Proof by Mathematical Induction.

Transcription

1 Proof by Mathematical Inuction. Mathematicians have very peculiar characteristics. They like proving things or mathematical statements. Two of the most important techniques of mathematical proof are proof by inuction an proof by contraiction. You may woner what constitutes a proper mathematical proof? It is a complicate question an we nee to go eeper into the structure of mathematics to be able to answer this question. The following Power Point presentation shows the history of mathematics an its structure. It looks at axioms an Eucliean geometry. It explains how mathematics is base on axioms an how we buil more an more vali theorems. It also looks at various methos of proofs. Finally it asks those philosophical questions you must have hear in TOK. Proof A proof is a careful argument that establishes a new fact or theorem, given certain assumptions or hypotheses. There are various kins of proof, some of which we mention here. Proof by euction A euctive proof consists of a sequence of statements or sentences each of which is euce from previous ones or from hypotheses using stanar mathematical properties. The final statement may be calle a theorem. Proof by contraiction Sometimes it is easier to argue by contraiction, i.e. to assume that the esire conclusion is false an erive a contraiction to some known fact. Counter-examples To show that a statement is false it is enough to give a single instance for which it oes not hol, calle a counterexample. METHODS OF PROOF MATHEMATICAL INDUCTION The term inuction involves eriving of a general statement or rule from one or more particular cases. A statement may be true for a few special cases. Does it follow, then, that it is true in all cases? Consier the following two examples:. It is true that 60 is ivisible by,, 3, 4, 5 an 6. This establishes a pattern. Does it follow that 60 is ivisible by all positive integers? = + 3= = = 4

2 . The sum of the first two o numbers is a perfect square, so too is the sum of the first three o numbers, an of the first five o numbers. Does it follow that the sum of the first n numbers is a perfect square? The pattern seems to inicate that it is. We shall proof shortly that it is. The nee for proof If we see something that works a few times in a row, we're convince that it works forever. Regions of a Circle Consier a circle with n points on it. How many regions will the circle be ivie into if each pair of points is connecte with a chor? points regions = 3 points 4 regions = 4 points 8 regions = 3 5 points 6 regions = 4 At this point, probably everyone woul be convince that with 6 points there woul be 3 regions, but it's not prove, it's just conjecture. The conjecture is that the number of regions when n points are connecte is n -. Will fining the number of regions when there are six points on the circle prove the conjecture? No. If there are inee 3 regions, all you have one is shown another example to support your conjecture. If there aren't 3 regions, then you have prove the conjecture wrong. In fact, if you go ahea an try the circle with six points on it, you'll fin out that there aren't 3 regions. You can never prove a conjecture is true by example. You can prove a conjecture is false by fining a counter-example. To prove a conjecture is true, you nee some more formal methos of proof. One of these methos is the principle of mathematical inuction. Proof by inuction The metho of proof by inuction is applie to propositions, P(n), containing natural numbers n. Step : Verify that the statement is true for a specific case, usually n= P() is true. Step : Assume that the statement is true for some integer, say n = k, an then prove that it must be true for n = k +. In other wors prove Pk ( ) Pk ( + ). Step 3: If it is true for n =, then it must be true for n =, an then being true for

3 n =, it must be true for n = 3 an so on. So it is true for all natural numbers. P(n) is true for every n. Example Prove, by mathematical inuction, that the sum of the first n o integers is n (n ) = n Solution Step : When n = LHS = RHS = = LHS = RHS An the statement is true for n =. Step : Assume the statement is true for n= k (k ) = k Step 3 Base on the assumption in Step, show that the following is true, by replacing k with k+ [ ] (k ) + ( k+ ) = ( k+ ) We procee as follows: [ ] (k ) + ( k+ ) = ( k+ ) = k ( fromstep) LHS k k k k ( k ) RHS Conclusion: = + + = + + = + = We showe it is true for n = an we showe that it follows n = k true n = k + true, so the statement is true for any positive integer by the principle of mathematical inuction. Principle of Mathematical Inuction (common language). Show something works the first time.. Assume that it works for this time, 3. Show it will work for the next time. 4. Conclusion, it works all the time 3

4 Principle of Mathematical Inuction (mathematical language) To prove a particular proposition Pn ( ) for n Z +.. Show true for n =. P () true.. Assume true for n = k. Pk ( ) true 3. Show true for n = k + given true for n = k. Pk ( ) true Pk ( + ) true 4. Conclusion: Statement is true for all n >= 5. Write the final statement: Since P () is true an Pk ( ) true implies Pk+ ( ) true, therefore the statement is true by the principle of mathematical inuction. The key wor in step is assume. You are not trying to prove it's true for n = k, you're going to accept on faith that it is, an show it's true for the next number, n = k +. If it later turns out that you get a contraiction, then the assumption was wrong. The Principle of Mathematical Inuction can be informally state by saying: "We can establish the truth of a proposition if we can show that it follows from smaller instances of the same proposition, as long as we can establish the truth of the smallest instance (or instances) explicitly." (Grossman) In inuction, the truth percolates up through the layers to prove the whole proposition. If we know the first instance of a proposition is true an the secon one is erive from the st one, making it true, an the pattern repeats itself, then we know that our final result will also be true. But if the first, or any of the other instances following the first, turn out to be false then the final result will be wrong. Falling Dominoes Let's look at a row of ominoes, line up an reay to be pushe over. We know from experience that if we push over one omino, the rest of the ominoes will fall over. 4

5 How can we prove this? A. We know from experience that if we push over one omino, it shoul fall over. B. We also know that if a omino is falling an has been place correctly, it will knock over its neighbour. Intuitively, it shoul be very clear that the fall shoul cascae all the way up to the last omino. That is, if the next to the last omino falls, the last omino also falls. But now we nee to think about increasing the rigor of this argument by oing a real proof. A Formal Proof Let's look at the behaviour of the ominoes.. Assume that there's some omino k which oesn't fall over. i.e. k is the first omino which oesn't work right (maybe it's out of line, too far away, or has a ifferent gravitational constant).. Since k is the first such omino, the omino right before k must have fallen over. 5

6 3. But we know from B that a falling omino always knocks its neighbour over. 4. So omino k will fall over an we have a contraiction. What we have shown above is that because. We can knock over the first omino.. an a falling omino knocks over its neighbour. 3. then all the ominoes will fall over. Now, if we think of each omino as an instance of a proposition. If a given instance is true, the corresponing omino will fall over. Given a sequence of instances (row of ominoes): If we can prove:. The proposition is true in the first instance.. An if a given instance is true, the next one in the sequence will also be true. 3. Then the proposition will be true in all instances. This is calle a Proof by Inuction. Two more examples are shown in the following Power Point: Mathematical inuction applies to propositions involving ivisibility an to propositions involving matrix equations. It is illustrate in the examples below. Example. 6

7 Prove that Solution Let f(n) = 4 n+ + 3 is a multiple of 5 for any positive integer n. 4 n Step : Prove f() is a multiple of 5. But f() = = 35 (which is a multiple of 5). Step : We assume f(k) is a multiple of 5. We want to prove that f(k + ) is a multiple of 5. First we try to simplify: f(n + ) - f(n) = 4( n+ ) ( 4 n+ + ) = 3 4n+ 5 4n+ = 4n+ 4 4n+ ( ) = 5 Therefore: f(k + ) f(k) = 4 5 k+ 4k+ So, f(k + ) = 5 + f( k). Multiple of 5 Assume to be a multiple of 5 So, f(k + ) is a multiple of 5. Therefore by inuction, f(n) must be a multiple of 5 for all positive integers n. TRY THIS Prove that 3n n + is ivisible by 3 for any positive integer n. ANSWER Let f(n) = 3n n + If we here calculate, f(n + ) + f(n) = 3(n+ ) n+ + + ( 3n n + ) = 3n n + + 3n n + = 3n n n n + 7

8 = 3n+ ( 3 + ) + 5 n+ (5 + ) = 9 3n n+ Now we are reay to prove the result. Step : First prove true when n = : = = 57 (which is ivisible by 3). Step : Assume that the result is true when n = k, i.e. that f(k) is ivisible by 3. We showe above that f(k + ) + f(k) = 9 3k k+ i.e. that f(k + ) = 9 3k k+ f(k) Divisible by 3 Divisible by 3 Divisible by 3 (by assumption) So f(k + ) must be ivisible by 3 (as require). Therefore the result is true for n Z +. Example Prove that for all natural numbers... n =. 6 nn ( + ) n+ Solution Step Is this true for n =? = + = True for n =. Step Assume it is true for n = k: 8

9 k =. 6 kk ( + ) k+ Step 3 Base on the assumption in Step, show that the following is true, by replacing k with k+ k = 6 kk ( + ) ( k+ )( k+ + ) k+ + k k+ k k+ + = k+ ( k+ )( k+ ) k+ Fin the common enominator on the LHS: kk ( + ) LHS = + ( k+ )( k+ ) ( k+ )( k+ ) k + k+ = ( k+ )( k+ ) Factorize the enominator now: ( k+ )( k+ ) k+ LHS = = = RHS ( k+ )( k+ ) k+ LHS = RHS, so the result is true for n = k +. True for n = an P( k) true P( k + ) true, therefore true for all positive integer values of n. Example nπ Using mathematical inuction, prove that (cos x ) = cos x +, for all positive integer values of n. n x Solution n nπ Let p n be the statement cos x = cos x + for all positive integer values of n. x n n For n =, (cos x) = sin x = x π cos x + Therefore p is true. Assume the formula is true for n = k, that is, k x k kπ (cos x) = cos x + 9

10 Then k kπ (cos x) = cos x + x x x k + + x k k + + x k k + + x k kx (cos x) = sin x + kx π (cos x) = cos x + + ( k + )π (cos x) = cos x + which is p n when n = k +. So if p n is true for n = k then it is true for n = k + an by the principle of mathematical inuction p n is true for all positive integer values of n. Lesson Summary Mathematical inuction is a metho of proving statements involving the natural numbers,, 3,.... The iea is that (i) we prove the statement when n = an (ii) show that if the statement is true for some integer k then it is true for the integer immeiately above. From this we can conclue that the statement is true for all n =,, 3,.... The Principle of Mathematical Inuction. Let P(n) be a statement epening on an arbitrary positive integer n. Suppose that we can o the following two steps: (i) Verify that P() is true, (ii) Assume that P(k) is true an show that if P(k) is true then P(k + ) is true. Then the statement P(n) is true for all positive integers n. The statement P(n) is calle the inuctive hypothesis, step (i) is calle starting the inuction an step (ii) is calle the inuctive step. Note that the Principle of Inuction is intuitively obvious: if P() is true an P(k) P(k+), that is the truth of P(k) implies the truth of P(k+) for all n =,, 3,..., then P() P() P(3) )... P(n) )... by applying (ii) with n =,, 3,... in turn, so P(n) is true for all n. 0