Method of proofs

 Consider the statements: “Humans have two eyes”  It implies the “universal quantification”  If a is a Human then a has two eyes.

 Mathematical theorem:  “If x > y, where x and y are positive real numbers, then x 2 > y 2 ”  It stated that, if “x > y” is TRUE, then “x 2 > y 2 ” must be TRUE and the case where “x 2 > y 2 ” is FALSE will NEVER occurs.  How to prove this statement is valid?

 A definition is an exact, unambiguous explanation of the meaning of a mathematical word or phrase. (We do not need to prove definition) ◦ E.g. An integer n is even if n = 2a for some integer a 2  Z.  A theorem is a statement that is true and has been proved to be true. ◦ E.g.: Every absolutely convergent series converges.

 Definition 3. The integer n is even if there exists an integer k such that n = 2k and it is odd if there exists an integer k such that n = 2k + 1. Example: 6 is an even integer, because you can find k=3 such that 6 = 2  3. 9 is an odd integer, because you can find k=4 such that 9 = 2 

 A proof of a theorem is a written verification that shows that the theorem is definitely and certainly true.

 Direct proof  Indirect proof  Proof by contradiction

P  Q  Shows that if P is true, then Q must be true. Suppose that P is true. Therefore, Q is true.

 If x is an even integer, then x 2 - 6x + 5 is odd.  Proof:

 P  Q is logically equivalent to its contrapositive P  Q   Q  P  So, in indirect proof, instead of proving P  Q, we prove  Q  P.  E.g. Prove that “If n is odd, then n 2 is odd”  Prove that “If n 2 is even, then n is even”

P  Q  Steps: Suppose that  Q (is true). Therefore  P.

 If a man accused of holding up a bank can prove that he was some place else at the time the crime was committed, he will certainly be acquitted.  Assume that he committed the crime.  Then at the time of the crime, I would have had to be at the scene of the crime.

 In fact, at the time of the crime I was in a meeting with 20 people far from the crime scene, as they will testify.  This contradicts the assumption that I committed the crime, since it is impossible to be in two places at one time. hence that assumption is false.

 To prove that a statement p is true.  Suppose that p is false (  p is true).  Show that it produces contradictory results c   c. So the assumption of  p must be false.

 Prove the theorem “There is no greatest integer”. Proof:  Suppose the opposite is true there is a greatest integer N. Then N  n for every integer n. Let M = N + 1. Now M is an integer since it is a sum of integers. Also M > N since M = N + 1.  Thus M is an integer that is greater than N. So N is the greatest integer and N is not the greatest integer, which is a contradiction.

 Given the implication p  q, suppose that the conclusion q is false (  q and p is true),  following a sequence of steps, we deduce the statement  p (false).  But  p is a contradiction to the supposition that p and  q are true. Therefore q must be true.

 Prove that “If 3n + 2 is odd, then n is odd”

 Arguing from examples ◦ E.g: The sum of any two even integers is even. Solution:  This is true because if m = 14 and n = 6, which are both even, then m + n = 20, which is also even.  It is not sufficient to show that the conclusion “m + n is even” is true for m=14 and n = 6. You must give an argument to show that the conclusion is true for any even integers m and n.

 Using the same letter to mean two different things.  Consider the following proof:  Suppose m and n are odd integers. Then by definition of odd, m = 2k + 1 and n = 2k + 1 for some integer k.  This is incorrect. Using the same symbol, k, in the expressions for both m and n implies that m = 2k + 1 = n.

 Jumping to a conclusion. ◦ Jump to conclusion without giving an adequate reason. Consider the following proof that the sum of any two even integers is even.  Suppose m and n are any even integers. By definition of even, m=2r and n = 2s for some integers r and s. Then m + n = 2r and 2s. So m + n is even.  The problem with this proof is the crucial calculation 2r + 2s = 2 (r + s) is missing.

 Begging the question ◦ Assume what is to be proved. Consider:  Prove that the product of any two odd integers is odd.

 Suppose that m and n are odd integers. If mn is odd, then mn=2k + 1 for some integer k. Also by definition of odd, m = 2a + 1 and n = 2b + 1 for some integers a and b. Then mn = (2a + 1)(2b + 1) = 2k + 1, which is odd by definition of odd.  It is wrong because it assumes that the conclusion mn is odd is true, and later assumes it to be true by setting (2a + 1)(2b + 1)=2k + 1.

 To disprove  xP(x)  Q(x) is false is to equivalent to show that  xP(x)   Q(x)  For example, show that “if n is even then n 2 is even”  If we can find at least an “even” number which square is “not even” then this statement is false.  The example that show a universal statement is false is called the counterexample

 Disprove the following statement by finding a counterexample:   a,b  R, if a 2 = b 2 then a = b. Solution: To disprove this statement, you need to find real numbers a and b such that a 2 =b 2 and a  b. For example, when a =-1 and b=1.

 We wish to establish the truth of  xP(x). That is, P(x) is true for at least one x in the domain.  One way to prove this statement is to find an x that makes P(x) is true.  Then, generalize the results using Existential Generalization rule.

 Existential generalization (EG) If Q(x) is true, then there  xQ(x) is true.

 Theorem: There exists an integer solution to the equation x 2 + y 2 = z 2. Proof: Choose x = 3, y = 4, z = 5 so that =5 2. According to EG, there exists integers such that x 2 + y 2 = z 2.

 Proof techniques are useful to prove or disprove the validity of a statement