Exercise · ZW-PROOF-BY-CONTRADICTION-006 · ★★☆ Intermediate
To prove «if \(p\) then \(q\)» by contradiction, assume:
Choose one option.
Show answer
- C. \(p\) and \(\neg q\)
Explanation
The negation of \(p\Rightarrow q\) is \(p\wedge\neg q\). We assume that and look for an absurdity.
Review the concept: Proof by contradiction
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
- ZW-PROOF-BY-CONTRADICTION-001 To prove p by contradiction, the first step is:
- ZW-PROOF-BY-CONTRADICTION-002 We want to prove «there are infinitely many primes» by contradiction. What is assumed?
- ZW-PROOF-BY-CONTRADICTION-003 If the complete list of primes were 2,3,5, what is N=2·3·5+1?
- ZW-PROOF-BY-CONTRADICTION-004 Assume √2=a/b with a/b in lowest terms and obtain a^2=2b^2. What follows immediately?
- ZW-PROOF-BY-CONTRADICTION-005 In the proof that √2 is irrational, where is the contradiction?
- ZW-PROOF-BY-CONTRADICTION-007 To prove that the sum of a rational r and an irrational x is irrational, assume r+x=s with s ration…
- ZW-PROOF-BY-CONTRADICTION-008 Which of these is a valid contradiction at the end of a proof?
- ZW-PROOF-BY-CONTRADICTION-009 To prove that a^2-b^2=2 has no integer solutions, assume a solution. Then (a-b)(a+b)=2; which argum…
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