4
$\begingroup$

I am interested in a function which would stop Mathematica reducing fraction.

If my input is 50/100, Mathematica outputs 1/2, I want it to output 50/100. Any nice function which does this?

I need this because I want to define a recursive function with this behavior:

f[0]=50/100
f[n_]:=(Numerator[f[n-1]]+1)/(Denominator[f[n-1]]+1)

I want to obtain f[1]=51/101, not f[1]=2/3, which I get.

A more complicated example:

f[0]=50/100
f[n_]:=If[Mod[n,2]==0,(Numerator[f[n-1]+1])/n,Numerator[f[n-1]/(Denominator[f[n-1]‌​]+1)

This gets really annoying if I store 50/100 as {50,100}.

Improve this question
$\endgroup$
  • $\begingroup$ I need it because I am defining a recursive function, where I need the denominator and nominator of the previous element of the function. i.e. let's say my function is this: f[0]=50/100 f[n_]=Nominator[f[n-1]]+1/Denominator[f[n-1]]+1 Then this is useless if mathematica reduces the fractions... EDIT: First time here, no clue how to make enters:P $\endgroup$ – matti0006 Sep 25 '13 at 19:53
  • 4
    $\begingroup$ How about storing f[0]={50,100}? (and there is an "edit" button right below your question, I think it would be nice to include the example) $\endgroup$ – Pinguin Dirk Sep 25 '13 at 19:55
  • $\begingroup$ That's a possibility, but it is slower than just not reducing the fraction I would say. It will probably work for what I am doing, but I would still like to know if there is a generic function to achieve this. $\endgroup$ – matti0006 Sep 25 '13 at 19:57
  • $\begingroup$ Did that! (10 characters) $\endgroup$ – matti0006 Sep 25 '13 at 20:00
  • $\begingroup$ Well yeah, this is just a quick thing I typed up, the thing I really want to do is a bit more complicated. Also HoldForm just keeps "Numerator" as a string. I'll edit my orginal post with brackets. $\endgroup$ – matti0006 Sep 25 '13 at 20:02
3
$\begingroup$

Try this one

f[0] = HoldForm[50]/HoldForm[100];

f[n_] := HoldForm[#] &[ReleaseHold@Numerator[f[n - 1]] + 1]/
    HoldForm[#] &[ReleaseHold@Denominator[f[n - 1]] + 1];

f[5]

55/105

The second example

f2[0] = HoldForm[50]/HoldForm[100];

f2[n_] := 
 If[Mod[n, 2] == 0, 
  HoldForm[#] &[ReleaseHold@Numerator[f[n - 1]] + 1]/HoldForm[#] &[n],
   Numerator[f[n - 1]]/HoldForm[#] &[ReleaseHold@Denominator[f2[n - 1]] + 1]]

f2[1]

50/101

More elegant approach consists in the definition of your own irreducible Fraction

Fraction /: MakeBoxes[Fraction[n1_, n2_], StandardForm] := MakeBoxes@HoldForm[n1/n2]

f[0] = Fraction[50, 100];
f[n_] := Fraction[#1 + 1, #2 + 1] & @@ f[n - 1];
f[5]

55/105

f2[0] = Fraction[50, 100];
f2[n_] := If[Mod[n, 2] == 0, Fraction[#1 + 1, n], Fraction[#1, #2 + 1]] & @@ f2[n - 1];
f2[1]

50/101

Improve this answer
$\endgroup$
  • $\begingroup$ I think I sort of understand what this does, but I'm not entirely sure how I would apply it to the 2nd example, would it still work? $\endgroup$ – matti0006 Sep 25 '13 at 20:28
  • $\begingroup$ @matti0006 Yes, it is! See my update. $\endgroup$ – ybeltukov Sep 25 '13 at 20:34
  • $\begingroup$ Awesome, this seems to do what I want. Not sure if it's actually faster than making a list in my case, but I'd never even heard of Holdform and it seems useful in other situations! $\endgroup$ – matti0006 Sep 25 '13 at 20:35
  • $\begingroup$ @matti0006 See also the second update! I think it is more elegant than numerous HoldForm/ReleaseHold. $\endgroup$ – ybeltukov Sep 25 '13 at 20:56
  • $\begingroup$ This looks a lot more elegant, could you maybe elaborate on MakeBoxes though, mathematica's help function is not giving me enough insight to see exactly what it does. Also, what does the /: in Fraction mean? (Sorry, not been using mathematica for too long:)) $\endgroup$ – matti0006 Sep 25 '13 at 20:59
1
$\begingroup$

As mentionend in the comments, as a kind of "workaround", you could use:

f[0] = {50, 100}; f[x_] := f[x - 1] + 1

(using that Plus is Listable) which yields for f[1]:

{51,101}

or using memoization:

f[0] = {50, 100}; f[x_] := f[x] = f[x - 1] + 1

On another note (or reading the comments, thanks @Kuba), you won't even need recurisivity here:

f[x_]:= f[0] + n

EDIT

For the extended example (I guess there might be a typo in your comment below), you could do:

ff[0] = {50, 100};
ff[x_] := If[EvenQ[x], {#1 + 1, x}, {#1, #2 + 1}] & @@ ff[x - 1];
ff /@ Range[10]

{{50, 101}, {51, 2}, {51, 3}, {52, 4}, {52, 5}, {53, 6}, {53, 7}, {54, 8}, {54, 9}, {55, 10}}

I leave it to you to judge what's nicer to look at (and faster...)

Improve this answer
$\endgroup$
  • 2
    $\begingroup$ I believe the real case would be interesting since here f[n_]:=(50+n)/(100+n) :) $\endgroup$ – Kuba♦ Sep 25 '13 at 20:05
  • $\begingroup$ uhm... I must be tired $\endgroup$ – Pinguin Dirk Sep 25 '13 at 20:06
  • $\begingroup$ This works in this simple case, as I said in the comments above, but as things get more complicated, this is less and less effective, say I want instead to do this: f[0]=50/100 f[n]=If[Mod[n,2]==0,(Numerator[f[n-1]+1])/n,Numerator[f[n-1]/(Denominator[f[n-1]]+1) Using suggested notation, this gets very unreadable with all the brackets, since you'll have to use f[n-1][[1]] and f[n-1][[2]] $\endgroup$ – matti0006 Sep 25 '13 at 20:06
  • $\begingroup$ I will add this to original question for better notation... $\endgroup$ – matti0006 Sep 25 '13 at 20:07
  • $\begingroup$ Edited my original post to show how problems can arise if you do something a bit more difficult, it is still possible to store it as a list (obviously), but it won't be very pretty code. $\endgroup$ – matti0006 Sep 25 '13 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.