1. 1 Advanced Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

2. 2 e.g.1 (Page 5) Let n be a non-negative power of 2. Suppose that T(n)  2T(n/2) + cn for some constant c. T(n) > 0 Prove that T(n) = O(n log n) Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n)

3. 3 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n nn0n0 T(n) k. n log n How can we find the value for n 0 ? How can we find the value for k? The above equation can be proved by our traditional induction if we know the value for n 0 and k. How can we prove it?

4. 4 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n log n……………..(*) ” We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. Consider P(1) RHS of (*) = k. 1 log 1 We want to see whether P(1) is true. (i.e., “T(1)  k. 1 log 1……………..(*)”) = k. 1. 0 = 0 In other words, according to (*), we need to prove T(1)  0 Since T(n) > 0, we have T(1) > 0 We conclude that P(1) is false P(1) is false. Thus, it is not possible to prove that T(1)  0 Since we don ’ t want “ n = 1 ” to happen, I want that n is larger (e.g., n = 2, 2 2, … ) n 0  1 Since n > n 0, we have “ n 0  1 ”

5. 5 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n log n……………..(*)” We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. Consider P(2) RHS of (*) = k. 2 log 2 We want to see whether P(2) is true. (i.e., “T(2)  k. 2 log 2……………..(*)”) = k. 2. 1 = 2k In other words, according to (*), we need to prove T(2)  2k Currently, we don ’ t know the value of k. P(1) is false. P(2) is true. In order to prove T(2)  2k, we should set a value of k such that k  T(2)/2 k  T(2)/2 Thus, P(2) is true. n 0  1

6. 6 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n log n……………..(*)” P(1) is false. P(2) is true. k  T(2)/2 We guess that P(2), P(2 2 ), P(2 3 ) … are also true. Thus, the base case is when n = 2. P(2) n 0  1

7. 7 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n Suppose that n > n 0. Let P(n) be “T(n)  k. n log n……………..(*)” P(1) is false. P(2) is true. k  T(2)/2 Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 2. Step 2(a): Assume that P(n/2) is true for n > 2. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k. (n/2) log (n/2) Consider T(n) We want to show that T(n)  k. n log n n 0  1  2T(n/2) + cn  2. (k. (n/2) log (n/2)) + cn = k. n log (n/2) + cn = k. n log n – k. n log 2 + cn = k. n log n – kn + cn Currently, we don ’ t know the value of k. In order to prove T(n)  k. n log n, we should set a value of k such that – kn + cn  0 (i.e., k  c) Thus, P(n) is true. P(n) is true. k  c

8. 8 e.g.1 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that T(n) = O(n log n) We want to prove that T(n) = O(n log n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n log n Suppose that n > n 0. Let P(n) be “T(n)  k. n log n……………..(*)” P(1) is false. P(2) is true. k  T(2)/2 Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 2. Step 2(a): Assume that P(n/2) is true for n > 2. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k. (n/2) log (n/2) We want to show that T(n)  k. n log n n 0  1 P(n) is true. k  c We prove that “ P(n/2)  P(n) ” is true for all n > 2 By Mathematical Induction,  n  2, T(n)  k. n log n We should set some values for n 0 and k. Since n 0  1, we set n 0 = 1. Since k  T(2)/2 and k  c, we set k = max{c, T(2)/2}

9. 9 e.g.2 (Page 11) Consider a traditional induction proof. Let n be a non-negative power of 2. Suppose that T(n)  2T(n/2) + cn for some constant c. T(n) > 0 Prove that T(n) = O(n log n) Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that  n > 1, T(n)  k. n log n where k = max{c, T(2)/2}  n > 1, T(n)  k. n log n where k = max{c, T(2)/2}

10. 10 e.g.2 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that  n > 1, T(n)  k. n log n where k = max{c, T(2)/2} Step 1: Prove that P(2) (i.e., the base case) is true. We want to show that “ T(2)  k. 2 log 2 where k = max{c, T(2)/2} ” Consider T(2)= T(2)/2. 2 (i.e., “T(2)  k. 2 where k = max{c, T(2)/2} ”)  k. 2 Thus, P(2) is true. (Since k = max{c, T(2)/2}, k  T(2)/2)

11. 11 e.g.2 Let n be a non-negative power of 2. T(n)  2T(n/2) + cn T(n) > 0 Prove that  n > 1, T(n)  k. n log n where k = max{c, T(2)/2} Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 2. Step 2(a): Assume that P(n/2) is true for n > 2. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k. (n/2) log (n/2) Consider T(n) We want to show that T(n)  k. n log n  2T(n/2) + cn  2. (k. (n/2) log (n/2)) + cn = k. n log (n/2) + cn = k. n log n – k. n log 2 + cn = k. n log n – kn + cn  k. n log n (Since k = max{c, T(2)/2}, k  c and thus –kn + cn  0) Thus, P(n) is true. We prove that “ P(n/2)  P(n) ” is true for all n > 2 By Mathematical Induction,  n  2, T(n)  k. n log n

12. 12 e.g.3 (Page 12) Traditional Induction Proof The values of n 0 and k are given Advanced Induction Proof The values of n 0 and k are to be found when we are proving. Advantage: We can use the original induction for the proof. Disadvantage: The values of n 0 and k look “ strange ”. We don ’ t know the reason why we have these “ strange ” values. Advantage: The values of n 0 and k are found during the proof. Thus, we know why the values of n 0 and k are generated. Disadvantage: It looks more complicated during the proof.

13. 13 e.g.4 (Page 13) Let n be a non-negative power of 3. Suppose that T(n)  T(n/3) + cn for some constant c. T(n) > 0 Prove that T(n) = O(n) Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n)

14. 14 e.g.4 We want to prove that T(n) = O(n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n nn0n0 T(n) k.nk.n How can we find the value for n 0 ? How can we find the value for k? The above equation can be proved by our traditional induction if we know the value for n 0 and k. How can we prove it? Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n)

15. 15 e.g.4 We want to prove that T(n) = O(n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n ……………..(*)” We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. Consider P(1) RHS of (*) = k. 1 We want to see whether P(1) is true. (i.e., “T(1)  k. 1 ……………..(*)”) In other words, according to (*), we need to prove T(1)  k Currently, we don ’ t know the value of k. P(1) is true. In order to prove T(1)  k, we should set a value of k such that k  T(1) k  T(1) Thus, P(1) is true. I guess n can be 1, 3, 3 2, … Since n > n 0, we have “ n 0  0 ” n 0  0 Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n) = k

16. 16 e.g.4 We want to prove that T(n) = O(n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n ……………..(*)” We guess that P(1), P(3 1 ), P(3 2 ) … are also true. Thus, the base case is when n = 1. P(1) P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n)

17. 17 e.g.4 We want to prove that T(n) = O(n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n Suppose that n > n 0. Let P(n) be “T(n)  k. n ……………..(*)” Step 2: Prove that “ P(n/3)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/3) is true for n > 1. Step 2(b): According to P(n/3), we deduce that P(n) is true. That is, T(n/3)  k. (n/3) Consider T(n) We want to show that T(n)  k. n  T(n/3) + cn  k. (n/3) + cn = k. n + (c – 2k/3)n Currently, we don ’ t know the value of k. In order to prove T(n)  k. n, we should set a value of k such that c – 2k/3  0 (i.e., k  3c/2) Thus, P(n) is true. P(n) is true. k  3c/2 P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n) = k. n – 2kn/3 + cn

18. 18 e.g.4 We want to prove that T(n) = O(n) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n Suppose that n > n 0. Let P(n) be “T(n)  k. n ……………..(*)” Step 2: Prove that “ P(n/3)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/3) is true for n > 1. Step 2(b): According to P(n/3), we deduce that P(n) is true. That is, T(n/3)  k. (n/3) We want to show that T(n)  k. n We prove that “ P(n/3)  P(n) ” is true for all n > 1 By Mathematical Induction,  n  1, T(n)  k. n We should set some values for n 0 and k. Since n 0  0, we set n 0 = 0. Since k  T(1) and k  3c/2, we set k = max{3c/2, T(1)} P(n) is true. k  3c/2 P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that T(n) = O(n)

19. 19 e.g.5 (Page 15) Consider a traditional induction proof. Let n be a non-negative power of 3. Suppose that T(n)  T(n/3) + cn for some constant c. T(n) > 0 Prove that T(n) = O(n) Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that  n > 0, T(n)  k. n where k = max{3c/2, T(1)}  n > 0, T(n)  k. n where k = max{3c/2, T(1)}

20. 20 e.g.5 Step 1: Prove that P(1) (i.e., the base case) is true. We want to show that “ T(1)  k. 1 where k = max{3c/2, T(1)} ” Consider T(1)= T(1). 1  k. 1 Thus, P(1) is true. Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that  n > 0, T(n)  k. n where k = max{3c/2, T(1)} (since k = max{3c/2, T(1)}, k  T(1))

21. 21 e.g.5 Step 2: Prove that “ P(n/3)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/3) is true for n > 1. Step 2(b): According to P(n/3), we deduce that P(n) is true. That is, T(n/3)  k. (n/3) Consider T(n) We want to show that T(n)  k. n  T(n/3) + cn  k. (n/3) + cn = kn/3 + cn = kn – 2kn/3 + cn = kn + (c – 2k/3)n  kn (Since k = max{3c/2, T(1)}, k  3c/2 and thus c – 2k/3  0) Thus, P(n) is true. We prove that “ P(n/3)  P(n) ” is true for all n > 1 By Mathematical Induction,  n  1, T(n)  k. n Let n be a non-negative power of 3. T(n)  T(n/3) + cn T(n) > 0 Prove that  n > 0, T(n)  k. n where k = max{3c/2, T(1)}

22. 22 e.g.6 (Page 16) Let n be a non-negative power of 2. Suppose that T(n)  4T(n/2) + cn for some positive constant c. T(n) > 0 Prove that T(n) = O(n 2 ) Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 )

23. 23 e.g.6 We first show an “inappropriate” proof After that, we show an “appropriate” proof Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 )

24. 24 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n 2 nn0n0 T(n) k.n2k.n2 How can we find the value for n 0 ? How can we find the value for k? The above equation can be proved by our traditional induction if we know the value for n 0 and k. How can we prove it? Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 )

25. 25 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n 2 Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n 2 ……………..(*)” We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. Consider P(1) RHS of (*) = k. 1 2 We want to see whether P(1) is true. (i.e., “T(1)  k. 1 2 ……………..(*)”) In other words, according to (*), we need to prove T(1)  k Currently, we don ’ t know the value of k. P(1) is true. In order to prove T(1)  k, we should set a value of k such that k  T(1) k  T(1) Thus, P(1) is true. I guess n can be 1, 2, 2 2 … Since n > n 0, we have “ n 0  0 ” n 0  0 Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) = k

26. 26 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n 2 Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k. n 2 ……………..(*)” We guess that P(1), P(2 1 ), P(2 2 ) … are also true. Thus, the base case is when n = 1. P(1) P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 )

27. 27 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n 2 Suppose that n > n 0. Let P(n) be “T(n)  k. n 2 ……………..(*)” Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/2) is true for n > 1. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k. (n/2) 2 Consider T(n) We want to show that T(n)  k. n 2  4T(n/2) + cn  4. k. (n/2) 2 + cn = 4. k. n 2 /4 + cn In order to prove T(n)  k. n 2, we should have the following cn  0 The inequality “ cn  0 ” is impossible. P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) = k. n 2 + cn Note that c is positive and n is positive

28. 28 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k  R such that  n > n 0, T(n)  k. n 2 Suppose that n > n 0. Let P(n) be “T(n)  k. n 2 ……………..(*)” The inequality “ cn  0 ” is impossible. P(1) is true. k  T(1) n 0  0 Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) Note that c is positive and n is positive What went wrong? Statement P(n) is too weak to be proved by induction. To fix this, let’s see if we can prove something that is actually stronger that we were originally trying to prove – namely T(n)  k 1. n 2 – k 2 n for some positive constants k 1 and k 2

29. 29 e.g.6 Now, we show an “appropriate” proof.

30. 30 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k 1, k 2  R such that  n > n 0, T(n)  k 1 n 2 – k 2 n nn0n0 T(n) k 1 n 2 – k 2 n How can we find the value for n 0 ? How can we find the value for k 1 and k 2 ? The above equation can be proved by our traditional induction if we know the value for n 0, k 1 and k 2. How can we prove it? Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 )

31. 31 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k 1, k 2  R such that  n > n 0, T(n)  k 1 n 2 – k 2 n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k 1 n 2 – k 2 n ……………..(*)” We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. Consider P(1) RHS of (*) = k 1. 1 2 – k 2. 1 We want to see whether P(1) is true. (i.e., “T(1)  k 1. 1 2 – k 2. 1 ……………..(*)”) In other words, according to (*), we need to prove T(1)  k 1 – k 2 Currently, we don ’ t know the values of k 1 and k 2. P(1) is true. In order to prove T(1)  k 1 – k 2, we should set values of k 1 and k 2 such that k 1 – k 2  T(1) k 1 – k 2  T(1) Thus, P(1) is true. I guess n can be 1, 2, 2 2, … Since n > n 0, we have “ n 0  0 ” n 0  0 Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) = k 1 – k 2

32. 32 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k 1, k 2  R such that  n > n 0, T(n)  k 1 n 2 – k 2 n Step 1: Prove that P( ? ) (i.e., the base case) is true. Suppose that n > n 0. Let P(n) be “T(n)  k 1 n 2 – k 2 n ……………..(*)” We guess that P(1), P(2 1 ), P(2 2 ) … are also true. Thus, the base case is when n = 1. P(1) P(1) is true. Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) k 1 – k 2  T(1) n 0  0

33. 33 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k 1, k 2  R such that  n > n 0, T(n)  k 1 n 2 – k 2 n Suppose that n > n 0. Let P(n) be “T(n)  k 1 n 2 – k 2 n ……………..(*)” Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/2) is true for n > 1. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k 1. (n/2) 2 –k 2 (n/2) Consider T(n) We want to show that T(n)  k 1 n 2 – k 2 n  4T(n/2) + cn  4. (k 1 (n/2) 2 –k 2 (n/2)) + cn = 4. (k 1. n 2 /4–k 2 n/2) + cn Currently, we don ’ t know the value of k. In order to prove T(n)  k 1 n 2 – k 2 n, we should set a value of k such that c – k 2  0 P(1) is true. Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) = k 1 n 2 – 2k 2 n + cn = k 1 n 2 – k 2 n – k 2 n+ cn (i.e., k 2  c) k 2  c P(n) is true. k 1 – k 2  T(1) n 0  0 = k 1 n 2 – k 2 n + (c– k 2 )n Thus, P(n) is true.

34. 34 e.g.6 We want to prove that T(n) = O(n 2 ) By the definition of big O, we need to show the following.  n 0  Z and k 1, k 2  R such that  n > n 0, T(n)  k 1 n 2 – k 2 n Suppose that n > n 0. Let P(n) be “T(n)  k 1 n 2 – k 2 n ……………..(*)” Step 2: Prove that “ P(n/2)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n/2) is true for n > 1. Step 2(b): According to P(n/2), we deduce that P(n) is true. That is, T(n/2)  k 1. (n/2) 2 –k 2 (n/2) We want to show that T(n)  k 1 n 2 – k 2 n P(1) is true. Let n be a non-negative power of 2. T(n)  4T(n/2) + cn where c:+ve T(n) > 0 Prove that T(n) = O(n 2 ) k 2  c P(n) is true. k 1 – k 2  T(1) n 0  0 We prove that “ P(n/2)  P(n) ” is true for all n > 1 By Mathematical Induction,  n  1, T(n)  k 1 n 2 – k 2 n We should set some values for n 0, k 1, k 2. Since n 0  0, we set n 0 = 0. Since k 2  c, we set k 2 = c. Since k 1 – k 2  T(1) (i.e., k 1  T(1) + k 2 ) we set k 1 = T(1) + c.

35. 35 e.g.7 (Page 19) Why was it easier to prove stronger statement T(n)  k 1 n 2 – k 2 n than to prove weaker statement T(n)  kn 2 ?

36. 36 e.g.7 Why was it easier to prove stronger statement T(n)  k 1 n 2 – k 2 n than to prove weaker statement T(n)  kn 2 ? Proving something about p(n) uses p(1)  …  p(n-1) The stronger that p(1)  …  p(n-1) are, the greater help they provide in proving p(n). Our first proof was that p(n-1) were too weak, and thus we were not able to use them to prove p(n). When we give an induction proof with a stronger statement, we are using a stronger inductive hypothesis.