Discrete Mathematics Tutorial 11 Chin [email protected].
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Transcript of Discrete Mathematics Tutorial 11 Chin [email protected].
Discrete MathematicsTutorial 11
Induction
• Suppose– P(1) is true, and– If P(n) is true, P(n+1) is also true
• Then P(n) is true for every n ≥ 1654321
Induction
• Suppose I proved that1. P(1) and P(2) are true.2. If P(n) is true, then P(n+2) is true.
• Can I conclude that– P(n) is true for every n?
• Suppose I proved that1. P(1) and P(2) are true.2. If P(n) is true, then P(n+2) is true.
• Two stack of dominoes– P(1)→P(3)→P(5)→…– P(2)→P(4)→P(6) →…
Induction
...7531
...8642
Induction
• Suppose I proved that1. P(1, 1) is true.2. If P(n, m) is true, then P(n+1, m+1) is true.
• Can I conclude that– P(n, m) is true for every n, m?
Induction
• Suppose I proved that1. P(1, 1) is true.2. If P(n, m) is true, then P(n+1, m+1) is true.
• What about P(2,1)? (1,1)
(2,1) (2,2)
(3,1) (3,3)
(4,1) (4,4)
(5,1) (5,5)
(6,1) (6,6)
(7,1) (7,7)
Induction
• Suppose I proved that1. P(1, 1) is true.2. If P(n, m) is true, then P(n+1, m) is true.3. If P(n, m) is true, then P(n-1, m+1) is true.
• Can I conclude that– P(n, m) is true for every n, m?
Induction
• Suppose I proved that1. P(1, 1) is true.2. If P(n, m) is true, then P(n+1, m) is true.3. If P(n, m) is true, then P(n-1, m+1) is true.
(1,1) (1,2) (1,3) (1,4) (1,5)
(2,1) (2,2) (2,3) (2,4)
(3,1) (3,2) (3,3)
(4,1) (4,2)
(5,1)
(6,1)
(7,1)
2,33,2
4,11,3
2,23,11,2
2,11,1
Induction
• Show that∑ i2 = n(2n+1)(n+1)/6n
i=1
Induction
• Show that∑ i2 = n(2n+1)(n+1)/6
i.e. 12 + 22 + 32 + … + n2 = n(2n+1)(n+1)/6
• Base case:– when n = 1– L.H.S. = 1– R.H.S. = 1 x 3 x 2 / 6 = 1
n
i=1
Induction
• Assume P(n) is true:∑ i2 = n(2n+1)(n+1)/6
i.e. 12 + 22 + 32 + … + n2 = n(2n+1)(n+1)/6• Then we need to show P(n+1) is also true:
∑ i2 = (n+1)(2(n+1)+1)((n+1)+1)/6
i.e. 12 + 22 + … + n2 + (n+1)2 = (n+1)(2n+3)(n+2)/6• How to use the induction assumption?
n
i=1
n+1
i=1
Induction
• Then we need to show P(n+1) is also true:∑ i2 =(n+1)(2n+3)(n+2)/6
i.e. 12 + 22 + … + n2 + (n+1)2 = (n+1)(2n+3)(n+2)/6• How to use the induction assumption?
∑ i2 = ∑ i2 + (n+1)2
∑ i2 = n(2n+1)(n+1)/6 + (n+1)2
n+1
i=1
n+1
i=1
n
i=1
n+1
i=1
Induction
• Now we have∑ i2 = n(2n+1)(n+1)/6 + (n+1)2
• Our goal is to make R.H.S. equal(n+1)(2n+3)(n+2)/6
• Just expand everything out and factorize
n+1
i=1
Induction
∑ i2
= n(2n + 1)(n + 1)/6 + (n + 1)2
= (2n3 + 3n2 + n)/6 + (n2 + 2n + 1)= ((2n3 + 3n2 + n) + (6n2 + 12n + 6)) /6= (2n3 + 9n2 + 13n + 6)/6
How to factor?
n+1
i=1
Induction
∑ i2
= (2n3 + 9n2 + 13n + 6)/6
= (n + 1)(2n2 + 7n + 6)/6= (n + 1)(2n + 3)(n + 2)/6
n+1
i=1
• How to factor?• Remember, our goal is to make R.H.S. equal
(n+1)(2n+3)(n+2)/6
• So P(n+1) is also true. Done.
Strong induction
• Suppose– P(1) is true, and– If P(s) is true for every s < n, P(n+1) is also true
• Then P(n) is true for every n ≥ 1
65 4
32 1
Strong induction
• Prove thatTo divide up a chocolate bar with m x n squares, we need at least mn - 1 splits
5 = 2x3 - 1 splits
Strong induction
• Prove thatTo divide up a chocolate bar with m x n squares, we need at least mn - 1 splits
• We prove a stronger statementP(s) := To divide up a chocolate bar with s squares, we need at least s - 1 splits
Strong Induction
• Prove thatP(s) := To divide up a chocolate bar with s squares, we need at least s-1 splits
• Base case P(1):– No need to split.
Strong Induction
• Prove thatP(s) := To divide up a chocolate bar with s squares, we need at least s-1 splits
• Induction step:– (If P(k) is true for every k < s)– Assume for any chocolate bar with 1 ≤ k < s squares,
we need at least k – 1 splits.
Strong induction
• Induction step (If P(k) is true for every k < s) :– Assume for any chocolate bar with 1 ≤ k < s squares,
we need at least k – 1 splits.
• Given a chocolate bar with s squaresSplit it into two smaller bars with i and j squares, i + j = s
j squares
i squares
s squares
Strong induction
• By induction assumption (P(k) is true for any k < s)
• Small chocolate bar with i < s squares– needs at least i-1 splits for one of the two bars
• Small chocolate bar with j < s squares– needs at least j-1 splits for one of the two bars
• Therefore we need (i-1) + (j-1) + 1 = s-1 splits– because i + j = s
• P(s) is true. Done.j-1 splits
i-1 splits
1 split
Well-ordering principle
• Every nonempty set of nonnegative integers has a least element
• This is equivalent to Mathematical Induction:– If MI is true, then we can prove WOP is also true.– If WOP is true, then we can prove MI is also true.
Well-ordering principle
• Prove by well-ordering principle that∑ i = n(n+1)/2n
i=1
Well-ordering principle
• Prove by well-ordering principle that∑ i = n(n+1)/2
• Prove by contradiction• Suppose there exist some m, ∑ i ≠ m(m+1)/2
• Let S be the set containing all such m– S is nonempty (exists at least one)– S is a set of nonnegative integers
n
i=1
m
i=1
Well-ordering principle
• Let S be the set containing all the m∑ i ≠ m(m+1)/2
– S is nonempty (exists at least one)– S is a set of nonnegative integers
• By well-ordering principle, – there exists a least element m’ in S– i.e. m’ is the smallest number such that
∑ i ≠ m’(m’+1)/2
m
i=1
m’
i=1
Well-ordering principle
• m’ is the smallest number such that∑ i ≠ m’(m’+1)/2
• This means for any 0 ≤ n < m’∑ i = n(n+1)/2
m’
i=1
n
i=1
Well-ordering principle
• This means for any 0 ≤ n < m’∑ i = n(n+1)/2
• This is true when n = 0, so – m’ > 0– m’-1 is non-negative and m’-1 < m’, so
∑ i = (m’-1)((m’-1)+1)/2
n
i=1
m’-1
i=1
Well-ordering principle
∑ i = (m’-1)((m’-1)+1)/2
∑ i = ∑ i + m’ = (m’-1)m’/2 + m’ = m’(m’+1)/2
• But∑ i ≠ m’(m’+1)/2
• Contradiction
m’-1
i=1
m’
i=1
m’
i=1
m’-1
i=1
Well-ordering principle
• Show thata2 + b2 = 3(s2 + t2)
has no non-zero integer solutions.
Well-ordering principle
• Suppose it has non-zero integer solutions.
• Let S be the collection of (a, b, s, t) such that – (a, b, s, t) ≠ (0, 0, 0, 0), and– a2 + b2 = 3(s2 + t2)
• S is nonempty, by well-ordering principle– there is a least element (a1, b1, s1, t1) such that
a12 + b1
2 = 3(s12 + t1
2)
Well-ordering principle
• S is nonempty, by well-ordering principle– there is a least element (a1, b1, s1, t1) such that
a12 + b1
2 = 3(s12 + t1
2)
• Right idea, but…– S is a set containing 4-tuples (a, b, s, t)– not a set of non-negative integers
• Cannot apply the well-ordering principle
Well-ordering principle
• Suppose there exists (a, b, s, t) ≠ (0, 0, 0, 0), a2 + b2 = 3(s2 + t2)
• Let S be the collection of |a| such that there exist b, s and t
a2 + b2 = 3(s2 + t2)• S is a nonempty set of non-negative integers• By the well-ordering principle– There exists an a1 in S such that |a1| is smallest
Well-ordering principle
• Let S be the collection of |a| such that there exist b, s and t
a2 + b2 = 3(s2 + t2)• By the well-ordering principle– There exists an a1 in S such that |a1| is smallest
• And for this smallest |a1|, there exist b1, s1 and t1,
a12 + b1
2 = 3(s12 + t1
2)
Well-ordering principle
• And for this smallest |a1|, there exist b1, s1 and t1,
a12 + b1
2 = 3(s12 + t1
2)
• This means– a1
2 + b12
is a multiple of 3
– a1 and b1 are both multiples of 3
• proof by contrapositive
– a1= 3a2 and b1 = 3b2 for some a2 and b2
Well-ordering principle
• And for this smallest |a1|, there exist b1, s1 and t1,
a12 + b1
2 = 3(s12 + t1
2)
• a1 = 3a2 and b1 = 3b2 for some a2 and b2
– (3a2)2 + (3b2)2 = 3(s12 + t1
2)
– 9a22 + 9b2
2 = 3(s12 + t1
2)
– 3(a22 + b2
2) = s12 + t1
2
– s12 + t1
2 = 3(a22 + b2
2)
• therefore (s1, t1, a2, b2) is also a solution
Well-ordering principle
• There is a least element (a1, b1, s1, t1)a1
2 + b12 = 3(s1
2 + t12)
• We have just showed – (s1, t1, a2= a1/3, b2 = b1/3) is also a solution
• Repeat the argument– (a2=a1/3, b2=b1/3, s2=s1/3, t2=t1/3) is also solution
• But |a2| < |a1|, contradiction.
Invariant Method
• The numbers 1, 2, 3, 4 and 5 are written on a board
• Repeat the following until there is only one number left:– pick any two of the numbers– erase them– write the absolute value of their difference on board
• Can the remaining number be 2?
Invariant Method
• Example
• 1 2 3 4 5 → 2 3 4 4• 2 3 4 4 → 2 4 1• 2 4 1 → 2 3• 2 3 → 1
Invariant Method
• Observe
• 1 2 3 4 5 (1 + 2 + 3 + 4 + 5 = 15)• 2 3 4 4 (2 + 3 + 4 + 4 = 13)• 2 4 1 (2 + 4 + 1 = 7)• 2 3 (2 + 3 = 5)• 1 (1)
• What is the pattern?
Invariant Method
• Observation– can only get an odd number in the final answer– total sum of the numbers on the board is always odd
• Suppose I pick m and n on the board– Change in total sum = m + n - (m - n) = 2n– the change in total sum must be even– Therefore the total sum is always odd
Invariant Method
• At the beginning the total sum is 15– which is odd
• Therefore the total sum is always odd
• Impossible to get 2 as final answer– In fact, impossible to get any even number as final
answer
End
• Questions?