A function in which addition and multiplication behave the same way

Question

Exponents have a well-known property:

$$x^ax^b = x^{a+b}$$

but

$$x^{a} + x^{b} \neq x^{a+b}$$

Similarly,

$$\log(a) + \log(b) = \log(ab) $$

But

$$\log(a)\log(b) \neq \log(ab)$$

So my question is this:

Is there a function $f$ on $\mathbb{R}$ or some infinite subset of $\mathbb{R}$ with the following properties

$$(1)\quad f(x)f(y) = f(x+y)$$ $$(2)\quad f(x)+f(y) = f(x+y)$$ ie $$(3)\quad f(x)+f(y) = f(x)f(y)$$

It seems that $(2)$ requires the function to be linear...

Answer 1

Your title expresses interest in "a function in which addition and multiplication behave the same way". That's condition (3) alone. Conditions (1) and (2) are unnecessarily-strong requirements that artificially restrict the possible solutions. Be that as it may ...

Let's invoke condition (3) with three arbitrary values, $x$, $y$, $z$.

$$\begin{align} f(x) + f(y) = f(x)\cdot f(y) \\ f(x) + f(z) = f(x)\cdot f(z) \end{align}$$ Subtracting, we get $$f(y) - f(z) = f(x)\cdot(\;f(y)-f(z)\;)\quad\to\quad\left(f(x)-1\right)\cdot\left(f(y)-f(z)\right) = 0$$ For all choices of $x$, $y$, $z$, at least one factor must vanish. We conclude that $f$ must be some constant; say, $k$. (The vanishing of the first factor requires specifically that $k=1$, but we'll go ahead and absorb this into the more-general statement.)

Then condition (3) reduces to $$k + k = k\cdot k \quad\to\quad k(k-2) = 0$$ so that $k = 0$ or $k = 2$. That is, we have two ways to satisfy condition (3):

$$f(x) \equiv 0 \qquad\text{or}\qquad f(x) \equiv 2$$

Imposing conditions (1) and (2) limits the solutions to just the first.

Answer 2

The only such function is $f\equiv 0$. $f(0) + f(0) = f(0 + 0) = f(0)$, so that $f(0) = 0$. But then $f(x) = f(x + 0) = f(x)f(0) = 0$.

Answer 3

From $(2), f(x)+f(0)=f(x+0)$, so $f(0)=0$. Then from $(1), f(x)f(0)=0=f(x)$, so only the zero function satisfies your requirements.

Answer 4

Take $y = 0$. Then we need to satisfy the second equation:

$$f(x) + f(0) = f(x+0)$$

From the second equation, we must have $f(0) = 0$.

Now take $y = -x$. We must now satisfy:

$$f(x) * f(-x) = f(x-x)$$

$$f(x) + f(-x) = f(0)$$

From the second equation, we need $f(x) = -f(-x)$. The first equation then becomes:

$$-f(x)^2 = f(0)$$

And we must have $f(x) = 0$ for all $x$. Therefore, only the function $f(x)=0$ satisfies your constraints.

Answer 5

you are only searching

$ f(x)+f(y) = f(x)f(y)$ <=> a+b = a b <=> $a-1\ne0, b = \frac{a}{a-1}$

It is like saying that for any x,y $f(y) = \frac{f(x)}{f(x)-1}$

for example for x = y, this leads to $\forall x, f(x) = \frac{f(x)}{f(x)-1}$ => $f(x) \in \{2,0\}$ . The solution $f(x)=2$ is not compatible with $4 = f(x)f(y) = f(x+y) = 2$ and $n \not=4$. Remains only $\forall x,f(x)=0 $

Answer 6

$$f(x)+f(y)=f(x)f(y)$$ implies $$f(x)+f(x) = f(x)f(x)$$ so $$2f(x)=\left(f(x)\right)^2$$ $$f(x)\left(f(x)-2\right)=0$$ So, for every $x$ must be either $f(x)=0$ or $f(x)=2$. (*)

However, if there were two numbers $y,z$ such that $f(y)=f(z)=2$, then $f(y+z)=f(y)+f(z)$ would be $4$, which contradicts (*). So there can be at most one such number $z$, that $f(z)=2$. (**)

However, if such $z$ exists then it can't be $1$ and $-1$ at the same time, hence at least one of $f(-1)$ and $f(1)$ is not $2$, so it must be $0$ by (*),
then at least one of $f(z-1) = f(z)+f(-1)$ and $f(z+1)=f(z)+f(1)$ equals $f(z)+0=f(z)$. That means at least two of $\{f(z-1), f(z), f(z+1)\}$ equal $2$, which contradicts (**).

Hence there's no such number for which $f$ output is $2$, and $f$ must be zero everywhere: $$f:x\mapsto 0$$ or $$f(x)\equiv 0$$