ProofsPart4

# ProofsPart4 - MATH 135 Fall 2006 Proofs Part IV We’ve...

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MATH 135 Fall 2006 Proofs, Part IV We’ve spent a couple of days looking at one particular technique of proof: induction. Let’s look at a few more. Direct Proof Here we start with what we’re given and proceed in a direct line to the conclusion. Example If n is an odd integer, then n 2 - 1 is divisible by 4. Proof Recall that every even integer can be written as 2 m for some integer m and every odd integer can be written as 2 k + 1 for some integer k . Since n is odd, n = 2 k + 1 for some k Z . Thus n 2 - 1 = (2 k + 1) 2 - 1 = 4 k 2 + 4 k + 1 - 1 = 4 k 2 + 4 k = 4( k 2 + k ) so is divisible by 4. Tips 1) Most steps should be justiﬁed, with the exception of straightforward algebraic steps. 2) Always introduce/deﬁne new variables and give their domain. 3) Start with what you’re given, not with what you want to prove. 4) Keep one eye forward and one behind; always be aware of where you’re going. Example If a , b , c form an arithmetic sequence, then ( b - c ) x 2 + ( c - a ) x + ( a - b ) = 0 has equal roots. Proof Since a , b , c form an arithmetic sequence, then b = a + d , c = a + 2 d for some d R . Thus our equation is - dx 2 + 2 dx - d = 0 - d ( x 2 - 2 x + 1) = 0 - d ( x - 1) 2 = 0 and so has two equal roots. Example If AB is a diameter of a circle and C is on the circle, then ACB = π 2 .

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Proof Let O be the centre of the circle and join CA , CO , and CB . Suppose ACO = x . Since AO , BO and CO are radii, then AO = BO = CO . Since AO = CO , then 4 ACO is isosceles, so CAO = ACO = x . Therefore, COA = π - ACO - CAO = π - 2 x . Also, COB = π - COA = π - ( π - 2 x ) = 2 x . Since CO = BO , then 4 BCO is isosceles, so BCO = CBO . Thus, looking at the angles in 4 COB , 2 x + 2 BCO = π , so BCO = π 2 - x . Therefore, ACB = ACO + BCO = x + π 2 - x = π 2 , as required. A C B O x Compound Statements If A and B are mathematical statements, we often see compound statements such as A and B A or B For “ A and B ” to be TRUE, both A and B must be TRUE. Otherwise (when one is FALSE or both are FALSE), “ A and B ” is FALSE. For “ A or B ” to be TRUE, either or both of A and B must be TRUE. Otherwise (when both are FALSE), “ A or B ” is FALSE. Example A =“2 is a prime number”, B =“5 is a perfect square” Is “ A and B ” TRUE or FALSE? Is “ A or B ” TRUE or FALSE? Aside Regarding Sets Recall that if A and B are sets, then A B is the set of elements that are in either A or B , and A B is the set of elements that are in both A and B . So
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ProofsPart4 - MATH 135 Fall 2006 Proofs Part IV We’ve...

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