Difference between revisions of "Proofs:Biconditionals"
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| − | 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 | + | 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. |
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| + | 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". | ||
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| + | : 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. | ||
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| + | : 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. | ||
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| + | Therefore, <math> P \iff Q </math>. | ||
Revision as of 11:35, 24 September 2021
A biconditional of two propositions P and Q takes the form " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P } if and only if ". This can also be written as , which is equivalent to . When proving a biconditional statement, we need to prove that and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \implies P } are true. Remember that the contrapositive of a conditional is logically equivalent to the conditional. Thus, " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \implies Q } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \implies P } " is logically equivalent to " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \implies Q } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg P \implies \neg Q } ", " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \implies P } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg Q \implies \neg P } ", or " Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg P \implies \neg Q } and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg Q \implies \neg P } ". Thus, we do have some options as to how to prove the two directions of a biconditional statement.
Example of a biconditional proof: "For Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in\Z, 3x + 3 } is odd if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x } is even". Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P } = "3x + 3 is odd" and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q } = "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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \implies P } 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \neg Q \implies \neg P } , which is logically equivalent to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \implies Q } , is true.
Therefore, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \iff Q } .
