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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...

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    $\begingroup$ The zero function obviously satisfies all of this, but I assume you mean non-trivial function. $\endgroup$
    – Alex Ortiz
    Sep 12, 2016 at 3:34
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    $\begingroup$ Yes, non-trivial $\endgroup$ Sep 12, 2016 at 3:37
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    $\begingroup$ Your question is well-formulated, but if you are unfamiliar with the term "homomorphism" then I might suggest that you read up on it (or keep the word in mind for the future). Although the sort of function you ask after does not exist outside of the zero-function, there are functions that "respect addition and multiplication" i.e. for which $f(x+y) = f(x) + f(y)$ and $f(x \cdot y) = f(x) \cdot f(y)$. Although such functions would need to be equipped with an appropriate domain and range (and suitable notions of addition and multiplication)... $\endgroup$ Sep 12, 2016 at 4:27
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    $\begingroup$ I can abuse "infinite subset of R" to make many more such but they all have various forms of degeneracy. $\endgroup$
    – Joshua
    Sep 12, 2016 at 21:10
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    $\begingroup$ Consider the case $x=y$. $\endgroup$
    – Bumblebee
    Sep 27, 2016 at 18:42

7 Answers 7

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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.

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    $\begingroup$ I gave the $100^{th}$ upvote. :-) $\endgroup$
    – SarGe
    Aug 21, 2020 at 13:56
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The only such function is $f\equiv 0$. $f(0) + f(0) = f(0 + 0) = f(0)$, so that $f(0) = 0$. But then $f(x) = f(x + 0) = f(x)f(0) = 0$.

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  • $\begingroup$ Since you all posted around the same time, I picked the one I liked best. You're the winner. Though I would have liked both you and Mr. Millikan to format your posts in a manner more like user2825632 $\endgroup$ Sep 12, 2016 at 3:52
  • $\begingroup$ Glad you like it ;) next time I'll take care to format a little more nicely $\endgroup$
    – Stahl
    Sep 12, 2016 at 3:57
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    $\begingroup$ @tenCupMaximum Just a question: why did you unaccept this answer? $\endgroup$
    – EKons
    Sep 14, 2016 at 11:55
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    $\begingroup$ I felt the new accepted answer was more complete. I had a lot of difficulty with the decision to change the accepted answer to that one especially after writing the above comment, and if the content of the new one had been similar to the first 3 that were posted, I wouldn't have bothered. It's true that I should have at least given an explanation... I'm sorry for any trouble it's caused, I know what it's like to lose an accepted answer, it's not any fun. $\endgroup$ Sep 14, 2016 at 18:12
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    $\begingroup$ No worries. You have every right to change an accepted answer, and the answer you did accept addresses a possibility mine did not. I was not unhappy with your decision; in fact, I thought it was the answer that should be accepted! $\endgroup$
    – Stahl
    Sep 14, 2016 at 18:45
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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.

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    $\begingroup$ My hopes and dreams dashed in minutes! Thanks for playing $\endgroup$ Sep 12, 2016 at 3:44
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    $\begingroup$ The solution to most problems like this is finding clever values to plug in fo the variables. $0$ and $1$ are the first to try. $x=y$ and $x=-y$ come next. Then look at the particular problem and see if other values seem worthwhile. $\endgroup$ Sep 12, 2016 at 3:47
  • $\begingroup$ Yes I'll admit I hadn't thought very hard before posting this ;) Not that laziness is any excuse for failing to see the obvious $\endgroup$ Sep 12, 2016 at 3:49
  • $\begingroup$ Easily the best answer. Why hasn't it been accepted? $\endgroup$ Sep 12, 2016 at 22:59
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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.

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$$f(x)+f(y)=f(x)f(y)$$ 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 $$f(x)\equiv 0$$

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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 $

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Any functions where both sum and product = 2 will work. zero will not work because it results in division by zero. $(x+y)/(x*y)-1=0$

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