proof-that-2-is-an-irrational-number-

Question Number 1865 by Denbang last updated on 18/Oct/15

p r o o f t h a t \sqrt{2} i s a n i r r a t i o n a l n u m b e r

Answered by 123456 last updated on 18/Oct/15

suppuse by absurf that \sqrt{2} \in Q, then

\exists (p, q) \in Z, q \neq 0 such that \sqrt{2} = \frac{p}{q}, (p, q) = 1

then

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

then p^{2} \equiv 0 (\mod 2) \Leftrightarrow p \equiv 0 (\mod 2)

wich imply that

p = 2 k, k \in Z

then

p^{2} = {(2 k)}^{2} = 4 k^{2} = 2 q^{2} \Leftrightarrow q^{2} = 2 k^{2}

q^{2} \equiv 0 (\mod 2) \Leftrightarrow q \equiv 0 (\mod 2)

so p \equiv 0 (\mod 2) and q \equiv 0 (\mod)

but this imply (p, q) \neq 1, absurd

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