Exercise · ZW-PROOF-BY-CONTRADICTION-009 · ★★★ Advanced
To prove that \(a^2-b^2=2\) has no integer solutions, assume a solution. Then \((a-b)(a+b)=2\); which argument gives the contradiction ?
Choose one option.
A \(a-b\) and \(a+b\) have different parity, so the product is even B \(a-b\) and \(a+b\) have the same parity, so the product is odd or a multiple of 4, never 2 C \(a\) and \(b\) must both be negative D \(a+b\) must be greater than \(a-b\) Show answer B. \(a-b\) and \(a+b\) have the same parity, so the product is odd or a multiple of 4, never 2← Back to the question
Explanation \(a-b\) and \(a+b\) differ by \(2b\), an even number, so they have the same parity. If both are odd the product is odd; if both are even it is a multiple of 4. It is never 2.
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