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Proof by mathematical induction

Proof by mathematical induction. Introduction. Proof by mathematical induction is an extremely powerful tool for proving mathematical statements As we know, proof is essential in Maths as although something may seem to work for a number of cases, we need to be sure it will work in every case

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Proof by mathematical induction

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  1. Proof by mathematical induction

  2. Introduction • Proof by mathematical induction is an extremely powerful tool for proving mathematical statements • As we know, proof is essential in Maths as although something may seem to work for a number of cases, we need to be sure it will work in every case • You have seen some of the formulae used in the series chapter – the had to be proven to work for every case before mathematicians could confidently use them

  3. Teachings for Exercise 6A

  4. Proof by mathematical induction How this works mathematically You can obtain a proof for the summation of a series, by using the induction method The way ‘proof by mathematical induction’ works is often likened to knocking dominoes over • If the dominoes are lined up, then you knock over the first one, every domino afterwards will fall down iPhone Dominoes • Mathematically, if we want to prove that something is true for all possible cases, we cannot do it numerically (as the numbers would just go on forever) • However, if we show that if one case is true, and so is the next case, then we can therefore show it is true for every case… BASIS  Show that the statement to be proven works for the case n = 1 ASSUMPTION  Assume that the statement is true for n = k (just replace the ns with ks!) INDUCTIVE • Show that if the statement is true for n = k, it is also true for n = k + 1 (ie – the next case) • This is harder to explain without an example. Essentially you find a way to express the next ‘case’ using k and show that it is equivalent to replacing k with ‘k + 1’ CONCLUSION • You have shown that if the statement is true for one case, it must be true for the next • As it was true for n = 1, it must therefore be true for n = 2, 3, 4 and so on, PROVING the statement! 6A

  5. Proof by mathematical induction You can obtain a proof for the summation of a series, by using the induction method We will start by proving statements relating to the sum of a series. Prove by mathematical induction that, for So we need to use the steps from before to prove this statement… This means ‘n can be any positive integer’ This is the formula for the sequence This is the formula for the sum of the first n terms of the sequence 6A

  6. Proof by mathematical induction BASIS You can obtain a proof for the summation of a series, by using the induction method We will start by proving statements relating to the sum of a series. Prove by mathematical induction that, for So we need to use the steps from before to prove this statement… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Replace n with 1 Replace n with 1 There will only be one term here, that we get by subbing n = 1 into the expression Calculate The statement given is therefore true for n = 1 6A

  7. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method We will start by proving statements relating to the sum of a series. Prove by mathematical induction that, for So we need to use the steps from before to prove this statement… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k Write out the first few terms in the sequence, and the last term, which will be in terms of k We are going to assume that this sequence is true for k, and hence the sum will be equal to k2 6A

  8. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method We will start by proving statements relating to the sum of a series. Prove by mathematical induction that, for So we need to use the steps from before to prove this statement… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra term  (sub in (k + 1)) for it! You can replace the first part as we assumed it was equal to k2 earlier 6A

  9. Proof by mathematical induction CONCLUSION You can obtain a proof for the summation of a series, by using the induction method We will start by proving statements relating to the sum of a series. Prove by mathematical induction that, for So we need to use the steps from before to prove this statement… BASIS ASSUMPTION INDUCTIVE CONCLUSION We assumed that for n = k, the sum of the series would be equal to k2 Using this assumption, we showed that the summation for (k + 1) is equal to (k + 1)2 • So if the statement is true for one value, it will therefore be true for the next value • As it is true for the next value, it will therefore be true for the value after that, and so on… • However, this all relies on the assumption being correct… • Remember for the BASIS step, we showed that the statement is true for n = 1? • Well because it is true for n = 1, it must therefore be true for n = 2, n = 3……… and so on! • The statement is therefore true for all values of n! 6A

  10. Proof by mathematical induction BASIS You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , So we are now going to prove one of the formulae you have learnt in chapter 5! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Replace n with 1 Replace n with 1 There will only be one term here, that we get by subbing n = 1 into the expression Calculate The statement given is therefore true for n = 1 6A

  11. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , So we are now going to prove one of the formulae you have learnt in chapter 5! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k Write out the first few terms in the sequence, and the last term, which will be in terms of k We are going to assume that this sequence is true for k, and hence the sum will be equal to the expression above 6A

  12. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , So we are now going to prove one of the formulae you have learnt in chapter 5! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra term  (sub in (k + 1)) for it! Replace the first part with the assumed formula from earlier! This requires more simplification which will be shown on the next slide!! 6A

  13. Proof by mathematical induction INDUCTIVE You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , So we are now going to prove one of the formulae you have learnt in chapter 5! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra term  (sub in (k + 1)) for it! Replace the first part with the assumed formula from earlier! Rewrite both as fractions over 6 Combine ‘Clever factorisation’ method! Expand and simplify the inner brackets Factorise the inner part 6A

  14. Proof by mathematical induction CONCLUSION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , So we are now going to prove one of the formulae you have learnt in chapter 5! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Explain why it proves the original statement For n = k For n = (k + 1) Rewrite some of the brackets Written in this way, you can see that the k’s in the first statement have all been replaced with ‘k + 1’s •  So the statement was true for n = 1 • We also showed that if it is true for one statement, it is true for the next • Therefore the formula has been proven! 6A

  15. Proof by mathematical induction BASIS You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , This looks more complicated, but you just follow the same process as you have seen already! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Replace n with 1 Replace n with 1 There will only be one term here, that we get by subbing n = 1 into the expression Calculate The statement given is therefore true for n = 1 6A

  16. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , This looks more complicated, but you just follow the same process as you have seen already! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k We are going to assume that this sequence is true for k, and hence the sum will be equal to the expression above Write out the first few terms in the sequence, and the last term, which will be in terms of k 6A

  17. Proof by mathematical induction ASSUMPTION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , This looks more complicated, but you just follow the same process as you have seen already! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra term  (sub in (k + 1)) for it! Replace the first part with the assumed formula from earlier! The simplification for this is difficult  You need to aim for the power of 2 to be ‘k + 1’ (as it was ‘k’ originally) 6A

  18. Proof by mathematical induction INDUCTIVE You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , This looks more complicated, but you just follow the same process as you have seen already! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show the statement is then true for (k + 1) ie) The next term 2 x 2k = 2k+1 (add the powers) In total, we have (k – 1) + (k + 1) 2k+1s Simplify the bracket Re-factorise the bracket 6A

  19. Proof by mathematical induction CONCLUSION You can obtain a proof for the summation of a series, by using the induction method Prove, by mathematical induction, that for , This looks more complicated, but you just follow the same process as you have seen already! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Explain why this shows the statement is true For n = k For n = (k + 1) Rewrite the first ‘k’ as ‘k + 1 – 1’ Written in this way, you can see that the k’s in the first statement have all been replaced with ‘k + 1’s •  So the statement was true for n = 1 • We also showed that if it is true for one statement, it is true for the next • Therefore the formula has been proven! • You will need to become familiar with manipulating powers in the way shown here! 6A

  20. Teachings for Exercise 6B

  21. Proof by mathematical induction BASIS You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers You follow the same steps as before! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Sub in n = 1 Calculate 20 is divisible by 4, so the statement is true for n = 1 6B

  22. Proof by mathematical induction ASSUMPTION You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers You follow the same steps as before! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show that the statement is then true for n = (k + 1) Multiply out the bracket 32k+2 = 32k x 32 (adding powers when multiplying) So we have 9 lots of 32k At this point we will combine the expressions for f(k) and f(k + 1) in order to prove that the statement is always divisible by 4 6B

  23. Proof by mathematical induction INDUCTIVE You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers You follow the same steps as before! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is then true for n = (k + 1) Subtract f(k) from f(k + 1), using the expressions above Group terms on the right side Take out 4 as a factor Add f(k) • This shows that f(k + 1) is just f(k) with an expression added on • We assumed f(k) was divisible by 4 • The expression to be added is divisible by 4 • So the answer must be divisible by 4, if f(k) is! • As the first case (n = 1) was divisible by 4, the statement must be true! CONCLUSION 6B

  24. Proof by mathematical induction BASIS You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible by 3 for all positive integers BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Sub in n = 1 Calculate 3 is divisible by 3, so the statement is true for n = 1 6B

  25. Proof by mathematical induction ASSUMPTION You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible by 3 for all positive integers BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show that the statement is then true for n = (k + 1) Multiply out the brackets Group up terms 6B

  26. Proof by mathematical induction INDUCTIVE You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ‘n3 – 7n + 9’ is divisible by 3 for all positive integers BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is then true for n = (k + 1) Subtract f(k) from f(k + 1), using the expressions above ‘Remove’ the brackets Group terms Take out 3 as a factor Add f(k) • This shows that f(k + 1) is just f(k) with an expression added on • We assumed f(k) was divisible by 3 • The expression to be added is divisible by 3 • So the answer must be divisible by 3, if f(k) is! • As the first case (n = 1) was divisible by 3, the statement must be true! CONCLUSION 6B

  27. Proof by mathematical induction BASIS You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible by 133 for all positive integers This example will require more manipulation as we work through it, but is essentially the same as the previous two… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Sub in n = 1 Calculate 133 is divisible by 133, so the statement is true for n = 1 6B

  28. Proof by mathematical induction ASSUMPTION You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible by 133 for all positive integers This example will require more manipulation as we work through it, but is essentially the same as the previous two… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show that the statement is then true for n = (k + 1) Simplify powers * Rewrite 11k+2 = 11 x 11k+1 122k+1 = 122 x 122k-1 Simplify They have been re-written in this way so that, on the next step, the 11s and 12s have the same powers as in the f(k) expression and therefore can be grouped up! * 6B

  29. Proof by mathematical induction INDUCTIVE You can use proof by induction to prove that an expression is divisible by a given integer Prove, by induction, that the expression ’11n+1 + 122n-1’ is divisible by 133 for all positive integers This example will require more manipulation as we work through it, but is essentially the same as the previous two… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is then true for n = (k + 1) Subtract f(k) from f(k + 1), using the expressions above Group terms Split the 143 into 2 parts The first 2 terms are just 10 lots of f(k) Add f(k) •  If f(k) is divisible by 133, so is 11f(k) • 133(122k-1) is divisible by 133 • Therefore f(k+1) will also be divisible by 133 • As f(1) was divisible by 133, the statement is therefore true! • Make sure you practise enough so you can spot how and when to manipulate in this way! 6B

  30. Teachings for Exercise 6C

  31. Proof by mathematical induction Example 1 You can use mathematical induction to produce a proof for a general term of a recurrence relation You will have seen recurrence relations in C1. A recurrence relation is a sequence where generating a term relies on a previous term. It is very important that you understand the notation! The next term in the sequence The current term The first term • This is telling you that the first number in the sequence is 3 • And to get the next number, you add on 5 • The sequence will be: 3, 8, 13, 18, 23…… and so on… • As this is an arithmetic sequence, we can use the formula from C1 for the ‘nth’ term.. (a + (n -1)d) Example 2 • This is telling you that the first number in the sequence is 1 • To get the next number, you multiply the current number by 3 and subtract 1… • The sequence will be: 1, 2, 5, 14, 41, 122 ……and so on… • This is NOT an arithmetic sequence, so we cannot use the arithmetic sequence method for the nth term • The ‘nth’ term for a sequence like this is far more complicated! • They can also be proven to be correct (once you think you know what they are!) by use of induction! 6C

  32. Proof by mathematical induction Using the recurrence relation Using the ‘nth’ term formula You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2. • Before we start, let’s generate the first 5 terms in both the ways shown above… • You do not need to do this on an exam, this is just to show you the two ways of generating the sequence give the same result! • So now, let’s prove this is the case! 6C

  33. Proof by mathematical induction BASIS You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2. So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3n – 2… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 and n = 2  There is a slight difference here. As we are given u1 already (n = 1), we need to also check the statement is true for n = 2… We already know u1 Sub in n = 1 Now use the recurrence relation to find u2 Calculate Now sub in n = 2 Calculate Calculate  The first 2 terms are both 1 and 7, so the statement is true for n = 1 and 2 6C

  34. Proof by mathematical induction ASSUMPTION You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2. So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3n – 2… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume that the statement is true for n = k INDUCTIVE  Use the recurrence relation to create an expression for uk+1 Replace uk with the assumed expression above Multiply out the brackets Simplify 6C

  35. Proof by mathematical induction INDUCTIVE You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n – 2. So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3n – 2… BASIS ASSUMPTION INDUCTIVE CONCLUSION  Use the recurrence relation to create an expression for uk+1 The ‘k’ terms have all become ‘k + 1’ terms CONCLUSION • If the statement is true for ‘k’, it is also true for ‘k + 1’ • We showed in the basis that it is true for n = 1 and n = 2 • Therefore the statement is true for all 6C

  36. Proof by mathematical induction You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35: Prove by induction that un = 2n+1 + 3n+1 This sequence is slightly different to what you have seen! BASIS ASSUMPTION INDUCTIVE CONCLUSION The next term in the sequence The current term The previous term The first term The secondterm • So for this sequence, the next term is based on the current term AND the term before that! • This is why you have been given the first 2 terms… 6C

  37. Proof by mathematical induction BASIS You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35: Prove by induction that un = 2n+1 + 3n+1 This sequence is slightly different to what you have seen! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1, n = 2 and n = 3 We already know u1 and u2 Calculate Sub in u2 and u1 to find u3 Calculate Calculate Calculate  The first 3 terms are 13, 35 and 97 for both sequences, so the statement has been shown to be true up to n = 3 6C

  38. Proof by mathematical induction ASSUMPTION You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35: Prove by induction that un = 2n+1 + 3n+1 This sequence is slightly different to what you have seen! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume that the statement is true for n = k AND n = k + 1 INDUCTIVE  Use the recurrence relation to create an expression for uk+2 Sub in the assumed expressions for uk+1 and uk from before Split the bracketed parts up Rewrite all as powers of ‘k + 2’ (see below) 6 = 2 x 3 6 = 3 x 2 The 3 adds 1 to the power The 2 adds 1 to the power 6C

  39. Proof by mathematical induction ASSUMPTION You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35: Prove by induction that un = 2n+1 + 3n+1 This sequence is slightly different to what you have seen! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume that the statement is true for n = k AND n = k + 1 INDUCTIVE  Use the recurrence relation to create an expression for uk+2 Sub in the assumed expressions for uk+1 and uk from before Split the bracketed parts up Rewrite all as powers of ‘k + 2’ Group the ‘like’ terms The 2 and 3 add 1 to the powers of 2 and 3 respectively 6C

  40. Proof by mathematical induction CONCLUSION You can use mathematical induction to produce a proof for a general term of a recurrence relation Given that un+2 = 5un+1 – 6un, and u1 = 13 and u2 = 35: Prove by induction that un = 2n+1 + 3n+1 This sequence is slightly different to what you have seen! BASIS ASSUMPTION INDUCTIVE CONCLUSION These can both be written differently • As you can see, k is replaced with (k + 1), and then with (k + 2) • So we have shown that IF the statement is true for n = k and n = k + 1, then it must also be true for n = k + 2 • As we showed in the basis that the statement is true for n = 1 and n = 2, then it must therefore be true for n = 3 • And consequently it is then true for all values of n! 6C

  41. Teachings for Exercise 6D

  42. Proof by mathematical induction BASIS You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: As always, follow the same pattern as with the other induction questions! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Replace n with 1 Replace n with 1 Calculate Calculate So the statement is true for n = 1 6D

  43. Proof by mathematical induction ASSUMPTION You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: As always, follow the same pattern as with the other induction questions! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show using the assumption, that the statement will also be true for n = k + 1 • Replace the power ‘k’ term with the assumed matrix • The second matrix doesn’t need the power! • Now we need to multiply these matrices using the skills from chapter 4! 6D

  44. Proof by mathematical induction INDUCTIVE You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: As always, follow the same pattern as with the other induction questions! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show using the assumption, that the statement will also be true for n = k + 1 • Replace the power ‘k’ term with the assumed matrix • The second matrix doesn’t need the power! Work out each term Simplify (remember to manipulate the powers) 6D

  45. Proof by mathematical induction CONCLUSION You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: As always, follow the same pattern as with the other induction questions! BASIS ASSUMPTION INDUCTIVE CONCLUSION  We assumed that:  Using this, we showed that: • As you can see, all the ‘k’ terms have been replaced with ‘k + 1’ terms • Therefore, IF the statement is true for one term, it will also be true for the next term, and so on… • As we already showed that the statement is true for n = 1, it is therefore true for all values of n!

  46. Proof by mathematical induction BASIS You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: More complicated, but the same process! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show that the statement is true for n = 1 Replace n with 1 Replace n with 1 Calculate Calculate So the statement is true for n = 1

  47. Proof by mathematical induction ASSUMPTION You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: More complicated, but the same process! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Assume the statement is true for n = k INDUCTIVE  Show using the assumption, that the statement will then be true for n = k + 1 • Replace the power ‘k’ term with the assumed matrix • The second matrix doesn’t need the power! • Now we need to multiply these matrices using the skills from chapter 4!

  48. Proof by mathematical induction INDUCTIVE You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: More complicated, but the same process! BASIS ASSUMPTION INDUCTIVE CONCLUSION  Show using the assumption, that the statement will also be true for n = k + 1 • Replace the power ‘k’ term with the assumed matrix • The second matrix doesn’t need the power! Simplify terms (probably a good idea to do in stages…) Simplify fully This is the answer to the multiplication 6D

  49. Proof by mathematical induction CONCLUSION You can use proof by induction to prove general statements involving matrix multiplication Use mathematical induction to prove that: More complicated, but the same process! BASIS ASSUMPTION INDUCTIVE CONCLUSION  We assumed that:  Using this, we showed that: Each part of the matrix can be written differently… (you will see why in a moment!) • If you compare this to the original matrix, you can see that all the ‘k’ terms have been replaced with ‘k + 1’ terms • So we have shown that if the statement is true for n = k, it will also be true for n = k + 1 • As it was true for n = 1, it is also true for all positive values of k! 6D

  50. Summary • We have seen how to use proof by induction • We have seen how to use it in situations regarding the summation of a series, tests of divisibility, recurrence relationships and matrices • The four steps will always be the same, you will need to practice the ‘clever manipulation’ behind some of the inductive steps though!

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