Mathematical Induction 2. Assume the statement is true for any - PowerPoint PPT Presentation

Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide01.html Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide02.html Mathematical Induction prev | slides | next prev | slides | next Using mathematical induction in the proof of a statement consists of a two-step process: 1. Show that the statement is true for a particular value of some parameter n . This is the basis step . Mathematical Induction 2. Assume the statement is true for any particular value of n and show that it is also true for n +1. This is the inductive step . That’s it! What can be confusing about induction is how it differs from the fallicy of circular reasoning that we discussed earlier. 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 of 1 09/24/2003 07:28 AM 1 of 1 09/24/2003 07:28 AM Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide03.html Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide04.html Mathematical Induction Mathematical Induction prev | slides | next prev | slides | next Example: Prove that 1+2+3+...+ n = n ( n +1)/2 is valid for positive Next comes the inductive step. The key to this, (and what makes it integers n . different from circular reasoning) is that we will assume that the statement is true for some value of n and show that if that is true Proof: Begin with the basis step. We need to prove that this then it is true for n +1. statement is true for one particular value of n . The best one to use (because it makes the rest of our work easier) is usually the smallest We therefore assume that 1+2+3+...+ n = n ( n +1)/2 and need to show value of n allowed by the statement, in this case n =1. that this implies that 1+2+3+...+ n +( n +1) = ( n +1)( n +2)/2. We merely substitute this value of n into the statement and Note that the inductive step is an application of modus ponens: p demonstrate that a true statement results. together with p q implies that q is true. 1 = 1×(1+1)/2 = 2/2 = 1 We’ll start with 1+2+3+...+ n = n ( n +1)/2 and add n +1 to both sides. Thus, we know that for one particular value, n =1, that the statement 1 2 3 4 5 6 7 8 9 is true. 1 2 3 4 5 6 7 8 9 1 of 1 09/24/2003 07:28 AM 1 of 1 09/24/2003 07:28 AM

Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide05.html Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide06.html Mathematical Induction Mathematical Induction prev | slides | next prev | slides | next The changes in each line are highlighted. The "statement" in a mathematical induction proof is called the inductive hypothesis . In the last example the inductive hypothesis was 1+2+3+...+ n +( n +1)= n ( n +1)/2+( n +1) = n ( n +1)/2+2( n +1)/2 P ( n ) = 1 + 2 + 3 + ... + n = n ( n + 1)/2 = ( n +2)( n +1)/2 = ( n +1)( n +2)/2 the rule of inference describing mathematical induction is The last line represents the statement we were trying to obtain. [ P (1) n ( P ( n ) P ( n +1))] n P ( n ) Because of the basis step we know that the statement we are trying (Actually this is only correct when n is a positive integer - if n to prove is true for n =1. Because of the inductive step we know that comes from another set then this would need to be modified.) it must then be true for n =2. Similarly if it’s true for n =2 then it must be true for n =3, and so 1 2 3 4 5 6 7 8 9 on... Thus we see that the statement must be true for all positive values of n , which is the desired result. 1 2 3 4 5 6 7 8 9 1 of 1 09/24/2003 07:28 AM 1 of 1 09/24/2003 07:28 AM Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide07.html Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide08.html Mathematical Induction Mathematical Induction prev | slides | next prev | slides | next Mathematical Induction can be used on a variety of problems, Let’s start with ( n +1) 3 -( n +1) and see that this is divisible by 3. including those that don’t involve the sum or product of a sequence. ( n +1) 3 -( n +1) = n 3 +3 n 2 +3 n +1- n -1 For example, use mathematical induction to show that 3 divides = n 3 - n +3 n 2 +3 n n 3 - n whenever n is a positive integer. = ( n 3 - n )+3( n 2 + n ) Proof: The basis step requires that we show that 3 divides n 3 - n The first term here is divisible by 3 because of the induction when n =1. Substituting 1 for n gives 1 3 -1 = 0, which is clearly hypothesis, and the second term is divisible by 3 since it is a divisible by 3. multiple of 3. This completes the inductive step and also the proof. The inductive hypothesis is "if 3 divides n 3 - n then 3 divides 1 2 3 4 5 6 7 8 9 ( n +1) 3 -( n +1)." 1 2 3 4 5 6 7 8 9 1 of 1 09/24/2003 07:28 AM 1 of 1 09/24/2003 07:28 AM

Mathematical Induction http://localhost/~senning/courses/ma229/slides/induction/slide09.html Mathematical Induction prev | slides | next Problems like the following can also be solved using induction. Show that all values of postage greater than 11 cents can be obtained using only 4 cent and 5 cent stamps. Basis Step: A 12 cent postage can be formed with three 4 cent stamps. Inductive Step: Assume that a postage of value n can be formed. Since n is greater than or equal to 12, there must be at least one 4 cent stamp present or at least three 5 cents stamp present. If there are at least three 5 cent stamps we can remove them and replace them with four 4 cent stamps. Otherwise, remove one 4 cent stamp and replace it with one 5 cent stamp. In this fashion we can always increase the postage by one cent. 1 2 3 4 5 6 7 8 9 1 of 1 09/24/2003 07:28 AM