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  x \epsilon \mathbb{R} , is said to be multiplicative if, for all  y \epsilon \mathbb{R} ,  x \epsilon \mathbb{R} ,

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)  f(x^{n}) =   \left \{ f(x) \right \}^{n} 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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(Original post by bigmansouf)
Pure Mathematics 1 by Backhouse
Ex:2f q13
The real function f, defined for all  x \epsilon \mathbb{R} , is said to be multiplicative if, for all  y \epsilon \mathbb{R} ,  x \epsilon \mathbb{R}
This sentence seems to be cut off.
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bigmansouf
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(Original post by RDKGames)
This sentence seems to be cut off.
sorry i hae corrected it
The real function f, defined for all  x \epsilon \mathbb{R} , is said to be multiplicative if, for all  y \epsilon \mathbb{R} ,  x \epsilon \mathbb{R}

f(xy)= f(x) f(y)
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RDKGames
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(Original post by bigmansouf)
sorry i hae corrected it
The real function f, defined for all  x \epsilon \mathbb{R} , is said to be multiplicative if, for all  y \epsilon \mathbb{R} ,  x \epsilon \mathbb{R}

f(xy)= f(x) f(y)
OK, then for the first one, let y = 0 and so you get that f(x \cdot 0) = f(x) f(0). Thus f(0) = f(x) f(0)

Can you move on from here?
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bigmansouf
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(Original post by RDKGames)
OK, then for the first one, let y = 0 and so you get that f(x \cdot 0) = f(x) f(0). Thus f(0) = f(x) f(0)

Can you move on from here?
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 y = 0
f(x \cdot 0) = f(x) f(0).
f(0) = f(x) f(0)
 {f(0)-f(x)f(0)}=0
f(0)(1-f(x))=0
 f(0)=0 or 1-f(x)=0
 f(0)=0 or f(x)=1

for the part b
let y = 1
f(x \cdot 1) = f(x) f(1).
f(x) = f(x) f(1)
 {f(x)-f(x)f(1)}=0
f(x)(1-f(1))=0
 f(x)=0 or 1-f(1)=0
 f(0)=0 or f(1)=1


sorry but could you please help me with the last part

c)  f(x)^{n} =   \left \{ f(x) \right \}^{n} for all positive integers n.
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RDKGames
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(Original post by 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 y = 0
f(x \cdot 0) = f(x) f(0).
f(0) = f(x) f(0)
 {f(0)-f(x)f(0)}=0
f(0)(1-f(x))=0
 f(0)=0 or 1-f(x)=0
 f(0)=0 or f(x)=1

for the part b
let y = 1
f(x \cdot 1) = f(x) f(1).
f(x) = f(x) f(1)
 {f(x)-f(x)f(1)}=0
f(x)(1-f(1))=0
 f(x)=0 or 1-f(1)=0
 f(0)=0 or f(1)=1


sorry but could you please help me with the last part

c)  f(x)^{n} =   \left \{ f(x) \right \}^{n} for all positive integers n.
Pretty sure that’s supposed to be:

 f(x^n) = [f(x)]^n

Is that what you meant?
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bigmansouf
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(Original post by RDKGames)
Pretty sure that’s supposed to be:

 f(x^n) = [f(x)]^n

Is that what you meant?
yes sorry that is what i meant
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RDKGames
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(Original post by bigmansouf)
yes sorry that is what i meant
f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1})

But by the same principle,

f(x^{n-1}) = f(x)f(x^{n-2}).

So f(x^n) = f(x) f(x) f(x^{n-2}).

Try using this key argument to complete the question.
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bigmansouf
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(Original post by RDKGames)
f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1})

But by the same principle,

f(x^{n-1}) = f(x)f(x^{n-2}).

So f(x^n) = f(x) f(x) f(x^{n-2}).

Try using this key argument to complete the question.
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
so i try to use proof via induction to solve it



prove

 f(x^n)=\left \{ f(x) \right \}^n for all positive integers n


first show that the step hold for n= 1

 f(x^1)=\left \{ f(x) \right \}^1
 \therefore
 f(x)=\left \{ f(x) \right \}

 f(x) =  f(x)



Suppose it holds for n = k

 f(x^n)=\left \{ f(x) \right \}^n

 f(x^k)=\left \{ f(x) \right \}^k
 f(x^1 . x^{k-1})=\left \{ f(x) \right \}^{1} . \left \{ f(x) \right \}^{k-1}
 f(x^1 . x^{k-1})= f(x)^1 . f(x)^{k-1}
 f(x^1) . f(x^{k-1})= f(x)^1 . f(x)^{k-1}
 f(x) . f(x^{k-1})= f(x) . f(x)^{k-1}

( f(x) . f(x^{k-1})) - (f(x) . f(x)^{k-1})=0

f(x)[( f(x^{k-1})) - ( f(x)^{k-1})]=0

 f(x)=0 or  [( f(x^{k-1})) - ( f(x)^{k-1})]=0


 f(x) = 0  is invalid since n represents all positive integers



 f(x^{k-1}) = f(x)^{k-1}
since  f(x^{k-1}) = f(x)^{k-1}

thus  f(x^{k}) = f(x)^{k}


then let n = k+1

 f(x^{k+1})=\left \{ f(x) \right \}^{k+1}

 f(x^{k+1})=\left \{ f(x) \right \}^{k+1}
 f(x^{k+1})= f(x)^{k+1}
 f(x^{k}).f(x^{1}) = f(x)^{k}.f(x)^{1}

 [f(x^{k}).f(x^{1})]- [f(x)^{k}.f(x)^{1}] = 0
 [f(x^{k}).f(x)]- [f(x)^{k}.f(x)] = 0
 f(x)[f(x^{k})-f(x)^{k}] = 0
 f(x) =0 or  [f(x^{k})-f(x)^{k}]=0


 f(x) = 0 is invalid since n represents all positive integers



 [f(x^{k})-f(x)^{k}]=0
 f(x^{k})=f(x)^{k}


since  f(x^{k}) = f(x)^{k}

thus  f(x^{k+1}) = f(x)^{k+1}


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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bigmansouf
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(Original post by RDKGames)
f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1})

But by the same principle,

f(x^{n-1}) = f(x)f(x^{n-2}).

So f(x^n) = f(x) f(x) f(x^{n-2}).

Try using this key argument to complete the question.
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
so i try to use proof via induction to solve it
prove
 f(x^n)=\left \{ f(x) \right \}^n for all positive integers n

first show that the step hold for n= 1
 f(x^1)=\left \{ f(x) \right \}^1
 \therefore
 f(x)=\left \{ f(x) \right \}
 f(x) =  f(x)

Suppose it holds for n = k
 f(x^n)=\left \{ f(x) \right \}^n
 f(x^k)=\left \{ f(x) \right \}^k
 f(x^1 . x^{k-1})=\left \{ f(x) \right \}^{1} . \left \{ f(x) \right \}^{k-1}
 f(x^1 . x^{k-1})= f(x)^1 . f(x)^{k-1}
 f(x^1) . f(x^{k-1})= f(x)^1 . f(x)^{k-1}
 f(x) . f(x^{k-1})= f(x) . f(x)^{k-1}
( f(x) . f(x^{k-1})) - (f(x) . f(x)^{k-1})=0
f(x)[( f(x^{k-1})) - ( f(x)^{k-1})]=0
 f(x)=0 or  [( f(x^{k-1})) - ( f(x)^{k-1})]=0
 f(x) = 0  is invalid since n represents all positive integers
 f(x^{k-1}) = f(x)^{k-1}
since  f(x^{k-1}) = f(x)^{k-1}
thus  f(x^{k}) = f(x)^{k}

then let n = k+1
 f(x^{k+1})=\left \{ f(x) \right \}^{k+1}
 f(x^{k+1})=\left \{ f(x) \right \}^{k+1}
 f(x^{k+1})= f(x)^{k+1}
 f(x^{k}).f(x^{1}) = f(x)^{k}.f(x)^{1}
 [f(x^{k}).f(x^{1})]- [f(x)^{k}.f(x)^{1}] = 0
 [f(x^{k}).f(x)]- [f(x)^{k}.f(x)] = 0
 f(x)[f(x^{k})-f(x)^{k}] = 0
 f(x) =0 or  [f(x^{k})-f(x)^{k}]=0
 f(x) = 0 is invalid since n represents all positive integers
 [f(x^{k})-f(x)^{k}]=0
 f(x^{k})=f(x)^{k}
since  f(x^{k}) = f(x)^{k}
thus  f(x^{k+1}) = f(x)^{k+1}

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

f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1}) - 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 n with n-1 everywhere in (*). You get that f(x^{n-1}) = f(x) f(x^{n-2}). Substitute this result into (*) and you end up with

f(x^n) = f(x)[f(x) f(x^{n-2})] = [f(x)]^2 f(x^{n-2}) ... (**)

We can repeat the trick. Replace every n in (*) with n-2. We have that f(x^{n-2}) = f(x) f(x^{n-3}). Substitute this into (**) and you end up with

f(x^n) = [f(x)]^2 [f(x)f(x^{n-3})] = [f(x)]^3f(x^{n-3})

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 x on the RHS will become zero, and the power of f will be precisely n. In this stage, we end up with

f(x^n) = [f(x)]^nf(x^0) ... (***)

but we know that for a general non-zero function f(x), we must have f(x^0) = f(1) = 1.

Hence (***) reduces to f(x^n) = [f(x)]^n

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
This is fine.

Suppose it holds for n = k
I am really unsure what you think you are doing here.

The word "suppose" means assume something. If you assume something is true, you don't need to manipulate it right there and then!

This step should literally only say:

"Suppose the statement holds for n=k; so we have f(x^k) = [f(x)]^k"

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:

f(x^{k+1}) = f(x^k \cdot x) = f(x^k)f(x)

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

We have...

f(x^k)f(x) = [f(x)]^kf(x) = [f(x)]^{k+1}

Hence f(x^{k+1}) = [f(x)]^{k+1}


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.
Good wording.

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
f(xy) = f(x)f(y) is what you want. The simplest example is literally just

f(x) = x.

We have; f(xy) = xy = f(x)f(y).
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bigmansouf
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(Original post by 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 f(x) is multiplicative.

f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1}) - 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 n with n-1 everywhere in (*). You get that f(x^{n-1}) = f(x) f(x^{n-2}). Substitute this result into (*) and you end up with

f(x^n) = f(x)[f(x) f(x^{n-2})] = [f(x)]^2 f(x^{n-2}) ... (**)

We can repeat the trick. Replace every n in (*) with n-2. We have that f(x^{n-2}) = f(x) f(x^{n-3}). Substitute this into (**) and you end up with

f(x^n) = [f(x)]^2 [f(x)f(x^{n-3})] = [f(x)]^3f(x^{n-3})

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 x on the RHS will become zero, and the power of f will be precisely n. In this stage, we end up with

f(x^n) = [f(x)]^nf(x^0) ... (***)

but we know that for a general non-zero function f(x), we must have f(x^0) = f(1) = 1.

Hence (***) reduces to f(x^n) = [f(x)]^n

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

This step should literally only say:

"Suppose the statement holds for n=k; so we have f(x^k) = [f(x)]^k"

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:

f(x^{k+1}) = f(x^k \cdot x) = f(x^k)f(x)

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

We have...

f(x^k)f(x) = [f(x)]^kf(x) = [f(x)]^{k+1}

Hence f(x^{k+1}) = [f(x)]^{k+1}




Good wording.



f(xy) = f(x)f(y) is what you want. The simplest example is literally just

f(x) = x.

We have; f(xy) = xy = f(x)f(y).
thank you
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