CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett
Today’s Topics: 1. Mathematical Induction Proof Examples 2
1. Mathematical Induction 3
Mathematical induction 4
Induction “For all integers n >= a, P(n).” Base case - push first domino Inductive step – n th domino pushes the n+1 th 5
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=____. ____________________________________________ Inductive step: Assume [or “Suppose”] that __________________. WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 6
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=____. ____________________________________________ Inductive step: Assume [or “Suppose”] that __________________. WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 7 A.1 B.2 C.k D.k+1 E.Other
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1___. ___1=1*(1+1)/2_________________________________ Inductive step: Assume [or “Suppose”] that __________________. WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 8
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1___. ____ 1=1*(1+1)/2 ________________________________ Inductive step: Assume [or “Suppose”] that __________________. WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 9 For the inductive step, we want to prove that IF the theorem is true for some n >=[basis], THEN the theorem is true for n+1. How do we prove an implication p→q? A.Assume p, WTS ¬q (“p and not q”). B.Assume p, WTS q. C.Assume q, WTS p. D.Assume p→q, show it does not lead to contradiction.
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1__ _. ____ 1=1*(1+1)/2 ________________________________ Inductive step: Assume [or “Suppose”] that __________________. WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 10 A.Assume n>=1 B.Assume that for all n>=1 ----> C.Assume that for some n>=1,->
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1___. ____ 1=1*(1+1)/2 ________________________________ Inductive step: Assume [or “Suppose”] that for some n>1, WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 11
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1___. ____ 1=1*(1+1)/2 ________________________________ Inductive step: Assume [or “Suppose”] that for some n>1, WTS: ____________________. _____________________________________________ So the inductive step holds, completing the proof. 12 A.The negation is true. B.The theorem is true for some integer k+1. C.The theorem is true for n+1. D.The theorem is true for some intege n>=1
Thm: Proof (by mathematical induction): Basis step: Show the theorem holds for n=1___. ____ 1=1*(1+1)/2 ________________________________ Inductive step: Assume [or “Suppose”] that for some n>1, WTS: The theorem is true for n+1. _____________________________________________ So the inductive step holds, completing the proof. 13
Thm: 14
Mathematical induction Want to prove “For all integers n >= a, P(n).” Base case: verify for n=a (usually by a simple direct calculation) Main step: Prove that P(n) P(n+1) 15
Mathematical induction 16 P(1)P(2)P(3)P(4)P(5) …
Mathematical induction 17 P(1)P(2)P(3)P(4)P(5) …
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: Inductive case: Assume… WTS… Proof… 18
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: Inductive case: Assume… WTS… Proof… 19 A.Theorem is true for all n. B.Thereom is true for n=0. C.Theorem is true for n>0. D.Theorem is true for n=1.
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume… WTS… Proof… 20
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume… WTS… Proof… 21 A.Theorem is true for some set B.Thereom is true for all sets C.Theorem is true for all sets of size n. D.Theorem is true for some set of size n.
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume theorem is true for all sets of size n. WTS… Proof… 22
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume theorem is true for all sets of size n. WTS… Proof… 23 A.Theorem is true for some set of size >n. B.Thereom is true for all sets of size >n. C.Theorem is true for all sets of size n+1. D.Theorem is true for some set of size n+1.
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume theorem is true for all sets of size n. WTS: Theorem is true for all sets of size n+1. Proof… 24
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume theorem is true for all sets of size n. WTS: Theorem is true for all sets of size n+1. Proof. Try by yourself first! 25
Another example Theorem: if |A|=n then |P(A)|=2 n. Proof by induction on n. Base case: n=0. So A= . Then P(A)={ } so |P(A)|=1=2 0. Inductive case: Assume theorem is true for all sets of size n. WTS: Theorem is true for all sets of size n+1. Proof… 26 Let A={a 1,…,a n+1 } be a set of size n+1. Define B={S A: a n+1 S} and C={S A: a n+1 S}. P(A)=B C and B C= hence |P(A)|=|B|+|C|. Also, |B|=|C| since we can define a bijective function f:B C by f(S)=S {a n+1 }. So, |P(A)|=2|B|. However, B=P({a 1,…,a n }) so by induction, |B|=2 n. We conclude that |P(A)|=2 2 n = 2 n+1. QED.