Exercise · ZW-PROOF-BY-CONTRADICTION-007 · ★★☆ Intermediate
To prove that the sum of a rational \(r\) and an irrational \(x\) is irrational, assume \(r+x=s\) with \(s\) rational. What contradiction arises?
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Explanation
The difference of two rationals is rational, so \(x=s-r\) would be rational, against the hypothesis that \(x\) is irrational.
A proof by contradiction of a statement \(p\) assumes \(p\) is false and deduces something absurd from it, so that \(p\) must be true.
\[\neg p\Rightarrow(r\wedge\neg r)\ \therefore\ p\]
More practice problems: proof by contradiction