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My objective for this post is to set parallels between logical, set, and numerical unary and binary operations. That is, I want to relate:

Union($\cup$) --> Or($\vee$) --> Addition (+)

Intersection($\cap$) --> And($\wedge$) --> Multiplication(*)

The definitions of the union and intersection include the and and or operators, so their relationship is clear. Addition and multiplication were set in their positions because multiplication can be seen as "stronger" than addition, as an intersection can be seen as a "stronger" operation than a union (In my eyes).

In set theory and Boolean algebra, De-Morgan's law states:

Sets: $(A\cup B)^c=A^c \cap B^c \text{ and } (A\cap B)^c=A^c \cup B^c$

Logic: $\neg (A\vee B)=\neg A \wedge \neg B \text{ and } \neg (A\wedge B)=\neg A \vee \neg B$

However, numerical operations are different. I want to find a function $f(x)$ such that:

$f(a+b)=f(a)f(b) \text{ and } f(ab)=f(a)+f(b)$

This $f(x)$ would be the "complement" or "negation" of the reals.

The exponential ($r^x$) satisfies the first equation but not the second, and the logarithm $\log_r (x)$ satisfies the second equation but not the first, but I want a function that satisfies both at once. I've tried taking the convolution of those functions (assuming a range of $(0,\infty)$), but the resulting function did not hold the desired properties. Other combinations, such as multiplication and addition, also do not work. Using this unknown function $f(x)$, we can include in our table:

$(A-B)^c$ --> Implication($\Rightarrow$) --> $a+f(b)$

Difference($A-B$) --> Non-implication($\not\Rightarrow$) --> $f(a)\cdot b$

$(A\backslash B)^c$ --> XOR($\not\Leftrightarrow$) --> $bf(a)+af(b)$

Symmetric Difference($A\backslash B$) --> Biconditional(XNOR*)($\Leftrightarrow$) --> $f(bf(a)+af(b))$ *This association is not completely correct, but okay for this case

Another couple of associations I made, which may be wrong:

$\subset$ --> $\Leftarrow$ --> $<$

$\subseteq$ --> $\Leftarrow$ --> $\leq$

$\supset$ --> $\Rightarrow$ --> $>$

$\supseteq$ --> $\Rightarrow$ --> $\geq$

My three questions are:

  1. What is this unknown function $f(x)$?
  2. Can we find any comparisons for other types of binary/unary operators, and are there better operations to use in the reals for this example with a “complement” function that satisfies De-Morgan’s law for this specific pair of operations (except for use of logic operators with numbers, eg. $3\wedge 5$)?
  3. Can we add any other operations on sets, logic, or numbers to these associations of operators?

Please correct me if I get any of these details wrong!

The domain of $f$ only has to include to positive real numbers. While 0 and negative numbers would be nice to include, logarithm properties make that extremely difficult.

This question is already answered for question 1 (the function must be 0), but answers for questions 2 and 3 would be welcome!

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    $\begingroup$ You probably definitely want to assume $f$ isn't constantly zero, but are there any other constraints to impose? $\endgroup$ Commented Oct 10 at 15:38
  • $\begingroup$ See artofproblemsolving.com/community/c4h607240p3608710 on AoPS. $\endgroup$ Commented Oct 10 at 15:40
  • $\begingroup$ The answers given so far show that the answer depends highly on the domain of $f$. Please update your question with the exact assumptions you are making. $\endgroup$ Commented Oct 10 at 19:16

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If $0$ is in the domain of your function the only solution is the constant function to $0$.

Since $f(0)=f(0 \cdot 0) = f(0)+f(0)$ we must have $f(0)=0$. But now $f(x) = f(x+0)=f(x)f(0)=f(x)0=0$ so $f$ is constantly zero. There are no other solutions.

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    $\begingroup$ Nitpicking: Unless I am mistaken, this works only if $0$ is in the domain of $f$ (which is not given in the question). $\endgroup$ Commented Oct 10 at 15:43
  • $\begingroup$ Oh, fair point. Not sure that's a nitpick. $\endgroup$ Commented Oct 10 at 15:43
  • $\begingroup$ A similar argument works if $1$ is in the domain of $f$ because then $f(1) = 0$. $\endgroup$ Commented Oct 10 at 15:48
  • $\begingroup$ @jjagmath: $f(x) = f(x-1 + 1) = f(x-1) f(1)$ ... ? $\endgroup$ Commented Oct 10 at 19:09
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    $\begingroup$ @jjagmath: Yes, the answer to this problem seems to depend highly on tthe domain of $f$. Unfortunately, OP did not update their question with the exact assumption. $\endgroup$ Commented Oct 10 at 19:13
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Let $\alpha$ in the domain of $f$ such that $f(\alpha)=\beta\neq 0$. I assume that $f$ is defined on $\bigoplus_{k\ge 1} \mathbb N_{>0} \alpha^k$.

Then $f(\alpha^2)=2\beta$ so $f(\alpha+\alpha^2)=2\beta^2$ and $f((\alpha+\alpha^2)^2)=4\beta^2$ which must be equal to $$f(\alpha^2+2\alpha^3+\alpha^4)=2\beta\times (3\beta)^2\times 4\beta=72\beta^4$$ so $18\beta^2=1$. But the same calculation is valid for $f(\alpha^2)=2\beta$, so $18(2\beta)^2=1$, which is impossible.

Therefore, such a function must be zero everywhere.

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  • $\begingroup$ It’s slightly disappointing that holds true. Hopefully there’s other operations that allow for a non-zero f(x). (That is, De-Morgan’s law holds for a different pair of operations in the real numbers) $\endgroup$ Commented Oct 11 at 1:32

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