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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 '16 at 3:34
  • $\begingroup$ Yes, non-trivial $\endgroup$ – quietContest Sep 12 '16 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)... $\endgroup$ – Benjamin Dickman Sep 12 '16 at 4:27
  • $\begingroup$ I can abuse "infinite subset of R" to make many more such but they all have various forms of degeneracy. $\endgroup$ – Joshua Sep 12 '16 at 21:10
  • $\begingroup$ Consider the case $x=y$. $\endgroup$ – Bumblebee Sep 27 '16 at 18:42
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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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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 $\endgroup$ – quietContest Sep 12 '16 at 3:52
  • $\begingroup$ Glad you like it ;) next time I'll take care to format a little more nicely $\endgroup$ – Stahl Sep 12 '16 at 3:57
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    $\begingroup$ @tenCupMaximum Just a question: why did you unaccept this answer? $\endgroup$ – EKons Sep 14 '16 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. $\endgroup$ – quietContest Sep 14 '16 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 '16 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 $\endgroup$ – quietContest Sep 12 '16 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. $\endgroup$ – Ross Millikan Sep 12 '16 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 $\endgroup$ – quietContest Sep 12 '16 at 3:49
  • $\begingroup$ Easily the best answer. Why hasn't it been accepted? $\endgroup$ – MathematicsStudent1122 Sep 12 '16 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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