Lecture 11 21 127 Concepts of Math 09 21 2012 Biconditional Statements and Logical Equivalence The phrase if and only if means exactly that if and only if P if Q clearly means If Q then P i e Q P P only if Q means It cannot be the case that P holds and Q fails That is the only case where P Q fails is False Thus this corresponds to P Q Together P Q is logically the same as saying P Q Q P We will use to assert that two statements are logically equivalent Remember applies to numbers or sets only The symbol applies to two statements with truth values We use in definitions to say that some concept is logically equivalent to its definition For example we would define even by saying Given any x Z we say x is even k Z x 2k We also use to say two statements are logically equivalent that they have the same truth value whatever value that might be For example we might write x R x 0 1 0 x and this is a True statement We also use to make such a claim that isn t as concrete as the above example For instance we can say that for any mathematical statements P Q R however they are composed that P Q R P Q R and also P Q P Q Associative Laws Let P Q R be any mathematical statements Then P Q R P Q R and P Q R P Q R Distributive Laws Let P Q R be any mathematical statements Then P Q R P Q P R and P Q R P Q P R 1 DeMorgan s Laws Let P and Q be any mathematical statements Then P Q P Q and P Q P Q We can prove all of these logical equivalences via truth tables Contrapositive Given a conditional statement P Q the contrapositive is Q P It is true that P Q Q P because Q P Q P Q P and P Q is logically equivalent to Q P That is a conditional statement and its contrapositive are guaranteed to have the same truth value Converse Given a conditional statement P Q the converse is Q P It is not guaranteed that a conditional statement and its converse have the same truth value it could go either way Thus when considering a conditional statement it is always an interesting question to ask Does the converse hold Logical Negations To negate a conditional statement we need to remember this logically equivalent form P Q P Q Thus P Q P Q P Q P Q To negate a biconditional statement we just write it as a conjuction of two conditional statements P Q P Q Q P P Q Q P 2 Proof Strategies Proving claims x S P x Direct proof Define a specific example y Prove that y S Prove that P y holds true Indirect proof AFSOC that for every y S P y holds Find a contradiction Examples Solving a system of linear equations Fix a b c d e f R with ad bc 6 0 Then one can simultaneously solve ax by e 1 cx dy f 2 for some x y R Define S x y to be the statement x and y simultaenously satisfy both equations 1 and 2 above Then we claim x y R S x y First do a bunch of scratch work to try to construct the solution Then write a proof that defines your objects and then shows why they work Proof Since ad bc 6 0 by assumption we may define x de bf ad bc and y af ce ad bc and know that x y R Then ade abf abf bce ade bce e ad bc e ad bc ad bc ad bc cde bcf adf cde adf bcd f ad bc cx dy f ad bc ad bc ad bc ax by so S x y holds 3 Theorem Suppose n N and we have n real numbers a1 a2 an R Then one of those numbers is at least as large as their average That is n B n aB 1X ai n i 1 Proof AFSOC none of the numbers are at least as large as the average That is suppose n i n ai Define the constant S Pn i 1 ai so that ai 1X ai n i 1 S n Then we can sum all of the ai s and observe that S n X ai i 1 n X S i 1 n n S S n This shows that the real number S is strictly less than itself S S This is a contradiction Therefore the original assumption was false and the claim follows This is a version of the Pigeonhole Principle 4 Proving claims x S P x Direct proof Let y S be arbitrary and fixed Prove that P y holds true Indirect proof AFSOC that y S such that P y holds Find a contradiction Examples AGM inequality x y R xy x y 2 2 Proof Let x y R be arbitrary and fixed We know 0 x y 2 Multiplying out and rearranging we get 2xy x2 y 2 Adding 2xy to both sides we get 4xy x2 2xy y 2 Factoring we get 4xy x y 2 Dividing by 4 and putting that into the square we get xy 2 is irrational a b Z a b 6 x y 2 2 2 Proof AFSOC a b Z ab 2 We may assume that ab is given in reduced form so that a and b have no common factors If not we could just reduce the fraction and obtain such a form Let such an a b Z be given 2 This means ab 2 so ab2 2 Thus 2b2 a2 so a2 is even Thus a is even Note You should prove this We will see it later Thus k Z a 2k Let such a k be given so a2 4k 2 Then 2b2 4k 2 so b2 2k 2 Thus b2 is even and so b is even This shows both a and b are even so in fact they have a common factor of 2 This contradictions our assumption that a and b have no common factors Therefore the original assumption must be false and the claim follows 5 Proving claims P Q Direct proof Prove that P holds true or else prove that Q holds true Indirect proof 1 Suppose that P holds Prove that Q holds Indirect proof 2 AFSOC that P Q holds Find a contradiction Examples Suppose that x R and x2 x Prove that x 1 or x 0 Proof Let x R be arbitrary and fixed and suppose that x2 x If it holds that x 0 we would be done so suppose otherwise That is suppose x 0 By assumption x2 x Since x 0 we can divide both sides by x This yields x 1 6 Proving claims P Q Direct proof …