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Proof by induction There are different ways to prove in Maths, induction uses a ‘domino effect’: Step 1: test that an algebraic statement is true for one.

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Presentation on theme: "Proof by induction There are different ways to prove in Maths, induction uses a ‘domino effect’: Step 1: test that an algebraic statement is true for one."— Presentation transcript:

1 Proof by induction There are different ways to prove in Maths, induction uses a ‘domino effect’: Step 1: test that an algebraic statement is true for one case, when n = 1 Step 2: assume it is true for a general case, when n = k Step 3: show that the statement holds for the next case, n = k + 1 Step 4: conclude that as the first case is true, and that subsequent cases are true, the statement must be true for all cases This technique is applied to 4 different types of statement in FP1: Proving the summation of a series Proving divisibility Eg prove that is divisible by 4 Eg prove that Proving a general term for a recurrence relation Proving a matrix power Eg Given that Eg prove that prove that

2 WB35(a) Prove by induction that, for any positive integer n,
LHS = RHS, true for n = 1 Assume that using assumption as required true for n = k + 1 if true for n = k. True for n = 1, true for all n.

3 WB36(a) Prove by induction that,
LHS = RHS, true for n = 1 Assume that using assumption Now try Ex 6A as required true for n = k + 1 if true for n = k. True for n = 1, true for all n.

4 Ex 6A Q7 Q6 Q9

5 Recurrence relationship
WB31 A sequence of numbers is defined by Prove by induction that, for n  ℤ, nth term rule nth term rule as given nth term rule true for n = 1 Recurrence relationship Assume that by recurrence relationship using assumption expanding as required true for n = k + 1 if true for n = k. True for n = 1, true for all n.

6 Recurrence relationship
WB32 A sequence of numbers u1, u2, u3, u4, . . ., is defined by Prove by induction that, for n  ℤ+, nth term rule Recurrence relationship nth term rule true for n = 1 Assume that by recurrence relationship using assumption Now try Ex 6C as required true for n = k + 1 if true for n = k. True for n = 1, true for all n.

7 WB33 Prove by induction, that for n  ℤ+,
true for n = 1 Assume that true for n = k + 1 if true for n = k. True for n = 1, true for all n. Now try Ex 6D

8 Ex 6D Q2 Q4 Q3

9 (b) is divisible by 12 true for n = 1 Assume that for some A  ℤ To prove divisibility, use the fact that: so true for n = k + 1 if true for n = k. True for n = 1, true for all n.

10 WB34 (a) Show that b) Hence, or otherwise, prove by induction that, for n  ℤ+, f(n) is divisible by 8. true for n = 1 Assume that for some A  ℤ Now try Ex 6B using (a) true for n = k + 1 if true for n = k. True for n = 1, true for all n.

11 The divisibility method must be adapted at times to achieve the required proof:
By using prove by induction that, for n  ℤ+, f(n) is divisible by 9. Eg Assume that for some A  ℤ Why consider instead of the usual ? This is not obviously divisible by 9 - the standard method fails

12 If the method is adapted, the proof may require some extra explanation
Eg By using prove by induction that, for n  ℤ+, f(n) is divisible by 5. Assume that for some A  ℤ using earlier result This tells us that But as 2 and 5 are coprime, f(k+1) must be divisible by 5 for some B  ℤ This statement validates the proof and must be included Coprime numbers share no factors

13 The general term given a recurrence relation
Proving: Example Method Matrix powers Summation of a series The general term given a recurrence relation If given given then that Divisibility is divisible by 4


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