C3 Functions Difficult Question Please help

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.

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?

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?

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.

$f(x^n) = f(x \cdot x^{n-1}) =f(x) f(x^{n-1})$

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

first show that the step hold for n= 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.

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

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; $f(xy) = xy = f(x)f(y)$.