I wanna get a result of i/j . Like 1/3, 1/5, 1/7, 3/5, 3, 5/3... Something like that. But none of my code works

Going through your attempts one by one:

Attempt 1

j = {1, 3, 5, 7, 9};
i = {1, 3, 5, 7, 9};
jase[i_, j_] := i/j

What you are doing here is three things:

  • You set the value of j to {1, 3, ...}
  • You set the value of i to {1, 3, ...}
  • You define a function jase that takes two arguments named i and j (no connection to the variables from before) that returns i/j. So e.g. jase[3,4] will return 3/4

To fix it, a first attempt could be to call jase[i,j]:

jase[i, j]
(* {1, 1, 1, 1, 1} *)

As we can see, this is not what we want. The problem is that {...}/{...} simply divides corresponding elements. To get all possible combinations, you can use Outer (as shown by @kglr):

Outer[jase, i, j]
(* {{1, 1/3, 1/5, 1/7, 1/9}, {3, 1, 3/5, 3/7, 1/3},
    {5, 5/3, 1, 5/7, 5/9}, {7, 7/3, 7/5, 1, 7/9},
    {9, 3, 9/5, 9/7, 1}} *)

Attempt 2

i != j;
Table[i/j, j ∈ {1, 3, 5, 7, 9}; i ∈ {1, 3, 5, 7, 9}]

You're doing two things:

  • i != j simply returns itself without doing anything. You're suppressing that output with the ;

  • The syntax for the Table expression is wrong. To specify that i and j should be elements of the list {1, 3, ...}, use {i, {1, 3, ...}}:

    Table[i/j, {j, {1, 3, 5, 7, 9}}, {i, {1, 3, 5, 7, 9}}]
(* {{1, 3, 5, 7, 9}, {1/3, 1, 5/3, 7/3, 3}, 
    {1/5, 3/5, 1, 7/5, 9/5}, {1/7, 3/7, 5/7, 1, 9/7}, 
    {1/9, 1/3, 5/9, 7/9, 1}} *)

Attempt 3

jase[i_, j_] := 
Module[{}, i ≠ j; Element[{i, j}, List[1, 3, 5, 7, 9]]; i/j]

This one defines a function jase with two arguments, that does three things:

  • Evaluate the expression i ≠ j (which does nothing on its own) and discard its result
  • Evaluate the expression Element[{i, j}, List[1, 3, 5, 7, 9]] (which does nothing on its own), and discard its result
  • Evaluate the expression i/j (which stays unevaluated) and return its result

So calling jase[1, 2] gives for example:

jase[1, 2]
(* 1/2 *)

I don't think this one can be easily fixed

Discarding i==j

You'll notice that all the examples above return a bunch of ones, since the cases where i==j are not discarded. There are several ways to do this, see e.g. the answers of @kglr and @ChrisK. A solution that is more explicit could be the following:

jase[i_, j_] := i/j
jase[i_, i_] := Nothing

Outer[jase, Range[1, 9, 2], Range[1, 9, 2]]
(* {{1/3, 1/5, 1/7, 1/9}, {3, 3/5, 3/7, 1/3}, {5, 5/3, 5/7, 5/9}, {7, 7/3, 7/5, 7/9}, {9, 3, 9/5, 9/7}} *)

This is essentially your first attempt with an additional case for jase: The second definition says "if both arguments are the same, then return Nothing". And Nothing is automatically removed from lists.


How's this?

range = {1, 3, 5, 7, 9};
DeleteCases[Flatten[Table[i/j, {i, range}, {j, range}]], 1]
(* {1/3, 1/5, 1/7, 1/9, 3, 3/5, 3/7, 1/3, 5, 5/3, 5/7, 5/9,
  7, 7/3, 7/5, 7/9, 9, 3, 9/5, 9/7} *)


range = Range[1, 9, 2];

# / DeleteCases[range, #] & /@ range // Flatten

{1/3, 1/5, 1/7, 1/9, 3, 3/5, 3/7, 1/3, 5, 5/3, 5/7, 5/9, 7, 7/3, 7/5, 7/9, 9, 3, 9/5, 9/7}

Additional alternatives:

Flatten @ MapIndexed[Drop] @ Outer[Divide, range, range]

MapIndexed[Apply[Sequence] @* Drop] @ Outer[Divide, range, range]

Tuples[foo[range, range]] /. foo[a_, a_] -> Nothing /. foo -> Divide

Join @@ Table[i/j, {i, range}, {j, DeleteCases[range, i]}]

Distribute[{range, range}, List, List, DeleteCases[{##}, 1] &, Divide]

Array[Divide, {1, 1} Length@range, {MinMax@range}, DeleteCases[Flatten[{##}], 1] &]

and a Halloween special:

☺ = range; 
## & @@@ (#/(☺ /. # -> (## &[])) & /@ ☺)

Tags:

List Manipulation

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