That seems a bit underspecified, or open-ended, at the moment.

For example, pi could be 1p1 (or o.1) or pi could be any of a number
of algorithms: https://en.wikipedia.org/wiki/Category:Pi_algorithms

Meanwhile, there's also the precision aspect - that could also be
specified in a variety of ways.

In other words, you might be asking for something like this:
   7%~x:<.0.5+7*1p1
22r7

Or you might have something different in mind, like perhaps this:
   ~.@(#~ (= <./)@:|@:-&1p1),%/~i.100x
22r7

Or, significantly slower, but almost tolerable:
   ~.@(#~ (= <./)@:|@:-&1p1),%/~i.1000x
355r113

Or, ... any of a variety of other mechanisms... (and even brute force
search techniques could be made to be significantly more efficient - a
moment's thought would show that incorporating the rounding technique,
for example, could have increased the speed of that last search by
something near a factor of 1000).

Anyways, if you can say a bit more about what you were aiming for,
maybe we could do a better job of getting there?

Thanks,

-- 
Raul

On Wed, Mar 9, 2016 at 11:18 AM, Don Guinn <[email protected]> wrote:
> How about rounding to a rational of some precision like pi rounded to 22r7 ?
> On Mar 9, 2016 8:52 AM, "Kip Murray" <[email protected]> wrote:
>
>> Thanks, Roger and Raul.  Not understanding  GCD +.  I had a Rube Goldberg
>> solution involving  Format ":  and  Do ".   --Kip
>>
>> On Wednesday, March 9, 2016, Roger Hui <[email protected]> wrote:
>>
>> > den and num illustrate different ways of computing the same thing,
>> > hopefully in so doing improves understanding.  If you want speed 2&x: is
>> > going to be faster.
>> >
>> > Also, watch this:
>> >
>> >    2 x: o. 1
>> > 1285290289249 409120605684
>> >    x: o. 1
>> > 1285290289249r409120605684
>> >
>> >    2 x:!.0 o. 1
>> > 884279719003555 281474976710656
>> >    x:!.0 o. 1
>> > 884279719003555r281474976710656
>> >
>> >
>> >
>> >
>> > On Wed, Mar 9, 2016 at 7:20 AM, Mike Day <[email protected]
>> > <javascript:;>> wrote:
>> >
>> > > Roger, are these (among) recommended and preferred methods for
>> > > recovering num & den,  or do they just show an elegant way of
>> > > avoiding 2 x: ?
>> > >
>> > > Thanks,
>> > >
>> > > Mike
>> > >
>> > >
>> > > On 09/03/2016 04:32, Roger Hui wrote:
>> > >
>> > >>     x=: %/?2$10^8x
>> > >>     x
>> > >> 69904549r40669028
>> > >>
>> > >>     den=: [: % 1 +. ]
>> > >>     num=: * den
>> > >>
>> > >>     den x
>> > >> 40669028
>> > >>     num x
>> > >> 69904549
>> > >>
>> > >>
>> > >>
>> > >>
>> > >> On Tue, Mar 8, 2016 at 7:28 PM, Kip Murray <[email protected]
>> > <javascript:;>>
>> > >> wrote:
>> > >>
>> > >> That's great!  It's still a nice puzzle to write your own.  --Kip
>> > >>>
>> > >>> On Tuesday, March 8, 2016, Raul Miller <[email protected]
>> > <javascript:;>> wrote:
>> > >>>
>> > >>>     2 x: 6r4
>> > >>>> 3 2
>> > >>>>
>> > >>>> --
>> > >>>> Raul
>> > >>>>
>> > >>>>
>> > >>>> On Tue, Mar 8, 2016 at 10:04 PM, Kip Murray <[email protected]
>> > <javascript:;>
>> > >>>> <javascript:;>> wrote:
>> > >>>>
>> > >>>>> How do you find the numerator and denominator in lowest terms of a
>> > >>>>>
>> > >>>> rational
>> > >>>>
>> > >>>>> fraction?  For example,
>> > >>>>>
>> > >>>>>      nd 6r4
>> > >>>>>   3 2
>> > >>>>>
>> > >>>>> --Kip Murray
>> > >>>>>
>> > >>>>>
>> > >>>>>
>> > >>>>> --
>> > >>>>> Sent from Gmail Mobile
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