"Solving for $f(2002)$ in $f(f(x)+f(y))=x+y$

[Albert1]

A function $f$, defined as $f(f(x)+f(y))=x+y,$ for all $x,y\in N$ find $f(2002)$

 

[Joppy]

Any hints for a dummy like me :(

 

[MarkFL]

A function $f$, defined as $f(f(x)+f(y))=x+y,$ for all $x,y\in N$ find $f(2002)$

My not-so-rigorous solution :

Spoiler

Let us assume that $f$ is linear, that is:

$$f(x+y)=f(x)+f(y)\tag{1}$$

$$f(ax)=a\cdot f(x)\implies f(x)=kx\tag{2}$$

Using (1), our functional equation becomes:

$$f(f(x))+f(f(y))=x+y$$

Now, let $y=x$, and the functional equation becomes:

$$f(f(x))=x$$

Using the implication from (2), we find:

$$f(kx)=x$$

$$k^2x=x$$

$$k=\pm1$$

Hence:

$$f(x)=\pm x\implies f(2002)=\pm2002$$

 

[Albert1]

My not-so-rigorous solution :

Spoiler

Let us assume that $f$ is linear, that is:

$$f(x+y)=f(x)+f(y)\tag{1}$$

$$f(ax)=a\cdot f(x)\implies f(x)=kx\tag{2}$$

Using (1), our functional equation becomes:

$$f(f(x))+f(f(y))=x+y$$

Now, let $y=x$, and the functional equation becomes:

$$f(f(x))=x$$

Using the implication from (2), we find:

$$f(kx)=x$$

$$k^2x=x$$

$$k=\pm1$$

Hence:

$$f(x)=\pm x\implies f(2002)=\pm2002$$

Spoiler

the answer is :
$f(2002)=2002$
hint:
prove for all $k\in N,f(k)=k$

 

[MarkFL]

Spoiler

You do see that $f(x)=-x$ also satisfies the functional equation?

 

[Albert1]

Spoiler

You do see that $f(x)=-x$ also satisfies the functional equation?

Spoiler

note the defnition of a function
$f(x)=\pm x, x\in N$ is not a function
for example :$f(f(1)+f(1))=1+1=2=f(2)$
how can it be possible $f(1)=1,also\,\, f(1)=-1$

 

[MarkFL]

Spoiler

I am saying to consider them as separate cases, both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation (which I probably should not have condensed as I did). However, this means there are (at least) 2 possible values for $$f(2002)$$. :D

 

[I like Serena]

Spoiler

I am saying to consider them as separate cases, both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation (which I probably should not have condensed as I did). However, this means there are (at least) 2 possible values for $$f(2002)$$. :D

Spoiler

Erm... if we consider the $f(x)=-x$, we have that $f(f(x)+f(y)) = f(-x-y)$, which is undefined, since $-x-y \notin \mathbb N$.

From the problem statement it's not immediately clear if that's acceptable, or if we do require that it is defined.
If we don't require that it's defined, then any function with $f(x) < 0$ for all $x\in\mathbb N$ is a solution, since effectively there are no constraints at all.
And if we do require it to be defined, that rules out the solution $f(x)=-x$...

 

[MarkFL]

Spoiler

Erm... if we consider the $f(x)=-x$, we have that $f(f(x)+f(y)) = f(-x-y)$, which is undefined, since $-x-y \notin \mathbb N$.

From the problem statement it's not immediately clear if that's acceptable, or if we do require that it is defined.
If we don't require that it's defined, then any function with $f(x) < 0$ for all $x\in\mathbb N$ is a solution, since effectively there are no constraints at all.
And if we do require it to be defined, that rules out the solution $f(x)=-x$...

Spoiler

I see it now...I was thinking only $x$ and $y$ need be natural numbers, but so does $-(x+y)$ if we consider $f(x)=-x$. I'll go stand in the corner now. (Sadface)

 

[mathbalarka]

Spoiler

Erm... if we consider the $f(x)=-x$, we have that $f(f(x)+f(y)) = f(-x-y)$, which is undefined, since $-x-y \notin \mathbb N$.

From the problem statement it's not immediately clear if that's acceptable, or if we do require that it is defined.
If we don't require that it's defined, then any function with $f(x) < 0$ for all $x\in\mathbb N$ is a solution, since effectively there are no constraints at all.
And if we do require it to be defined, that rules out the solution $f(x)=-x$...

Spoiler

To be super-nitpicky, in the original post $f$ is not required to be a function $\Bbb N \to \Bbb N$ though, so it might as well have taken negative integer arguments (the functional equation is required to be satisfied only for natural number arguments though). But probably you're right that that's what is meant.

 

[Joppy]

So we need to make the following assumptions (in the case that the answer is indeed $f(2002) = 2002$);

$f : \mathbb{N} \rightarrow \mathbb{N}$,

Linearity : $f(x + y) = f(x) + f(y)$, $f(ax) = af(x)$

and of course, $x, y \in \mathbb{N}$ as stated.

Is this correct?

Can we then just say that.. $f(f(x) + f(y)) = f(x + y) = x + y$, then clearly, $f(2002) = 2002$? Or am i doing something illegal.

This is the case that f(x) = x, f(y) = y.

 

[Albert1]

So we need to make the following assumptions (in the case that the answer is indeed $f(2002) = 2002$);

$f : \mathbb{N} \rightarrow \mathbb{N}$,

Linearity : $f(x + y) = f(x) + f(y)$, $f(ax) = af(x)$

and of course, $x, y \in \mathbb{N}$ as stated.

Is this correct?

Can we then just say that.. $f(f(x) + f(y)) = f(x + y) = x + y$, then clearly, $f(2002) = 2002$? Or am i doing something illegal.

This is the case that f(x) = x, f(y) = y.

Spoiler

may be I should make it clear :
$f:\ N \rightarrow\,\, N,\,\, and \,\,f(f(x)+f(y))=x+y, \,\, (for \,\,all \,\, x,y\in N)\,\,.find\,\, f(2002)$
now here is the question , I didn't say $f$ is linear .
this is a "not-so-rigorous solution "

 

[I like Serena]

Spoiler

may be I should make it clear :
$f:\ N \rightarrow\,\, N,\,\, and \,\,f(f(x)+f(y))=x+y, \,\, (for \,\,all \,\, x,y\in N)\,\,.find\,\, f(2002)$
now here is the question , I didn't say $f$ is linear .
this is a "not-so-rigorous solution "

Spoiler

A little more rigorous.

Let's assume that $0\in N$, then we have:

$f(f(0)+f(0))=0+0 \quad\Rightarrow\quad f(2f(0))=0 \quad\Rightarrow\quad f(0+0)=f(f(2f(0))+f(2f(0)))=2f(0) + 2f(0) \quad\Rightarrow\quad f(0)=0$

Lemma
Let $a=f(1)$. Then for all $k\in N$: $f(k)=ka$ and $f(ka)=k$.
Proof
By induction.
Base case: $f(0)=0=0a$ and $f(0a)=f(0)=0$.
Induction step: we assume the lemma holds up to $k$, then it follows that:
$f(k+1)=f(f(ka)+f(a))=ka+a=(k+1)a$ and $f((k+1)a)=f(ka+a)=f(f(k)+f(1))=k+1$.
Qed.

Since we have $f(k)=ka$ it follows that $f$ is linear, and as MarkFL has already shown, that implies:
$k=f(ka)=ka^2 \quad\Rightarrow\quad a=1$.

Therefore $f(2002)=2002$.

 

[I like Serena]

Can we then just say that.. $f(f(x) + f(y)) = f(x + y) = x + y$, then clearly, $f(2002) = 2002$? Or am i doing something illegal.

Given linearity, shouldn't that be $f(f(x) + f(y)) = f(f(x + y)) = x + y$, or more generally $f(f(x))=x$? (Wondering)

 

[Albert1]

Spoiler

A little more rigorous.

Let's assume that $0\in N$, then we have:

$f(f(0)+f(0))=0+0 \quad\Rightarrow\quad f(2f(0))=0 \quad\Rightarrow\quad f(0+0)=f(f(2f(0))+f(2f(0)))=2f(0) + 2f(0) \quad\Rightarrow\quad f(0)=0$

Lemma
Let $a=f(1)$. Then for all $k\in N$: $f(k)=ka$ and $f(ka)=k$.
Proof
By induction.
Base case: $f(0)=0=0a$ and $f(0a)=f(0)=0$.
Induction step: we assume the lemma holds up to $k$, then it follows that:
$f(k+1)=f(f(ka)+f(a))=ka+a=(k+1)a$ and $f((k+1)a)=f(ka+a)=f(f(k)+f(1))=k+1$.
Qed.

Since we have $f(k)=ka$ it follows that $f$ is linear, and as MarkFL has already shown, that implies:
$k=f(ka)=ka^2 \quad\Rightarrow\quad a=1$.

Therefore $f(2002)=2002$.

Spoiler

Why assume that $0\in N ?$
Never mind if $f$ is linear or not.(only use the given condition)
key point is how to prove $f(1)=1 $
you can not say $f(f(1)+f(0))=1+0=1$,since $0 \notin N$
we should prove $f(1)=1,f(2)=2,f(3)=3,----$and then use inductive method
suppose $f(1)=1$ (here you must prove)
$f(f(1)+f(1))=f(2f(1))=f(2)=1+1=2$
$f(f(1)+f(2))=f(1+2)=f(3)=1+2=3$
---------
$f(1)=1, why\,?$

 

[I like Serena]

Spoiler

Why assume that $0\in N ?$
Never mind if $f$ is linear or not.(only use the given condition)
key point is how to prove $f(1)=1 $
you can not say $f(f(1)+f(0))=1+0=1$,since $0 \notin N$
we should prove $f(1)=1,f(2)=2,f(3)=3,----$and then use inductive method
suppose $f(1)=1$ (here you must prove)
$f(f(1)+f(1))=f(2f(1))=f(2)=1+1=2$
$f(f(1)+f(2))=f(1+2)=f(3)=1+2=3$
---------
$f(1)=1, why\,?$

Spoiler

It's ambiguous whether $0\in N$ or not. See e.g. wiki.
The problem statement didn't mention it, so I took the liberty to include $0$, since the proof becomes a bit simpler if we assume $0$ is included. ;)

And no, I'm not using that $f$ is linear - I'm proving it as a spin-off.

 

[Albert1]

Prove:$f(1)=1$

Spoiler

suppoe $f(1)=c$
$f(2c)=f(c+c)=f(f(1)+f(1))=1+1=2----(1)$
we must prove $c=1$
if $c=1+a>1, \,\,and\,\, f(a)=b\,\, (a,b\in N)$
$f(2b)=f(b+b)=f(f(a)+f(a))=2a$
$f(2c)=f(2+2a)=f(f(2c)+f(2b))=2c+2b>2---(2)$
from $(1)(2)\,\,c>1$ is impossible
$\therefore c=1,\rightarrow f(1)=1$
now use inductive method
$f(f(1)+f(1))=f(2f(1))=f(2)=1+1=2$
$f(f(1)+f(2))=f(1+2)=f(3)=1+2=3$
---------
let $f(k)=k$
we have $f(k+1)=f(f(k)+f(1))=k+1$
the proof is done
therefore $f(2002)=2002$