# C3 Functions Difficult Question Please help

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bigmansouf

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Pure Mathematics 1 by Backhouse

Ex:2f q13

The real function f, defined for all , is said to be multiplicative if, for all , ,

f(xy)= f(x) f(y)

Q: Prove that if f is multiplicative function then

a) either f(0) =0 or f(x)=1

b) either f(1) = 1 or f(x) = 0

c) = for all positive integers n.

Give example of a non- constant multiplicative function

My attempt:

I have tried to apply what I have learn from the chapter of functions from this book but there is no bit on multiplicative functions, I have only learnt about odd & even functions. I tried but failed please help

let f(x) = even functions thus f(a) = f(-a)

f(y) = odd functions thus - f(b) = f(-b)

f(ab) = f(a)f(b)

= f(-a) (-f(b)) = -f(-a)f(-b) = -f(a)f(-b)

hence i am stuck please help

Ex:2f q13

The real function f, defined for all , is said to be multiplicative if, for all , ,

f(xy)= f(x) f(y)

Q: Prove that if f is multiplicative function then

a) either f(0) =0 or f(x)=1

b) either f(1) = 1 or f(x) = 0

c) = for all positive integers n.

Give example of a non- constant multiplicative function

My attempt:

I have tried to apply what I have learn from the chapter of functions from this book but there is no bit on multiplicative functions, I have only learnt about odd & even functions. I tried but failed please help

let f(x) = even functions thus f(a) = f(-a)

f(y) = odd functions thus - f(b) = f(-b)

f(ab) = f(a)f(b)

= f(-a) (-f(b)) = -f(-a)f(-b) = -f(a)f(-b)

hence i am stuck please help

Last edited by bigmansouf; 2 years ago

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RDKGames

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#2

(Original post by

Pure Mathematics 1 by Backhouse

Ex:2f q13

The real function f, defined for all , is said to be multiplicative if, for all ,

**bigmansouf**)Pure Mathematics 1 by Backhouse

Ex:2f q13

The real function f, defined for all , is said to be multiplicative if, for all ,

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bigmansouf

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#3

(Original post by

This sentence seems to be cut off.

**RDKGames**)This sentence seems to be cut off.

The real function f, defined for all , is said to be multiplicative if, for all ,

f(xy)= f(x) f(y)

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RDKGames

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#4

(Original post by

sorry i hae corrected it

The real function f, defined for all , is said to be multiplicative if, for all ,

f(xy)= f(x) f(y)

**bigmansouf**)sorry i hae corrected it

The real function f, defined for all , is said to be multiplicative if, for all ,

f(xy)= f(x) f(y)

Can you move on from here?

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bigmansouf

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#5

(Original post by

OK, then for the first one, let and so you get that . Thus

Can you move on from here?

**RDKGames**)OK, then for the first one, let and so you get that . Thus

Can you move on from here?

let

.

for the part b

let

.

sorry but could you please help me with the last part

c) = for all positive integers n.

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RDKGames

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#6

(Original post by

Sorry for the late reply Thank you for the help I took dow nfrom notes did part a and b but i have tried part c but with no success

let

.

for the part b

let

.

sorry but could you please help me with the last part

c) = for all positive integers n.

**bigmansouf**)Sorry for the late reply Thank you for the help I took dow nfrom notes did part a and b but i have tried part c but with no success

let

.

for the part b

let

.

sorry but could you please help me with the last part

c) = for all positive integers n.

Is that what you meant?

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#8

(Original post by

yes sorry that is what i meant

**bigmansouf**)yes sorry that is what i meant

But by the same principle,

.

So .

Try using this key argument to complete the question.

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#9

(Original post by

But by the same principle,

.

So .

Try using this key argument to complete the question.

**RDKGames**)But by the same principle,

.

So .

Try using this key argument to complete the question.

Firstly i would like to ask about the method you posted - it is classified under a particular technique such as a proof contradiction or proof of induction so i can look it up and study it more. It is a bit difficult to understand it

so i try to use proof via induction to solve it

**prove**

for all positive integers n

**first show that the step hold for n= 1**

**Suppose it holds for n = k**

or

is invalid since n represents all positive integers

since

thus

**then let n = k+1**

or

is invalid since n represents all positive integers

since

thus

So we have shown that if it is true for some n=k it is also true for n=k+1. We have shown that it is true for n=1, therefore by the principle of mathematical induction it is true for all the positive integers n.

Please can you tell me if I am right

and the last point of the question is to give an example of a non-constant multiplicative function but i tried to read wikipedia but it is a bit difficult to understand please can you share some light?

thank you

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#10

**RDKGames**)

But by the same principle,

.

So .

Try using this key argument to complete the question.

Firstly i would like to ask about the method you posted - it is classified under a particular technique such as a proof contradiction or proof of induction so i can look it up and study it more. It is a bit difficult to understand it

so i try to use proof via induction to solve it

**prove**

for all positive integers n

**first show that the step hold for n= 1**

**Suppose it holds for n = k**

or

is invalid since n represents all positive integers

since

thus

**then let n = k+1**

or

is invalid since n represents all positive integers

since

thus

So we have shown that if it is true for some n=k it is also true for n=k+1. We have shown that it is true for n=1, therefore by the principle of mathematical induction it is true for all the positive integers n.

Please can you tell me if I am right and the last point of the question is to give an example of a non-constant multiplicative function but i tried to read wikipedia but it is a bit difficult to understand please can you share some light?

thank you

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#11

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#11

(Original post by

thank you

Firstly i would like to ask about the method you posted - it is classified under a particular technique such as a proof contradiction or proof of induction so i can look it up and study it more. It is a bit difficult to understand it

**bigmansouf**)thank you

Firstly i would like to ask about the method you posted - it is classified under a particular technique such as a proof contradiction or proof of induction so i can look it up and study it more. It is a bit difficult to understand it

- this should be clear to you, I hope, I refer to this relation as (*). I literally just split up the function due to its multiplicative property.

Now replace with everywhere in (*). You get that . Substitute this result into (*) and you end up with

... (**)

We can repeat the trick. Replace every in (*) with . We have that . Substitute this into (**) and you end up with

And so on... the jump in logic is that you need to realise this process will go on all the way until the power of on the RHS will become zero, and the power of will be precisely . In this stage, we end up with

... (***)

but we know that for a general non-zero function , we must have .

Hence (***) reduces to

But induction is a good alternative as well.

so i try to use proof via induction to solve it

**first show that the step hold for n= 1****Suppose it holds for n = k**

The word "suppose" means

__something. If you assume something is true, you don't need to manipulate it right there and then!__

*assume*This step should literally only say:

"Suppose the statement holds for ; so we have "

and move onto the next part...

**then let n = k+1**

This is

**not**how you do induction.

You, for some reason, began with the result you are

**trying to prove**. It makes no sense to do that?

You need to instead begin by saying:

and now use the assumption you made in the previous part!

We have...

Hence

So we have shown that if it is true for some n=k it is also true for n=k+1. We have shown that it is true for n=1, therefore by the principle of mathematical induction it is true for all the positive integers n.

and the last point of the question is to give an example of a non-constant multiplicative function but i tried to read wikipedia but it is a bit difficult to understand please can you share some light?

thank you

thank you

.

We have; .

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#12

(Original post by

It's not really a particular technique of proof that I've used. I merely employed the property that we are outright told; that is multiplicative.

- this should be clear to you, I hope, I refer to this relation as (*). I literally just split up the function due to its multiplicative property.

Now replace with everywhere in (*). You get that . Substitute this result into (*) and you end up with

... (**)

We can repeat the trick. Replace every in (*) with . We have that . Substitute this into (**) and you end up with

And so on... the jump in logic is that you need to realise this process will go on all the way until the power of on the RHS will become zero, and the power of will be precisely . In this stage, we end up with

... (***)

but we know that for a general non-zero function , we must have .

Hence (***) reduces to

But induction is a good alternative as well.

This is fine.

I am really unsure what you think you are doing here.

The word "suppose" means

This step should literally only say:

"Suppose the statement holds for ; so we have "

and move onto the next part...

This is

You, for some reason, began with the result you are

You need to instead begin by saying:

and now use the assumption you made in the previous part!

We have...

Hence

Good wording.

is what you want. The simplest example is literally just

.

We have; .

**RDKGames**)It's not really a particular technique of proof that I've used. I merely employed the property that we are outright told; that is multiplicative.

- this should be clear to you, I hope, I refer to this relation as (*). I literally just split up the function due to its multiplicative property.

Now replace with everywhere in (*). You get that . Substitute this result into (*) and you end up with

... (**)

We can repeat the trick. Replace every in (*) with . We have that . Substitute this into (**) and you end up with

And so on... the jump in logic is that you need to realise this process will go on all the way until the power of on the RHS will become zero, and the power of will be precisely . In this stage, we end up with

... (***)

but we know that for a general non-zero function , we must have .

Hence (***) reduces to

But induction is a good alternative as well.

This is fine.

I am really unsure what you think you are doing here.

The word "suppose" means

__something. If you assume something is true, you don't need to manipulate it right there and then!__*assume*This step should literally only say:

"Suppose the statement holds for ; so we have "

and move onto the next part...

This is

**not**how you do induction.You, for some reason, began with the result you are

**trying to prove**. It makes no sense to do that?You need to instead begin by saying:

and now use the assumption you made in the previous part!

We have...

Hence

Good wording.

is what you want. The simplest example is literally just

.

We have; .

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