Math, asked by srikarachary, 8 months ago

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

Answers

Answered by sumanroy8213
1

Answer:

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

√2+√5

Answered by Anonymous
16

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]

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