Discrete Mathematics: Proof by Induction and Set Sizes
Today's topics cover proof by induction, set sizes equality, and paying with 3 and 5 cent coins in a discrete math class. Learn about the theorems and techniques used to prove sets are of equal size.
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Clicker frequency: CA CSE 20 DISCRETE MATH Prof. Shachar Lovett http://cseweb.ucsd.edu/classes/wi15/cse20-a/
Todays topics Proof by induction: Set sizes equality Paying using 3 and 5 cent coins Section 3.6 in Jenkyns, Stephenson
Set sizes We saw last time the following theorem, useful to prove that two sets have equal size Theorem: let X,Y be finite sets. Assume there is a function ?:? ? which is both one-to-one and onto. Then |X|=|Y|. We will prove it using induction on |X|.
Set sizes Theorem: let X,Y be finite sets. Assume there is a function ?:? ? which is both one-to-one and onto. Then |X|=|Y|. Prove by induction on n=|X|.
Set sizes Theorem: let X,Y be finite sets. Assume there is a function ?:? ? which is both one-to-one and onto. Then |X|=|Y|. Prove by induction on n=|X|. Base case: n=0, so ? = . If Y is not empty, then f is not onto, so we must have Y = and hence |X|=|Y|=0.
Set sizes Theorem: let X,Y be finite sets. Assume there is a function ?:? ? which is both one-to-one and onto. Then |X|=|Y|. Prove by induction on n=|X|. Base case: n=0, so ? = . If Y is not empty, then f is not onto, so we must have Y = and hence |X|=|Y|=0. Inductive step: assume it is true for all sets of size n. Prove it is true when |X|=n+1.
Set sizes Inductive step: let |?| = ? + 1. Choose ? ? and let ? = ? ? ?. Define: ? = ? {? },? = ? {? }. We have ? = ?. We will prove that ? = ? and hence ? = ? + 1. First, we claim that ? ? ,? ? ? . If this is false for some ? ? , then since ? ? ? we must have ? ? ? ? = ? , so ? ? = ? = ? ? which is impossible since f is one-to-one. So, we can define ? :? ? by ? ? = ? ? . It is one-to-one because f is, and is onto because they only element missing is ? ? which is not in ? . Hence we can apply the inductive assumption and conclude that ? = ? and hence ? = ? + 1.
3-cent and 5-cent coins We will prove the following theorem Theorem: For all prices p 8 cents, the price p can be paid using only 5-cent and 3-cent coins 1851-1889 1866-today
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=____. Inductive step: Assume [or Suppose ] that WTS that So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=________. Inductive step: Assume [or Suppose ] that A. 0 cents B. 1 cent C. 2 cents D. 3 cents E. Other/none/more than one WTS that So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that WTS that So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that A. Theorem is true for p=8. B. Theorem is true for some p>8. C. Theorem is true for some p 8. D. Theorem is true for some p>0. E. Theorem is true for all p>8. WTS that So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that theorem is true for some p 8. WTS that So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that theorem is true for some p 8. WTS that A. Theorem is true for p=8. B. Theorem is true for some p>8. C. Theorem is true for p+1. D. Theorem is true for p+8. So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that theorem is true for some p 8. WTS that theorem is true for price p+1. So the inductive step holds, completing the proof.
Thm: For all prices p 8 cents, the price p can be paid using only 5-cent and 3- cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for price p=8. Inductive step: Assume [or Suppose ] that theorem is true for some p 8. WTS that theorem is true for price p+1. ??? So the inductive step holds, completing the proof.
3-cent and 5-cent coins Inductive step: Assume price p 8 can be paid using only 3-cent and 5-cent coins. Need to prove that price p+1 can be paid using only 3-cent and 5- cent coins. Main idea: reduce from price p+1 to price p.
Making change If we have 100 5-cent coins, and 100 3-cent coins (for a total of p = $8.00), how can we modify the number of 5- cent and 3-cent coins so that we can make the p+1 price (p+1 = $8.01)? A. 40 5-cent coins + 200 3-cent coins B. 39 5-cent coins + 202 3-cent coins C. 99 5-cent coins + 102 3-cent coins
Making change If we have 100 5-cent coins, and 100 3-cent coins (for a total of p = $8.00), how can we modify the number of 5- cent and 3-cent coins so that we can make the p+1 price (p+1 = $8.01)? A. 40 5-cent coins + 200 3-cent coins B. 39 5-cent coins + 202 3-cent coins C. 99 5-cent coins + 102 3-cent coins
Turning our modification scheme into a generic algorithm If we have n 5-cent coins, and m 3-cent coins (for a total of p = 5n+3m), how can we modify the number of 5-cent and 3-cent coins so that we can make the p+1 price (p+1 = 5n+3m+1)? A. n+1 5-cent coins + m-2 3-cent coins B. n-1 5-cent coins + m+2 3-cent coins C. n+1 5-cent coins + m+2 3-cent coins D. No generic way
Turning our modification scheme into a generic algorithm If we have n 5-cent coins, and m 3-cent coins (for a total of p = 5n+3m), how can we modify the number of 5-cent and 3-cent coins so that we can make the p+1 price (p+1 = 5n+3m+1)? A. n+1 5-cent coins + m-2 3-cent coins B. n-1 5-cent coins + m+2 3-cent coins C. n+1 5-cent coins + m+2 3-cent coins D. No generic way
What if we dont have any 5-cent coins to subtract?? If we have 0 5-cent coins, and m 3-cent coins (for a total of p = 3m), how can we modify the number of 5-cent and 3-cent coins so that we can make the p+1 price (p+1 = 3m+1)? A. You can t B. You can [explain to your group how]
What if we dont have any 5-cent coins to subtract?? If we have 0 5-cent coins, and m 3-cent coins (for a total of p = 3m), how can we modify the number of 5-cent and 3-cent coins so that we can make the p+1 price (p+1 = 3m+1)? Remove three 3-cent coins, add two 5-cent So: two 5-cent coins, m-3 3-cent points, for a total of 2*5+3*(m-3)=3m+1=p+1
That algorithm relies on being able to subtract three 3-cent coins. What if we don t have that many? (only 1 or 2?) A. Uh-oh, our proof can not work as we ve done it so far B. That could never happen [explain why not] C. That could happen, and we need to make a 3rd (or more) case(s) to handle it
Thm: For all prices p >= 8 cents, the price p can be paid using only 5-cent and 3-cent coins. Proof (by mathematical induction): Basis step: Show the theorem holds for p=8 (by example, e.g. p=3+5) Inductive step: Assume [or Suppose ] that the theorem holds for some p 8. WTS that the theorem holds for p+1. p 8. Assume that p=5n+3m where n,m 0 are integers. We need to show that p+1=5a+3b for integers a,b 0. Partition to cases: Case I: n 1. In this case, p+1=5*(n-1)+3*(m+2). Case II: m 3. In this case, p+1=5*(n+2)+3*(m-3). Case III: n=0 and m 2. Then p=5n+3m 6 which is a contradiction to p 8. So the inductive step holds, completing the proof.
We created an algorithm! Our proof actually allows us to algorithmically find a way to pay p using 3-cent and 5-cent coins Algorithm for price p: start with x=8=3+5 For x=8...p, in each step adjust the number of coins according to the modification rules we ve constructed to maintain price x
Algorithm pseudo-code PayWithThreeCentsAndFiveCents: Input: price p 8. Output: integers n,m 0 so that p=5n+3m 1. Let x=8, n=1, m=1 (so that x=5n+3m). 2. While x<p: a) x:=x+1 b) If n 1, set n:=n-1, m:=m+2 c) Otherwise, set n:=n+2, m:=m-3 3. Return (n,m)
Algorithm pseudo-code PayWithThreeCentsAndFiveCents: Input: price p 8. Output: integers n,m 0 so that p=5n+3m 1. Let x=8, n=1, m=1 (so that x=5n+3m). 2. While x<p: a) x:=x+1 b) If n 1, set n:=n-1, m:=m+2 c) Otherwise, set n:=n+2, m:=m-3 3. Return (n,m) Invariant: x=5n+3m Invariant: x=5n+3m We proved that n,m 0 in this process always; this is not immediate from the algorithm code
Algorithm run example x=8: n=1, m=1 While x<p: a) b) c) 8= Invariant: x=5n+3m x:=x+1 If n 1, set n:=n-1, m:=m+2 Otherwise, set n:=n+2, m:=m-3 9= 10 = 11= 12 =
Algorithm properties Theorem: Algorithm uses at most two nickels (i.e n 2) Proof: by induction on p Try to prove it yourself first! x=8: n=1, m=1 While x<p: a) b) c) Invariant: x=5n+3m x:=x+1 If n 1, set n:=n-1, m:=m+2 Otherwise, set n:=n+2, m:=m-3
x=8: n=1, m=1 While x<p: a) b) c) Invariant: x=5n+3m Algorithm properties x:=x+1 If n 1, set n:=n-1, m:=m+2 Otherwise, set n:=n+2, m:=m-3 Theorem: Algorithm uses at most two nickels (i.e n 2). Proof: by induction on p Base case: p=8. Algorithm outputs n=m=1. Inductive hypothesis: p=5n+3m where n 2. WTS p+1=5a+3b where a 2. Proof by cases: Case I: n 1. So p+1=5(n-1)+3(m+2) and a=n-1 2. Case II: n=0. So p+1=5*2+3(m-3). a=2. In both cases p+1=5a+3b where a 2. QED
Next class Strong induction Read section 3.6.1 in Jenkyns, Stephenson
