Question

PROVE THAT √2 IS AN IRRATIONAL number by the method of contradicon​

Answer 1

Answer:

PROVE THAT √2 IS AN IRRATIONAL number by the method of contradicon

√2+√5

Answer 2

let √2 be a rational number that is it can be expressed in the form of p/q where , p and q are integeres , q ≠0 and p and q are co primes .

$\sqrt{2} = \frac{p}{q}$

$sq. \: both \: sides$

$2 = \frac{ {p}^{2} }{ {q}^{2} }$

${p}^{2} = 2 {q}^{2}$

therefore , p is a multiple of 2

let p = 2m

${(2m)}^{2} = 2 {q}^{2}$

$4 {m}^{2} = 2 {q}^{2}$

${q}^{2} = \frac{4 {m}^{2} }{2}$

${q}^{2} = 2 {m}^{2}$

therefore , q is also a multiple of 2

but , p and q were co primes

therefore , our contradiction is wrong ..

$\sf {\boxed {\pink { [tex] \sqrt{2} \: is \: an \: irrational \: number$}}} [/tex]