How can I prevent Mathematica from simplifying fractions? What I would like to do is run a test on a list of fractions to see how many are in their reduced form.

For example, I have a list:

1/10, 2/10, 5/10

I do not want Mathematica to treat it as:

1/10, 1/5, 1/2

I would like it left exactly as it was originally entered for comparison, i.e. if one list contains 2/10 and the other list contains 1/5 they should be treated as different elements.

I tried converting the elements to strings but Mathematica still simplified the fractions...

What is the best way to maintain the entire list in the original form?

Something like:

a = {1/10, 2/10, 3/10, 4/10, 5/10};
Internal`RationalNoReduce[Numerator[a], Denominator[a]]

still gives

Internal`RationalNoReduce[{1, 1, 3, 2, 1}, {10, 5, 10, 5, 2}]
  • $\begingroup$ Related: mathematica.stackexchange.com/questions/69655/… $\endgroup$ – Michael E2 Jan 10 '15 at 23:51
  • 2
    $\begingroup$ See my answer here. $\endgroup$ – Chip Hurst Jan 11 '15 at 2:57
  • $\begingroup$ please see edit $\endgroup$ – Raksha Jan 11 '15 at 18:05
  • $\begingroup$ I reopened the question and posted an answer addressing your stated goal. Your focus on "an entire list" makes me wonder if you are importing a list of fractions; if this is the case it will be important to import them in a held form, and you will need to describe the external format in your question. $\endgroup$ – Mr.Wizard Jan 11 '15 at 19:23

Your expression will be evaluated unless it is entered in a held form, or a form that natively does not evaluate. For example:

a = Hold[1/10, 2/10, 3/10, 4/10, 5/10];

b = {{1,10}, {2,10}, {3,10}, {4,10}, {5,10}};

The second representation is a bit easier to work with as the first, which while not evaluating is nevertheless parsed differently from what you might expect:

a // FullForm
Hold[Times[1, Power[10, -1]], Times[2, Power[10, -1]], Times[3, Power[10, -1]], 
 Times[4, Power[10, -1]], Times[5, Power[10, -1]]]

Note that every fraction is not parsed as Rational but as Times and Power.

You can work with either format however. First the easy example (ref: Apply):

GCD @@@ b
{1, 2, 1, 2, 5}

A GCD of one means that the fraction is in reduced form, while any other value means it is not. You could for example Select the fractions that are or are not in reduced form:

Select[b, 1 == GCD @@ # &]
{{1, 10}, {3, 10}}

To work with the first format you could convert it to the second format with:

List @@ (a /. n_*(d_^-1) :> {n, d})
{{1, 10}, {2, 10}, {3, 10}, {4, 10}, {5, 10}}
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  • $\begingroup$ i think this is what I was looking for, thanks! :D $\endgroup$ – Raksha Jan 11 '15 at 19:42
  • $\begingroup$ @Solarmew Glad I could help. :-) By the way including in a question what you are actually trying to accomplish is a good idea and in this case necessary to distinguish it from prior questions. $\endgroup$ – Mr.Wizard Jan 11 '15 at 19:47

Use HoldForm




(* Out[68]= 1/5 *)
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  • $\begingroup$ Tyilo, but that only works for each element. Is there a way to apply it to the entire list without loops? $\endgroup$ – Raksha Jan 11 '15 at 4:07
  • $\begingroup$ @Solarmew yes, the way is to apply HoldForm to the entire list HoldForm[{1/10, 2/10, 5/10}] $\endgroup$ – user484 Jan 11 '15 at 5:28

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