Proofs:Contraposition - Revision history

Lila at 16:49, 1 October 2021

2021-10-01T16:49:22Z

Lila: /* Resources */

2021-09-24T15:59:53Z

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Lila: /* Resources */

2021-09-24T15:58:41Z

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Lila at 15:55, 24 September 2021

2021-09-24T15:55:48Z

Lila: Created page with "Let $P$ and $Q$ be propositions such that $P \implies Q$ . Then, the contrapositive of the conditional statement " $P \implies Q$
2021-09-24T15:39:55Z

Created page with "Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </m..."

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Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </math>" (read as "if P, then Q" or "P implies Q") is "<math> \neg Q \implies \neg P </math (read as "if not Q, then not P" or "not Q implies not P">.

==Resources==

← Older revision		Revision as of 16:49, 1 October 2021
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−	Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </math>" (read as "if P, then Q" or "P implies Q") is "<math> \neg Q \implies \neg P </math> (read as "if not Q, then not P" or "not Q implies not P"). The contrapositive is logically equivalent to the original conditional; that is, a conditional and its contrapositive always have the same truth values. For example, the contrapositive of "if <math> x > 0 </math>, then <math> x </math> is positive" is "if <math> x </math> is NOT positive, then <math> x \leq 0 </math>". These two statements are logically equivalent, and are both true. Sometimes, proving the contrapositive of a conditional is easier than proving the conditional itself. For example: Let x be an integer. If <math> x^2 </math> is even, then x is even. While we can attempt to prove this conditional statement directly, it is easier to show that if x is not even, then <math> x^2> </math> is not even, given that an odd number times an odd number must be odd (that is, not even).	+	Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </math>" (read as "if P, then Q" or "P implies Q") is "<math> \neg Q \implies \neg P </math> (read as "if not Q, then not P" or "not Q implies not P"). The contrapositive is logically equivalent to the original conditional; that is, a conditional and its contrapositive always have the same truth values. For example, the contrapositive of "if <math> x > 0 </math>, then <math> x </math> is positive" is "if <math> x </math> is NOT positive, then <math> x \leq 0 </math>". These two statements are logically equivalent, and are both true. Sometimes, proving the contrapositive of a conditional is easier than proving the conditional itself.
		+
		+	For example: Let x be an integer. If <math> x^2 </math> is even, then x is even. While we can attempt to prove this conditional statement directly, it is easier to show that if x is not even, then <math> x^2> </math> is not even, given that an odd number times an odd number must be odd (that is, not even).

	==Resources==		==Resources==

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 ==Resources==
 * [https://math.dartmouth.edu/~m22x17/misc/LaLonde2012_proof_by_contrapositive.pdf Proof by Contrapositive], Dartmouth University
 * [http://zimmer.csufresno.edu/~larryc/proofs/proofs.contrapositive.html Proof by Contrapositive], Fresno State University

← Older revision		Revision as of 15:58, 24 September 2021
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	==Resources==		==Resources==
		+	* [https://math.dartmouth.edu/~m22x17/misc/LaLonde2012_proof_by_contrapositive.pdf Proof by Contrapositive], Dartmouth University
		+	* [http://zimmer.csufresno.edu/~larryc/proofs/proofs.contrapositive.html Proof by Contrapositive], Fresno State University

← Older revision		Revision as of 15:55, 24 September 2021
Line 1:		Line 1:
−	Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </math>" (read as "if P, then Q" or "P implies Q") is "<math> \neg Q \implies \neg P </math (read as "if not Q, then not P" or "not Q implies not P">.	+	Let <math> P </math> and <math> Q </math> be propositions such that <math> P \implies Q </math>. Then, the contrapositive of the conditional statement "<math> P \implies Q </math>" (read as "if P, then Q" or "P implies Q") is "<math> \neg Q \implies \neg P </math> (read as "if not Q, then not P" or "not Q implies not P"). The contrapositive is logically equivalent to the original conditional; that is, a conditional and its contrapositive always have the same truth values. For example, the contrapositive of "if <math> x > 0 </math>, then <math> x </math> is positive" is "if <math> x </math> is NOT positive, then <math> x \leq 0 </math>". These two statements are logically equivalent, and are both true. Sometimes, proving the contrapositive of a conditional is easier than proving the conditional itself. For example: Let x be an integer. If <math> x^2 </math> is even, then x is even. While we can attempt to prove this conditional statement directly, it is easier to show that if x is not even, then <math> x^2> </math> is not even, given that an odd number times an odd number must be odd (that is, not even).

	==Resources==		==Resources==