[?] Keep x/y (Divide[x,y]) from turning into Times[x, Power[y,-1]]?

Questions
,
1

I would like to have a way to keep x/y in the Divide[x,y] notation, but it seems
to go immediately to the standard form

qTimes[x, Power[y,-1]]

There does not seem to be any attribute that controls this as there is
for properties such as commutative and associative, etc.
Similarly, you have

FullForm[2/(3x)]

as

Times[2, Times[Rational[1, 3], Power[x, -1]]]

when
I might want to distinguish between 2/(3x) and (2/3)(1/x) etc.
I know this is normally a feature, not a bug, but there are times when I’d like to have
more control over conversion to standard form. Is there any way to do this??
Much thanks. - Elaine

2

It depends on what you want to do with it. You may use Inactive[Divide][x, y], or define your own operator, with its rules:

 Format[myDivide[x_, y_]] := DisplayForm@FractionBox[x, y]
3

I guess you should make your own Divide with it’s own rules:

ClearAll[MyDivide]
MyDivide[x_, y_] MyDivide[a_, b_] ^:= MyDivide[x a, y b]
MyDivide[x_, y_] + MyDivide[a_, b_] ^:= MyDivide[b x + a y, b y]
Format[MyDivide[x_, y_]] := DisplayForm[FractionBox[x, y]]

You can now define your own definitions for multiply and plus et cetera, see example above…

4

Thanks all, but I’m sorry, I wasn’t very clear in my question. I meant I wanted to start from the standard infix notation x/y before converting to some internal notation such as MyDivide. I don’t want the end user (not me) to have to know about something such as MyDivide or Inactive. I want them to be able to write x/y and, after I apply a translation function, get MyDivide[x,y]. I also want them to be able to write x y^(-1) and have it stay that way, so I can’t just convert Times[x, Power[y, -1]] to MyDivide. Something like Inactive or Inactivate would be great if I could set it as an attribute, but that doesn’t seem possible, and if I wrap the whole expression, e.g. FullForm[Inactivate[x/y]], it’s already too late as I get Inactive[Times][x, Inactive[Power][y, -1]].

I know I could do string manipulations, but I’m trying to avoid having to write my own parser.

5

OK, I think I was going about this backwards. If I let Times[x, Power[y, -1]] be the standard representation for divide, then I can do ToExpression[StringReplace[], “^”->“myPower”] and end up with either x/y or x myPower[y, -1].