Download
# Proof by mathematical induction PowerPoint Presentation, PPT - DocSlides

danika-pritchard | 2018-12-08 | General

** Tags : **
proof-by-mathematical-induction
induction true
mathematical proof
true
induction
proof
mathematical
statement
term
prove
inductionyou
basis
show
divisible
sequence
inductive
expression
assumption
### Presentations text content in Proof by mathematical induction

Show

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. ID: 738633

- Views :
**0**

**Direct Link:**- Link:https://www.docslides.com/danika-pritchard/proof-by-mathematical-induction
**Embed code:**

Download this presentation

DownloadNote - The PPT/PDF document "Proof by mathematical induction" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.

Slide1

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodWe will start by proving statements relating to the sum of a series.

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodWe will start by proving statements relating to the sum of a series.

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+1 = 3u

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+2 = 5u

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+2 = 5u

Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

Proof by mathematicalinduction

Slide2IntroductionProof by mathematical induction is an extremely powerful tool for proving mathematical statementsAs 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 caseYou 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

Slide3Teachings for Exercise 6A

Slide4Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodThe 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 downiPhone 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…6A

How this works mathematically

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!

Slide5Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodWe 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…

6A

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

Slide6Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodWe 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…

BASISASSUMPTIONINDUCTIVECONCLUSION

6A

BASIS

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

Slide7Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodWe 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…

BASISASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

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 k

2

Slide8Prove by mathematical induction that, for

So we need to use the steps from before to prove this statement…

BASISASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

Assume the statement is true for n = k

INDUCTIVE

Show the statement is then true for (k + 1)

ie

) The next term

You can replace the first part as we assumed it was equal to k

2

earlier

The sequence will be the same, but with an extra term

(sub in (k + 1)) for it!

Slide9Prove by mathematical induction that, for

So we need to use the steps from before to prove this statement…

BASISASSUMPTIONINDUCTIVECONCLUSION

6A

We assumed that for n = k, the sum of the series would be equal to k

2

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!

CONCLUSION

Slide10Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

,

So we are now going to prove one of the formulae you have learnt in chapter 5!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

BASIS

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

Show that the statement is true for n = 1

Slide11Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

,

So we are now going to prove one of the formulae you have learnt in chapter 5!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

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

Slide12Proof by mathematical inductionYou can obtain a proof for the summation of a series, by using the induction methodProve, by mathematical induction, that for

,

So we are now going to prove one of the formulae you have learnt in chapter 5!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

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!!

Slide13,

So we are now going to prove one of the formulae you have learnt in chapter 5!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

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!

Rewrite both as fractions over 6

Combine

‘Clever factorisation’ method!

Expand and simplify the inner brackets

Factorise the inner part

Slide14,

So we are now going to prove one of the formulae you have learnt in chapter 5!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

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!

Slide15,

This looks more complicated, but you just follow the same process as you have seen already!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

BASIS

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

Show that the statement is true for n = 1

Slide16,

This looks more complicated, but you just follow the same process as you have seen already!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

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

Slide17,

This looks more complicated, but you just follow the same process as you have seen already!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

ASSUMPTION

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)

Slide18,

This looks more complicated, but you just follow the same process as you have seen already!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

INDUCTIVE

Show the statement is then true for (k + 1)

ie

) The next term

2 x 2

k

= 2

k+1

(add the powers)

In total, we have (k – 1) + (k + 1) 2

k+1

s

Simplify the bracket

Re-

factorise

the bracket

Slide19,

This looks more complicated, but you just follow the same process as you have seen already!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6A

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!

Slide20Teachings for Exercise 6B

Slide21Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that 32n

+ 11 is divisible by 4 for all positive integers

You follow the same steps as before!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6B

BASIS

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

Slide22Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that 32n

+ 11 is divisible by 4 for all positive integers

You follow the same steps as before!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6B

ASSUMPTION

Assume the statement is true for n = k

INDUCTIVE

Show that the statement is then true for n = (k + 1)

Multiply out the bracket

3

2k+2

= 3

2k

x 3

2

(adding powers when multiplying)

So we have 9 lots of 3

2k

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

Slide23Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that 32n

+ 11 is divisible by 4 for all positive integers

You follow the same steps as before!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6B

INDUCTIVE

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

Slide24Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that the expression ‘n3

– 7n + 9’ is divisible by 3 for all positive integers

BASISASSUMPTION

INDUCTIVECONCLUSION

6B

BASIS

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

Slide25Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that the expression ‘n3

– 7n + 9’ is divisible by 3 for all positive integers

BASISASSUMPTION

INDUCTIVECONCLUSION

6B

ASSUMPTION

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

Slide26Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, by induction, that the expression ‘n3

– 7n + 9’ is divisible by 3 for all positive integers

BASISASSUMPTION

INDUCTIVECONCLUSION

6B

INDUCTIVE

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

Slide27Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, 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…

BASISASSUMPTIONINDUCTIVE

CONCLUSION

6B

BASIS

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

Slide28Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, 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…

BASISASSUMPTIONINDUCTIVE

CONCLUSION

6B

ASSUMPTION

Assume the statement is true for n = k

INDUCTIVE

Show that the statement is then true for n = (k + 1)

Simplify powers

11

k+2

= 11 x 11

k+1

12

2k+1

= 12

2

x 12

2k-1

Simplify

Rewrite

*

*

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!

Slide29Proof by mathematical inductionYou can use proof by induction to prove that an expression is divisible by a given integerProve, 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…

BASISASSUMPTIONINDUCTIVE

CONCLUSION

6B

INDUCTIVE

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(12

2k-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!

Slide30Teachings for Exercise 6C

Slide31Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationYou 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!6C

Example 1

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!

Slide32Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+1 = 3u

n + 4, u1 = 1, prove by induction that un = 3

n – 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

Using the recurrence relation

Using the ‘nth’ term formula

Slide33

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+1 = 3u

n + 4, u1 = 1, prove by induction that un = 3

n – 2.So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3

n – 2…BASISASSUMPTION

INDUCTIVE

CONCLUSION

6C

BASIS

Show that the statement is true for n = 1 and n = 2

There is a slight difference here. As we are

given

u

1

already (n = 1), we need to also check the statement is true for n = 2…

The first 2 terms are both 1 and 7, so the statement is true for n = 1 and 2

We already know u

1

Now use the recurrence relation to find u

2

Calculate

Sub in n = 1

Calculate

Now sub in n = 2

Calculate

Slide34

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+1 = 3u

n + 4, u1 = 1, prove by induction that un = 3

n – 2.So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3

n – 2…BASISASSUMPTION

INDUCTIVE

CONCLUSION

6C

ASSUMPTION

Assume that the statement is true for n = k

INDUCTIVE

Use the recurrence relation to create an expression for u

k+1

Replace

u

k

with the assumed expression above

Multiply out the brackets

Simplify

Slide35n + 4, u1 = 1, prove by induction that un = 3

n – 2.So we are being asked to show that, for the sequence with this recurrence relation, that the nth term formula is 3

n – 2…BASISASSUMPTION

INDUCTIVE

CONCLUSION

6C

INDUCTIVE

Use the recurrence relation to create an expression for u

k+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

Slide36

Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+2 = 5u

n+1 – 6un, and u1 = 13 and u

2 = 35:Prove by induction that un

= 2n+1 + 3n+1

This sequence is slightly different to what you have seen!

BASIS

ASSUMPTION

INDUCTIVE

CONCLUSION

6C

The next term in the sequence

The current term

The

previous

term

The first term

The second

term

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…

Slide37Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+2 = 5u

n+1 – 6un, and u1 = 13 and u

2 = 35:Prove by induction that un

= 2n+1 + 3n+1

This sequence is slightly different to what you have seen!

BASIS

ASSUMPTION

INDUCTIVE

CONCLUSION

6C

BASIS

Show that the statement is true for n = 1, n = 2 and n = 3

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

We already know u

1

and u

2

Sub in u

2

and u

1

to find u

3

Calculate

Calculate

Calculate

Calculate

Slide38Proof by mathematical inductionYou can use mathematical induction to produce a proof for a general term of a recurrence relationGiven that un+2 = 5u

n+1 – 6un, and u1 = 13 and u

2 = 35:Prove by induction that un

= 2n+1 + 3n+1

This sequence is slightly different to what you have seen!

BASIS

ASSUMPTION

INDUCTIVE

CONCLUSION

6C

ASSUMPTION

Assume that the statement is true for n = k AND n = k + 1

INDUCTIVE

Use the recurrence relation to create an expression for u

k+2

Sub in the assumed expressions for u

k+1

and

u

k

from before

Split the bracketed parts up

Rewrite all as powers of ‘k + 2’

(see below)

6 = 3 x 2

The 2 adds 1 to the power

6 = 2 x 3

The 3 adds 1 to the power

Slide39n+1 – 6un, and u1 = 13 and u

2 = 35:Prove by induction that un

= 2n+1 + 3n+1

This sequence is slightly different to what you have seen!

BASIS

ASSUMPTION

INDUCTIVE

CONCLUSION

6C

ASSUMPTION

Assume that the statement is true for n = k AND n = k + 1

INDUCTIVE

Use the recurrence relation to create an expression for u

k+2

Sub in the assumed expressions for u

k+1

and

u

k

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

Slide40n+1 – 6un, and u1 = 13 and u

2 = 35:Prove by induction that un

= 2n+1 + 3n+1

This sequence is slightly different to what you have seen!

BASIS

ASSUMPTION

INDUCTIVE

CONCLUSION

6C

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!

Slide41Teachings for Exercise 6D

Slide42Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

As always, follow the same pattern as with the other induction questions!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6D

BASIS

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

Slide43Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

As always, follow the same pattern as with the other induction questions!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6D

ASSUMPTION

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!

Slide44Proof by mathematical inductionYou can use proof by induction to prove general statements involving matrix multiplicationUse mathematical induction to prove that:

As always, follow the same pattern as with the other induction questions!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

6D

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!

Work out each term

Simplify (remember to manipulate the powers)

Slide45As always, follow the same pattern as with the other induction questions!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

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!

Slide46More complicated, but the same process!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

BASIS

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

Slide47

More complicated, but the same process!BASIS

ASSUMPTIONINDUCTIVECONCLUSION

ASSUMPTION

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!

Slide48Simplify terms (probably a good idea to do in stages…)Proof by mathematical inductionYou 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

ASSUMPTIONINDUCTIVECONCLUSION

6D

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!

Simplify fully

This is the answer to the multiplication

Slide49

Proof by mathematical induction

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

6D

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!

Slide50SummaryWe have seen how to use proof by inductionWe have seen how to use it in situations regarding the summation of a series, tests of divisibility, recurrence relationships and matricesThe four steps will always be the same, you will need to practice the ‘clever manipulation’ behind some of the inductive steps though!

Today's Top Docs

Related Slides