prove-by-contradiction-that-2-is-irrational-

Question Number 78333 by Lontum Hans last updated on 16/Jan/20

prove by contradiction that \sqrt{2} is irrational .

Answered by MJS last updated on 16/Jan/20

\sqrt{2} \in Q \land \sqrt{2} > 0 \Rightarrow \sqrt{2} = \frac{p}{q}

p, q \in N \land \gcd (p, q) = 1 \Leftrightarrow p ∤ q \land q ∤ p

{(\sqrt{2} = \frac{p}{q})}^{2}

2 = \frac{p}{q} \Rightarrow p = 2 q \Rightarrow q ∣ p but q ∤ p \Rightarrow \sqrt{2} \notin Q

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