of that statement ...... √7 is an irrational number is a correct statement. √7 IS AN RATIONAL NUMBER .

Answer 2

Answer:

The statement $\sqrt{7}$is a rational number is not right.

Step-by-step explanation:

Let$\sqrt{p}$is a rational number and

$\sqrt{p} = \frac{a}{b}$

$p = \frac{ {a}^{2} }{ {b}^{2} }$

${a}^{2} = p \times {b}^{2}$

Therefore p divides ${a}^{2}$

But when a prime number divides the product of two numbers, it must divide atleast one of them.

Here ${a}^{2} = a \times a$

p divides a

Let $a = pk......(1)$

${p}^{2} {k}^{2} =p {b}^{2}$

So${b}^{2} = p {k}^{2}$

p divides ${b}^{2}$

But ${b}^{2} = b \times b$

Therefore p divides b

Thus, a and b have atleast one common multiple p But it arises the contradiction to our assumption that a and b are coprime.

Thus, our assumption is wrong and

$\sqrt{p}$ is irrational number.

So we can say $\sqrt{7}$

is a irrational number where 7 is a prime number.