Download
1 / 36
360 likes | 380 Views
Learn exhaustive proof & proof by cases using examples to prove theorems. Understand how to tackle complex mathematical problems step by step.
E N D
CSE15 Discrete Mathematics02/08/17 Ming-Hsuan Yang UC Merced
1.8 Proof methods and strategy • Proof by cases: pi→q for i=1,2,…,n • When it is not possible to consider all cases at the same time • Exhaustive proof: some theorems can be proved by examining a relatively small number of examples
Example • Prove (n+1)3≥3n if n is a positive integer with n≤4 • Proof by exhaustion as we only need to verify n=1, 2, 3 and 4. • For n=1, (n+1)3=8 ≥31=3 • For n=2, (n+1)3=27 ≥32=9 • For n=3, (n+1)3=64 ≥33=27 • For n=4, (n+1)3=125≥43=64
Example • An integer is a perfect power if it equals na, where a is an integer greater than 1 • Prove that the only consecutive positive integers not exceeding 100 that are perfect powers are 8 and 9 • Can prove this fact by examining positive integers n not exceeding 100 • First check whether n is a perfect power, and then check whether n+1 is a perfect power
Example perfect power: na, for a >1 • For positive integers • a=2, the squares ≤ 100: 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 • a=3, the cubes ≤ 100: 1, 8, 27, and 64 • a=4, the 4th powers n4 ≤ 100: 1, 16, and 81 • a=5, the 5th powers n5 ≤ 100: 1 and 32 • a=6, the 6th powers n6 ≤ 100: 1 and 64 • Look at the list of perfect powers, we see that the pair of n=8 and n+1=9 is the only two consecutive powers ≤ 100
Proof by cases • Prove that if n is an integer, then n2≥ n • We prove this by 3 cases: • n=0: trivial case as 02≥ 0 • n≥1: If n≥1 then n∙n ≥ n∙1 and thus n2≥ n • n≤-1: If n ≤ -1 then n2 ≥ 0>n and thus n2≥ n n=1 n=-1 n=0
Example • Show that |xy|=|x||y| for real numbers • x≥0, y≥0: xy ≥0 |xy|=xy=|x||y| • x≥0, y<0: xy<0 |xy|=-xy=x(-y)=|x||y| • x<0, y ≥0:xy<0 |xy|=-xy=(-x)y=|x||y| • x<0, y<0: xy>0 |xy|=xy=(-x)(-y)=|x||y|
Example • Formulate a conjecture about the decimal digits that occur at the final digit of the squares of an integer and prove the result • The smallest perfect squares are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225 and so on • Note that the digits that occur at the final digit of a squares are: 0, 1, 4, 5, 6, and 9 (and no 2, 3, 7, and 8) conjecture
Example • We can express an integer n as 10a+b were a and b are positive integers and 0≤b≤9 • n2=(10a+b)2=100a2+20ab+b2=10(10a2 +2b)+b2, so the final digit is the final digit of b2 • Note also that the final digit of (10-b)2=100-20b+b2. Thus, we only consider 6 cases • Case 1: if final digit of n is 1 or 9 (or b), then the last digit of n2 is 1
Example • Case 2: if the final digit of n is 2 or 8, then the final digit of n2 is 4 • Case 3: if the final digit of n is 3 or 7, then the final digit of n2 is 9 • Case 4: if the final digit of n is 4 or 6, then the final digit of n2 is 6 • Case 5: if the final digit of n is 5, then the final digit of n2 is 5 • Case 6: if the final digit of n is 0, then the final digit of n2 is 0
Example • Show that there are no solutions in integers x and y of x2+3y2=8 • x2>8when |x|≥3, and 3y2>8 when |y|≥2. The only values for x are -2,-1,0,1,2 and for y are -1, 0, 1 • So, possible values for x2 are, 0, 1, and 4. The possible values for 3y2 are 0 and 3 • No pair of x and y can be solution
Without loss of generality (WLOG) • In proof, sometimes we can apply the same argument for different cases • x≥0, y<0: xy<0 |xy|=-xy=x(-y)=|x||y| • x<0, y ≥0:xy<0 |xy|=-xy=(-x)y=|x||y| • By proving one case of a theorem, no additional argument is required to prove other specified cases
Without loss of generality Example: Show that if x and y are integers and both x∙yand x+y are even, then both x and y are even. Proof: Use a proof by contraposition. Suppose x and y are not both even. Then, one or both are odd. Without loss of generality, assume that x is odd. Then x = 2m + 1 for some integer k. Case 1: y is even. Then y = 2nfor some integer n, so x +y = (2m+ 1) + 2n = 2(m + n) + 1 is odd. Case 2: y is odd. Then y = 2n +1 for some integer n, so x ∙ y = (2m+ 1) (2n + 1) = 2(2m ∙ n +m + n) + 1 is odd. We only cover the case where x is odd because the case where y is odd is similar. The use phrase without loss of generality(WLOG) indicates this.
Common mistakes in exhaustive proof and proof by cases • Draw incorrect conclusions from insufficient number of examples • Need to cover every possible case in order to prove a theorem • Proving a theorem is analogous to showing a program always produces the desired output • No matter how many input values are tested, unless all input values are tested, we cannot conclude that the program always produces correct output
Example • Is it true that every positive integer is the sum of 18 4th powers of integers? • The 4th powers of integers: 0, 1, 16, 81, … • Select 18 terms from these numbers and add up to n, then n is the sum of 18 4th powers • Can show that integers up to 78 can be written as the sum as such • However, if we decided this was enough (or stop earlier), then we would come to wrong conclusion as 79 cannot be written this way
Example • What is wrong with this “proof” “Theorem”: If x is a real number, then x2 is a positive real number “Proof”: Let p1 be “x is positive” and p2 be “x is negative”, and q be “x2 is positive”. First show p1→q, and then p2→q. As we cover all possible cases of x, we complete this proof
Example • We missed the case x=0 • When x=0, the supposed theorem is false • If p is “x is a real number”, then we need to prove results with p1, p2, p3 (where p3 is the case that x=0)
Existence proof • A proof of a proposition of the form • Constructive proof: find one element a such that p(a) is true • Non-constructive proof: prove that is true in some other way, usually using proof by contradiction
Constructive existence proof • Show that there is a positive integer that can be written as the sum of cubes of positive integers in two different ways • By intuition or computation, we find that 1729=103+93=123+13 • We prove this theorem as we show one positive integer can be written as the sum of cubes in two different ways
Ramanujan • G. H. Hardy, when visiting Ramanujan, remarked that 1729, the number of the cab he took, was rather dull • Ramanujan replied “No, it is a very interesting number; it is the smallest number expressible as the sum of cubes in two different ways.”
Non-constructive existence proof • Show that there exist irrational numbers x and y such that xy is rational • We previously show that is irrational • Consider the number . If it is rational, we have two irrational number x and y with xy is rational (x= , y= ) • On the other hand if is not rational, then we let • We have not found irrational numbers x and y such that xy is rational • Rather, we show that either the pair x= , y= or the pair have the desired property, but we do not know which of these two pairs works.
Uniqueness proof • Some theorems assert the existence of a unique element with a particular property • Need to show • Existence: show that an element x with the desired property exists • Uniqueness: show that if y≠x, then y does not have the desired property • Equivalently, show that if x and y both have the desired property, then x=y • Showing that there is a unique element x such that p(x) is the same as proving the statement
Example • Show that if a and b are real numbers and a≠0, then there is a unique number r such that ar+b=0 • Note that the real number r=-b/a is a solution of ar+b=0. Consequently a real number r exists for which ar+b=0 • Second, suppose that s is a real number such that as+b=0. Then ar+b=as+b. Since a≠0, s must be equal to r. This means if s≠r, as+b≠0
Proof strategy • Can be challenging • First analyze what the hypotheses and conclusion mean • For conditional statements, usually start with direct proof, then indirect proof, and then proof by contradiction
Forward/backward reasoning • Direct proof: • start with premises, together with axioms and known theorems, • we can construct a proof using a sequence of steps that lead to conclusion • A type of forward reasoning • Backward reasoning: to prove q, we find a stement p that we can prove that p→q
Example • For two distinct positive real numbers x, y, their arithmetic mean is (x+y)/2, and their geometric mean is . Show that the arithmetic mean is always larger than geometric mean • To show , we can work backward by finding equivalent statements
Example • For two distinct real positive real numbers, x and y, (x-y)2>0 • Thus, x2-2xy+y2>0, x2+2xy+y2>4xy, (x+y)2>4xy. So, • We conclude that if x and y are distinct positive real numbers, then their arithmetic mean is greater than their geometric mean
Example • Suppose that two people play a game taking turns removing 1, 2, or 3 stones at a time from a pile that begins with 15 stones. The person who removes the last stone wins the game. • Show that the first player can win the game no matter what the second play does
Example • At the last step, the firstplayer can win if this player is left with a pile with 1, 2, or 3 stones • The second player will be forced to leave 1, 2 or 3 stones if this player has to remove stones from a pile containing 4 stones • The first player can leave 4 stones when there are 5, 6, or 7 stones left, which happens when the second player has to remove stones from a pile with 8 stones
Example • That means, there are 9, 10 or 11 stones when the first player makes this move • Similarly, the first player should leave 12 stones when this player makes the first move • We can reverse this argument to show that the first player can always makes this move to win (successively leave 12, 8, and 4 stones for 2nd player)
Adapting existing proof • Take advantage of existing proofs • Borrow some ideas used in the existing proofs • We proved is irrational. We now conjecture that is irrational. Can we adapt previous proof to show this? • Mimic the steps in previous proof • Suppose • Can we use this to show that 3 must be a factor of both c and d?
Example • We will use some results from number theory (discussed in Chapter 3) • As 3 is factor of c2, it must be a factor of c Thus, 9 is a factor of c2, which means 9 is a factor of 3d2 • Which implies 3 is a factor d2, and 3 is factor of d • This means 3 is factor of c and d, a contradiction
Looking for counterexamples • When confronted with a conjecture, try to prove it first • If the attempt is not successful, try to find a counterexample • Process of finding counterexamples often provides insights into problems
Example • We showed the statement “Every positive integer is the sum of two squares of integers” is false by finding a counterexample • Is the statement “Every positive integer is the sum of the squares of three integers” true? • Look for an counterexample: 1=02+02+12, 2=02+12+12, 3=12+12+12, 4=02+02+22, 5=02+12+22, 6=12+12+22, but cannot do so for 7
Example • The next question is to ask whether every positive integer is the sum of the squares of 4 positive integers • Some experiments provide evidence that the answer is yes, e.g., 7=12+12+12+22, 25=42+22+22+12, and 87=92+22+12+12 • It turns the conjecture “Every positive integer is the sum of squares of four integers” is true
Proof strategy in action • Formulate conjectures based on many types of possible evidence • Examination of special cases can lead to a conjecture • If possible, prove the conjecture • If cannot find a proof, find a counterexample • A few conjectures remain unproved • Fermat’s last theorem (a conjecture since 1637 until Andrew Wiles proved it in 1995)
