Proofs:Biconditionals - Revision history

Lila at 17:45, 24 September 2021

2021-09-24T17:45:48Z

Lila at 17:45, 24 September 2021

2021-09-24T17:45:14Z

Lila at 17:35, 24 September 2021

2021-09-24T17:35:44Z

Lila at 17:13, 24 September 2021

2021-09-24T17:13:19Z

Lila: Created page with "A biconditional of two propositions P and Q takes the form " $P$ if and only if $Q$ ". This can also be written as " $P \iff Q$ , which is e..."

2021-09-24T17:06:06Z

Created page with "A biconditional of two propositions P and Q takes the form "<math> P </math> if and only if <math> Q </math>". This can also be written as "<math> P \iff Q </math>, which is e..."

New page

A biconditional of two propositions P and Q takes the form "<math> P </math> if and only if <math> Q </math>". This can also be written as "<math> P \iff Q </math>, which is equivalent to "<math> P \implies Q </math> and <math> Q \implies P </math>"

← Older revision		Revision as of 17:45, 24 September 2021
Line 1:		Line 1:
	A biconditional of two propositions P and Q takes the form " <math> P </math> if and only if <math> Q </math> ". This can also be written as <math> P \iff Q </math>, which is equivalent to <math> P \implies Q \and Q \implies P </math>. When proving a biconditional statement, we need to prove that <math> P \implies Q </math> and <math> Q \implies P </math> are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " <math> P \implies Q </math> and <math> Q \implies P </math> " is logically equivalent to " <math> P \implies Q </math> and <math> \neg P \implies \neg Q </math> ", " <math> Q \implies P </math> and <math> \neg Q \implies \neg P </math> ", or " <math> \neg P \implies \neg Q </math> and <math> \neg Q \implies \neg P </math> ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.		A biconditional of two propositions P and Q takes the form " <math> P </math> if and only if <math> Q </math> ". This can also be written as <math> P \iff Q </math>, which is equivalent to <math> P \implies Q \and Q \implies P </math>. When proving a biconditional statement, we need to prove that <math> P \implies Q </math> and <math> Q \implies P </math> are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " <math> P \implies Q </math> and <math> Q \implies P </math> " is logically equivalent to " <math> P \implies Q </math> and <math> \neg P \implies \neg Q </math> ", " <math> Q \implies P </math> and <math> \neg Q \implies \neg P </math> ", or " <math> \neg P \implies \neg Q </math> and <math> \neg Q \implies \neg P </math> ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.
		+

	Example of a biconditional proof: "For <math> x\in\Z, 3x + 3 </math> is odd if and only if <math> x </math> is even". Let <math> P </math> = "3x + 3 is odd" and <math> Q </math> = "x is even".		Example of a biconditional proof: "For <math> x\in\Z, 3x + 3 </math> is odd if and only if <math> x </math> is even". Let <math> P </math> = "3x + 3 is odd" and <math> Q </math> = "x is even".

← Older revision		Revision as of 17:45, 24 September 2021
Line 8:		Line 8:

	Therefore, <math> P \iff Q </math>.		Therefore, <math> P \iff Q </math>.
		+
		+	==Resources==
		+	* [https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/A_Spiral_Workbook_for_Discrete_Mathematics_(Kwong)/02%3A_Logic/2.04%3A_Biconditional_Statements Biconditional Statements], Mathematics LibreTexts
		+	* [https://en.wikipedia.org/wiki/Logical_biconditional Logical Biconditional], Wikipedia

← Older revision		Revision as of 17:35, 24 September 2021
Line 1:		Line 1:
−	A biconditional of two propositions P and Q takes the form "<math> P </math> if and only if <math> Q </math>". This can also be written as "<math> P \iff Q </math>, which is equivalent to "<math> P \implies Q ~~</math>~~ and ~~<math>~~ Q \implies P </math>". When proving a biconditional statement, we need to prove that <math> P \implies Q </math> and <math> Q \implies P </math> are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, "<math> P \implies Q </math> and <math> Q \implies P </math>" is logically equivalent to "<math> P \implies Q </math> and <math> \neg P \implies \neg Q </math>", "<math> Q \implies P </math> and <math> \neg Q \implies \neg P </math>", or "<math> \neg P \implies \neg Q </math> and <math> \neg Q \implies \neg P </math>". Thus, we do have some options as to how to prove the two directions of a biconditional statement.	+	A biconditional of two propositions P and Q takes the form " <math> P </math> if and only if <math> Q </math> ". This can also be written as <math> P \iff Q </math>, which is equivalent to <math> P \implies Q \and Q \implies P </math>. When proving a biconditional statement, we need to prove that <math> P \implies Q </math> and <math> Q \implies P </math> are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " <math> P \implies Q </math> and <math> Q \implies P </math> " is logically equivalent to " <math> P \implies Q </math> and <math> \neg P \implies \neg Q </math> ", " <math> Q \implies P </math> and <math> \neg Q \implies \neg P </math> ", or " <math> \neg P \implies \neg Q </math> and <math> \neg Q \implies \neg P </math> ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.
		+
		+	Example of a biconditional proof: "For <math> x\in\Z, 3x + 3 </math> is odd if and only if <math> x </math> is even". Let <math> P </math> = "3x + 3 is odd" and <math> Q </math> = "x is even".
		+
		+	: If x is even, then 3x + 3 = 3(2k) + 3 for some integer k. 3(2k) + 3 = 3(2k + 1). 3 is odd and 2k+1 is odd, so their product is also odd. Thus, if x is even, 3x + 3 is odd. Therefore, the statement <math> Q \implies P </math> is true.
		+
		+	: If x is odd, then 3x + 3 = 3(2k + 1) + 3 for some integer k. 3(2k + 1) + 3 = 3(2k + 1 + 1) = 3(2k + 2) = 3(2)(k+1). Thus 3x + 3 is a multiple of 2 and is even if x is odd. So, the statement <math> \neg Q \implies \neg P </math>, which is logically equivalent to <math> P \implies Q </math>, is true.
		+
		+	Therefore, <math> P \iff Q </math>.

← Older revision		Revision as of 17:13, 24 September 2021
Line 1:		Line 1:
−	A biconditional of two propositions P and Q takes the form "<math> P </math> if and only if <math> Q </math>". This can also be written as "<math> P \iff Q </math>, which is equivalent to "<math> P \implies Q </math> and <math> Q \implies P </math>"	+	A biconditional of two propositions P and Q takes the form "<math> P </math> if and only if <math> Q </math>". This can also be written as "<math> P \iff Q </math>, which is equivalent to "<math> P \implies Q </math> and <math> Q \implies P </math>". When proving a biconditional statement, we need to prove that <math> P \implies Q </math> and <math> Q \implies P </math> are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, "<math> P \implies Q </math> and <math> Q \implies P </math>" is logically equivalent to "<math> P \implies Q </math> and <math> \neg P \implies \neg Q </math>", "<math> Q \implies P </math> and <math> \neg Q \implies \neg P </math>", or "<math> \neg P \implies \neg Q </math> and <math> \neg Q \implies \neg P </math>". Thus, we do have some options as to how to prove the two directions of a biconditional statement.

Proofs:Biconditionals - Revision history

Lila at 17:45, 24 September 2021

Lila at 17:45, 24 September 2021

Lila at 17:35, 24 September 2021

Lila at 17:13, 24 September 2021

Lila: Created page with "A biconditional of two propositions P and Q takes the form " P if and only if Q ". This can also be written as " P \iff Q , which is e..."

Lila: Created page with "A biconditional of two propositions P and Q takes the form " $P$ if and only if $Q$ ". This can also be written as " $P \iff Q$ , which is e..."