Question

proof root 2 is an irrational no​

Answer 1

Answer:

insted of

$\sqrt{7} put \: \sqrt{2}$

Answer 2

Step-by-step explanation:

:)..Let assume that √2 is rational number....

We know that...the rational number in the form of p/q where p nit equal to 0...

√2=p /q

squaring both side..

(√2)^2=(p/q)^2

2=p^2/q^2

2q^2=p^2

2 divides p^2

2 divides p. {2 is prime and 2 divides p^2=)2 divides p}

let p=2rfor some integer r

putting p=2r in (1),we get..

2q^2=4r^2

q^2=2r^2

2 divides q^2

2divides q. {2 is prime and 2 divides q^2=)2 divides q}

.Thus ,2 is common factor of p and q ..

But this contradicts the fact ...

hence our assumption is wrong...that √2 is rational no.

Hence √2 is irrational number....