Question

verify be the method of contradiction p : √7 is irrational .

Answer 1

p:√7=p/√7
so it not rational no(write p/q than p does not =0)
     so
than it is irrational no
i hope u understand

Answer 2

Answer:

We have to prove: √7 is an irrational number.

Let us assume that √7 is a rational number,

By the property of rational number,

√ 7 = p/q, where p and q are distinct integers and q≠0,

⇒ √7 q = p

By squaring both sides,

We get,

$7q^2=p^2$

Thus, $p^2$ is the multiple of 7,

⇒ p is the multiple of 7,

Let p = 7 k

Where k is any number,

⇒ $7q^2=(7k)^2$

⇒ $7q^2=49k^2$

⇒ $q^2=7k^2$

⇒ 7 is the multiple of $q^2$,

⇒ 7 is the multiple of q,

Therefore, p and q are not distinct numbers,

Which is a contradiction,

Hence, our assumption is wrong,

√7 is not a rational number,

⇒ √7 is an irrational number.

Hence, proved.