Mathematical Induction Everybody – do the wave! The Wave ● If done properly, everyone will eventually end up joining in. ● Why is that? ● Someone (me!) started everyone off. ● Once the person before you did the wave, you did the wave. The principle of mathematical induction states that if for some P(n) the following hold: P(0) is true …and it stays If it starts and true… true… For any n ∈ ℕ, we have P(n) → P(n + 1) then …then it's always true. For any n ∈ ℕ, P(n) is true. Induction, Intuitively ● It's true for 0. ● Since it's true for 0, it's true for 1. ● Since it's true for 1, it's true for 2. ● Since it's true for 2, it's true for 3. ● Since it's true for 3, it's true for 4. ● Since it's true for 4, it's true for 5. ● Since it's true for 5, it's true for 6. ● … Proof by Induction ● Suppose that you want to prove that some property P(n) holds of all natural numbers. To do so: ● Prove that P(0) is true. – This is called the basis or the base case. ● Prove that for all n ∈ ℕ, that if P(n) is true, then P(n + 1) is true as well. – This is called the inductive step. – P(n) is called the inductive hypothesis. ● Conclude by induction that P(n) holds for all n. Some Summations 20 = 1 = 21 – 1 20 + 21 = 1 + 2 = 3 = 22 – 1 20 + 21 + 22 = 1 + 2 + 4 = 7 = 23 – 1 20 + 21 + 22 + 23 = 1 + 2 + 4 + 8 = 15 = 24 – 1 20 + 21 + 22 + 23 + 24 = 1 + 2 + 4 + 8 + 16 = 31 = 25 – 1 Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. For our base case, we need to show P(0) is true, meaning the sum of the firstJust zeroJust aspowersas inin aa proofofproof two byisby 2 0 – 1. Since the 0 sum of the firstcontradiction contradictionzero powers oforor two contrapositive,contrapositive, is 0 = 2 – 1, we see P(0) is true. wewe shouldshould mentionmention thisthis proofproof For the inductive step, isassume by induction. that for some n ∈ ℕ that P(n) holds, meaning thatis 2by0 + induction. 21 + … + 2n-1 = 2n – 1. We need to show that P(n + 1) holds, meaning that the sum of the first n + 1 powers of two is numbers is 2n+1 – 1. Consider the sum of the first n + 1 powers of two. This is the sum of the first n powers of two, plus 2n. Using the inductive hypothesis, we see that 20 + 21 + … + 2n-1 + 2n = (20 + 21 + … + 2n-1) + 2n = 2n – 1 + 2n = 2(2n) – 1 n+1 = 2 – 1 Thus P(n + 1) is true, completing the induction. ■ Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. For our base case, we need to show P(0) is true, meaning the sum of the first zero powers of two is 20 – 1. Since the sum of the first zero powers of two is 0 = 20 – 1, we see P(0) is true. Now,Now, wewe statestate whatwhat For the inductiveproperty propertystep, assume PP(( nthatn)) we wefor aresomeare n ∈ ℕ that P(n) holds, meaning that 20 + 21 + … + 2n-1 = 2n – 1. We going to prove holds need to show that goingP(n + 1)to holds, prove meaning holds that the sum of the first n + 1 powersfor of all two n is ∈ ∈numbers ℕ ℕ. is 2n+1 – 1. for all n . Consider the sum of the first n + 1 powers of two. This is the sum of the first n powers of two, plus 2n. Using the inductive hypothesis, we see that 20 + 21 + … + 2n-1 + 2n = (20 + 21 + … + 2n-1) + 2n = 2n – 1 + 2n = 2(2n) – 1 n+1 = 2 – 1 Thus P(n + 1) is true, completing the induction. ■ Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. For our base case, we need to show P(0) is true, meaning the sum of the first zero powers of two is 20 – 1. Since the sum of the first zero powers of two is 0 = 20 – 1, we see P(0) is true. For the inductive step, assume that for some n ∈ ℕ that P(n) holds, meaning that 20 + 21 + … + 2n-1 = 2n – 1. We TheThe firstfirst stepstep ofof anan inductiveinductive need to show thatproof P (nis +to 1) show holds, P(0). meaning We that the sum of the first n +proof 1 powers is to of showtwo isP (0).numbers We is 2n+1 – 1. explicitly state what P(0) is, Consider the sumexplicitly of the statefirst nwhat + 1 Ppowers(0) is, of two. This is the sum of thethenthen first trytry n powerstoto proveprove of it.two,it. WeplusWe can can2n . Using the inductive hypothesis,proveprove P Pwe(0)(0) see usingusing that anyany proofproof 20 + 21 + … + 2techniquentechnique-1 + 2n = (2we'dwe'd0 + 2like.like.1 + … + 2n-1) + 2n = 2n – 1 + 2n = 2(2n) – 1 n+1 = 2 – 1 Thus P(n + 1) is true, completing the induction. ■ Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. For our base case, we need to show P(0) is true, meaning the sum of the first zero powers of two is 20 – 1. Since the sum of the first zero powers of two is 0 = 20 – 1, we see P(0) is true. For the inductive step, assume that for some n ∈ ℕ that P(n) holds, meaning that 20 + 21 + … + 2n-1 = 2n – 1. We need to show that P(n + 1) holds, meaning that the sum of the first n + 1 powers of two is 2n+1 – 1. Consider the sum of the first n + 1 powers of two. This is n the sum of theTheThe first goalgoal n ofpowersof thisthis ofstepstep two, isis plustoto prove prove2 . Using the inductive hypothesis, we see that 20 + 21“ +“ForFor … any +any 2 n n-1 ∈ +∈ ℕ ℕ2,n , if if= P P(2((nn0), ),+ then then21 + P P…((n n+ + + 2 1)” n1)”-1) + 2n = 2n – 1 + 2n ToTo dodo this,this, we'llwe'll choosechoose= 2(2 anann ) arbitrary arbitrary– 1 nn,, assumeassume that P(n) holds, then tryn+1 to prove P(n + 1). = 2 – 1 that P(n) holds, then try to prove P(n + 1). Thus P(n + 1) is true, completing the induction. ■ Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. For our base case, we need to show P(0) is true, meaning the sum of the first zero powers of two is 20 – 1. Since the sum of the first zero powers of two is 0 = 20 – 1, we see P(0) is true. For the inductive step, assume that for some n ∈ ℕ that P(n) holds, meaning that 20 + 21 + … + 2n-1 = 2n – 1. We need to show that P(n + 1) holds, meaning that the sum of the first n + 1 powers of two is 2n+1 – 1. Consider the sum of the first n + 1 powers of two. This is the sum of the first n powers of two, plus 2n. Using the inductive hypothesis,Here,Here, we'rewe're we explicitlyexplicitly see that statingstating PP((nn ++ 1),1), 20 + 21 +whichwhich … + is is2 nwhat-1what + 2 wenwe = want want(20 + to to2 1 prove.+prove. … + 2 Now,nNow,-1) + 2n n n wewe cancan useuse anyany proof=proof 2 – technique1technique + 2 wewe wantwant n to try= to2(2 prove) – 1 it. to try ton+1 prove it. = 2 – 1 Thus P(n + 1) is true, completing the induction. ■ Theorem: The sum of the first n powers of two is 2n – 1. Proof: By induction. Let P(n) be “the sum of the first n powers of two is 2n – 1.” We will show P(n) is true for all n ∈ ℕ. 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