3

Exact Question: Invent a single binary operator $*$ such that for every real numbers $a$ and $b$, the operations $a + b$, $a - b$, $a \times b$, $a \div b$ can be created by applying $*$ (multiple times), starting with only $a$'s and $b$'s

From my interpretation, you can apply $*$ recursively some number of times with carefully selected parameters to produce the desired outcome.

I thought the operator should be a combination of $a - b$ and $a \cdot b^{-1}$ Since $-$ and $\div$ can produce $+$ and $\times$ respectively

Please do not tell me the full answer. Give me a hint to point me towards the right path

BrianH
  • 158
  • 10
  • Hint: First derive all operations from subtraction and reciprocal and find a single binary operation that gets you these. – John Dec 21 '18 at 05:40
  • 1
    @John I realize that, as I've stated in my post as well. But could you give a hint on how to combine them into expression/operator? – BrianH Dec 21 '18 at 05:43
  • That would involve giving away what $\star$ is. But you said you only wanted a hint. – John Dec 21 '18 at 05:45

1 Answers1

0

I'm not sure that this is possible. You wish to define a binary operator $*:\mathbb{R}\times \mathbb{R}\rightarrow\mathbb{R}$ that has some properties, but in particular you must define what $0*0$ is equal to. This means that however $*$ is defined in terms of the four fundamental operations, it cannot involve division, since $0/0$ is an undefined quantity. But if we cannot involve division in the definition, then there is no way to get back $1/2$ from any number of applications (and groupings) of $*$ to $1$ and $2$, since $\mathbb{Z}$ is a ring under $+$ and $\times$.

Where does that leave us? The definition of $*$ we seek must use an operation beyond the four fundamental operations, and still must be able to recreate the effects of all four under some number of applications (and groupings)... which just strikes me as impossible.

DanielJack
  • 66
  • 1
  • What if we assume for the sake of simplicity and define * on $\mathbb{R} \setminus {0}$. I mean 0 + 0 = 0, 0 - 0 = 0, 0 $\cdot$ 0 = 0, 0/0 is irrelavent. So maybe glipmse over the fact? – BrianH Dec 21 '18 at 05:41
  • @BrianH That is a possibility; $0$ does present the obstructions I raise here, but I will have to think longer to come up with an answer in this restricted version of the problem. – DanielJack Dec 21 '18 at 05:43
  • @BrianH After thinking a bit more, I remain unconvinced that this is solvable. If division occurs anywhere in the definition of $*$ then whatever function of $a$ and $b$ we are dividing by must never be equal to $0$. Yes, such functions are easy to build, but to then get back values like $a+b$ at the end of the day? I just don't buy it. Where did you find this problem? Is it from a class? Textbook? You used the label contest math, so is there a particular year and contest this is from? – DanielJack Dec 21 '18 at 05:52
  • This is from a class called "Putnam Seminar" taught by Po Shen Loh, renown math olympiad coach (https://en.wikipedia.org/wiki/Po-Shen_Loh). I am not sure where he got the problem, but I believe there is an answer. He may have been a little less formal with the delivery of the problem... – BrianH Dec 21 '18 at 05:58
  • It is in A problem seminar by Donald J. Newman. – John Dec 21 '18 at 05:59
  • @John Can you link the problem and/or the solution? Because I still don't see how you can circumvent the issues I've raised. – DanielJack Dec 21 '18 at 06:01
  • Oh this has been asked before... https://math.stackexchange.com/questions/338370/is-there-an-numeric-arithmetic-with-a-single-operator – BrianH Dec 21 '18 at 06:01
  • @BrianH I can't say I like what they have done there, lol. Feels like cheating to me to have access to constants. Maybe there is a workaround as suggested, but even if you can synthesize $0$ and $1$ from some application of something, the fact that the operation is ill defined on $x*x$ kind of makes me mad. – DanielJack Dec 21 '18 at 06:08
  • Wait I am also kind of mad because the prompt says using only a's and b's, which are arbitrary. Taking 0 for granted seems wrong – BrianH Dec 21 '18 at 06:30