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I do not know how to code the following.

Suppose you have a list lst={x1,...,xn}and you need to find a number $Z$, to a given digits precision such that 

Z * lst={y1,..,yn}

in which all $y_i$ are integers. I am clueless.

Thanks in advance.

Edit: To be more precise, given a list of numbers lst={x1,...,xn}and a parameter $\varepsilon>0 $ I want to find the smallest real number $Z$ such that for each $i \leq n$ we have $\mid Z\cdot x_i-\text{Round}(z\cdot x_i) \mid<\varepsilon $. Here Round$(x)$ denotes the closest integer to $x$.

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  • $\begingroup$ Are your X floating point or rational? $\endgroup$
    – MikeY
    Commented Mar 16, 2020 at 23:02
  • $\begingroup$ the $x$ values are the result of a rational list times $\pi/180$ $\endgroup$
    – Victor Gustavo May
    Commented Mar 16, 2020 at 23:09
  • $\begingroup$ So you don't want an all integer list, you want an all "almost integer" list where they are all with $\varepsilon$ of an integer, correct? You want to find the minimum $Z$ that satisfies it? $\endgroup$
    – MikeY
    Commented Mar 17, 2020 at 2:47
  • $\begingroup$ Yes, precisely. That is what I am looking for $\endgroup$
    – Victor Gustavo May
    Commented Mar 17, 2020 at 13:52

3 Answers 3

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Here is a list of rationals:

n=20; listRats = RandomInteger[{1, 10}, n]/RandomInteger[{1, 10}, n];

You can find your integer the Least Common Multiple of all the denominators of the rationals, so that

LCM @@ Denominator[listRats]*listRats

is the desired list of integers.

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  • $\begingroup$ What if my list is not made up of rationals, but I wanted some degree precision, namely 10 digits. How could I transform my list into a list of rationals? so that I can use this $\endgroup$
    – Victor Gustavo May
    Commented Mar 16, 2020 at 23:20
  • $\begingroup$ You said your list was composed of rational numbers times $\pi$/180, so multiply by 180/$\pi$. Otherwise you can rationalize the list using Rationalize[list, inc]. $\endgroup$
    – bill s
    Commented Mar 16, 2020 at 23:27
  • $\begingroup$ Thanks, I did so. But my result is always an integer, which is not always the case for the minimum $z$ that works $\endgroup$
    – Victor Gustavo May
    Commented Mar 16, 2020 at 23:32
  • $\begingroup$ If it is not working, please show an example of how it fails. But really, look at the help for Rationalize, this is the function you want. $\endgroup$
    – bill s
    Commented Mar 17, 2020 at 0:51
  • $\begingroup$ It does work, but is not quite the answer to the problem. This is because this solution will only return the smallest integer $z$ such that z * lst is a integer list, what I need is the smallest positive real $z$. $\endgroup$
    – Victor Gustavo May
    Commented Mar 17, 2020 at 1:05
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qlst = Table[RandomInteger[{1, 10}]/RandomInteger[{1, 10}], 20]
z = Apply[LCM, qlst]
nlst = z*qlst
d = Apply[GCD, nlst]
lst = nlst/d
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Not a full answer, just a clear statement of the problem. It is actually pretty interesting.

Create a set of integer denominators (what really matters) in a 2D space.

SeedRandom[1234]; den = Table[RandomInteger[{5, 25}], {2}];
1/den
(* {1/6, 1/25} *)

Find the least common multiplier lcm, which is also your value of $Z$ when $\varepsilon=0$

lcm = LCM @@ den
(* 150 *)

Plot $Z$ as you vary it over the range {0, lcm}

ParametricPlot[z/den, {z, 0, lcm}, Frame -> True, 
         GridLines -> {Range[0, (lcm/den)[[1]]], 
                       Range[0, (lcm/den)[[2]]]
                       }]

enter image description here

You can see that it almost hits the {4,1} grid coord and others. We can plot that. Set $\varepsilon=0.1$

eps = 0.1;
Plot[Norm[z/den - Round[z/den], Infinity], 
    {z, 0, lcm}, 
    Epilog -> Line[{{0, eps}, {lcm, eps}}]
    ]

enter image description here

So $Z$ about 24 or so (eyeballing it) is your value.

Around each grid crossing is a $D$-dimensional cube centered on the crossing with side lengths $\varepsilon$. The problem is to find the first cube intersected by the vector, knowing that it is bounded by when the vector length = lcm.

It might be possible to brute force it, but lcm can explode as the number of terms increases. Again, interesting problem!

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