Sort with assumptionsvariable sized lists and using lists as variablesRetaining and reusing a one-to-one mapping from a sortSorting a matrix alphanumericallyRearranging a ListHow can I check if one expression implies another?Generating an Array of VectorsImporting, sorting and exporting listsDeleting Lonely Numbers From a ListApplying multiple functions to a single column in a tableFind positions in which list elements are equal

Control width of columns in a tabular environment

Extract substring according to regexp with sed or grep

What 1968 Moog synthesizer was used in the Movie Apollo 11?

Is divisi notation needed for brass or woodwind in an orchestra?

In the event of Brexit being postponed beyond the EU elections, will UK voters in EU countries be eligible to participate?

Why do Radio Buttons not fill the entire outer circle?

Is there any common country to visit for persons holding UK and Schengen visas?

Strange behavior in TikZ draw command

Pre-Employment Background Check With Consent For Future Checks

Not hide and seek

How do you justify more code being written by following clean code practices?

How can a new country break out from a developed country without war?

Boss fired me and is begging for me to come back - how much of a raise is reasonable?

Can a Knock spell open the door to Mordenkainen's Magnificent Mansion?

How do you say "Trust your struggle." in French?

Asserting that Atheism and Theism are both faith based positions

Does capillary rise violate hydrostatic paradox?

Why is indicated airspeed rather than ground speed used during the takeoff roll?

Calculate Pi using Monte Carlo

How does lowering the RF Gain help with SNR?

Why would five hundred and five same as one?

Derivative of an interpolated function

is this saw blade faulty?

Independent drivetrains on tandem bicycle



Sort with assumptions


variable sized lists and using lists as variablesRetaining and reusing a one-to-one mapping from a sortSorting a matrix alphanumericallyRearranging a ListHow can I check if one expression implies another?Generating an Array of VectorsImporting, sorting and exporting listsDeleting Lonely Numbers From a ListApplying multiple functions to a single column in a tableFind positions in which list elements are equal













4












$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.










share|improve this question









$endgroup$







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    18 mins ago















4












$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.










share|improve this question









$endgroup$







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    18 mins ago













4












4








4


1



$begingroup$


I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.










share|improve this question









$endgroup$




I have a list which looks like this



list = 0, Subscript[x,7], -Subscript[x,3]-Subscript[x,9], -Subscript[x,9];



and all the $x_i$'s are positive semidefinite (i.e. nonnegative) real numbers. I would like to be able to sort this into



sortedlist = -Subscript[x,3]-Subscript[x,9], -Subscript[x,9], 0, Subscript[x,7]



How do I achieve this? I tried



Assuming[Subscript[x,3] > 0 && Subscript[x,7] > 0 && Subscript[x,9] > 0, Sort[list]]



But this obviously does not work. In general, I'd like to be able to impose more constraints on the $x_i's$ when they're being sorted.







list-manipulation symbolic array sorting






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 39 mins ago









leastactionleastaction

22729




22729







  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    18 mins ago












  • 1




    $begingroup$
    An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
    $endgroup$
    – mikado
    18 mins ago







1




1




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
18 mins ago




$begingroup$
An interesting idea, but a symbolic list where you have an ordering for all the elements is rare
$endgroup$
– mikado
18 mins ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Here is a possibility:



sortWithAssumptions[list_, assum_] := Module[order,
order[a_, b_] := Simplify[a < b, assum];
Sort[list, order]
]


For your example:



sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
] //TeXForm



$left-x_3-x_9,-x_9,0,x_7right$




Another example:



sortWithAssumptions[
0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
] //TeXForm



$left-x_3-x_9,-x_9,0,x_7,x_9right$







share|improve this answer









$endgroup$












  • $begingroup$
    Thank you, Carl!
    $endgroup$
    – leastaction
    11 mins ago


















3












$begingroup$

How about:



list[[Ordering[list /. _Subscript -> 1]]]



-Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




So basically we sort it the way it would be sorted with all subscripts == 1.






share|improve this answer









$endgroup$












  • $begingroup$
    Thanks! Seems to work like a charm, but can you shed some light on why?
    $endgroup$
    – leastaction
    17 mins ago










  • $begingroup$
    @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
    $endgroup$
    – Kuba
    17 mins ago










  • $begingroup$
    I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
    $endgroup$
    – leastaction
    13 mins ago






  • 1




    $begingroup$
    @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
    $endgroup$
    – Kuba
    10 mins ago


















2












$begingroup$

Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

(* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





share|improve this answer









$endgroup$




















    2












    $begingroup$

    In this case, we can use RankedMin and FullSimplify to get the answer you seek



    Assuming[
    Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
    FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
    (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


    This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






    share|improve this answer











    $endgroup$












      Your Answer





      StackExchange.ifUsing("editor", function ()
      return StackExchange.using("mathjaxEditing", function ()
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      );
      );
      , "mathjax-editing");

      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "387"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193600%2fsort-with-assumptions%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
      order[a_, b_] := Simplify[a < b, assum];
      Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$












      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        11 mins ago















      2












      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
      order[a_, b_] := Simplify[a < b, assum];
      Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$












      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        11 mins ago













      2












      2








      2





      $begingroup$

      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
      order[a_, b_] := Simplify[a < b, assum];
      Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$







      share|improve this answer









      $endgroup$



      Here is a possibility:



      sortWithAssumptions[list_, assum_] := Module[order,
      order[a_, b_] := Simplify[a < b, assum];
      Sort[list, order]
      ]


      For your example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7right$




      Another example:



      sortWithAssumptions[
      0,Subscript[x,7],-Subscript[x,3]-Subscript[x,9],-Subscript[x,9], Subscript[x,9],
      Subscript[x,3]>0&&Subscript[x,7]>0&&Subscript[x,9]>0&&Subscript[x,7]<Subscript[x,9]
      ] //TeXForm



      $left-x_3-x_9,-x_9,0,x_7,x_9right$








      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 14 mins ago









      Carl WollCarl Woll

      70.8k394184




      70.8k394184











      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        11 mins ago
















      • $begingroup$
        Thank you, Carl!
        $endgroup$
        – leastaction
        11 mins ago















      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      11 mins ago




      $begingroup$
      Thank you, Carl!
      $endgroup$
      – leastaction
      11 mins ago











      3












      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$












      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        17 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        17 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        13 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba
        10 mins ago















      3












      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$












      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        17 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        17 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        13 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba
        10 mins ago













      3












      3








      3





      $begingroup$

      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.






      share|improve this answer









      $endgroup$



      How about:



      list[[Ordering[list /. _Subscript -> 1]]]



      -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7]




      So basically we sort it the way it would be sorted with all subscripts == 1.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 21 mins ago









      KubaKuba

      106k12209530




      106k12209530











      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        17 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        17 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        13 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba
        10 mins ago
















      • $begingroup$
        Thanks! Seems to work like a charm, but can you shed some light on why?
        $endgroup$
        – leastaction
        17 mins ago










      • $begingroup$
        @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
        $endgroup$
        – Kuba
        17 mins ago










      • $begingroup$
        I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
        $endgroup$
        – leastaction
        13 mins ago






      • 1




        $begingroup$
        @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
        $endgroup$
        – Kuba
        10 mins ago















      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      17 mins ago




      $begingroup$
      Thanks! Seems to work like a charm, but can you shed some light on why?
      $endgroup$
      – leastaction
      17 mins ago












      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba
      17 mins ago




      $begingroup$
      @leastaction just take a look at list /. _Subscript -> 1 and at Ordering[list /. _Subscript -> 1].
      $endgroup$
      – Kuba
      17 mins ago












      $begingroup$
      I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
      $endgroup$
      – leastaction
      13 mins ago




      $begingroup$
      I see, so you basically assigned the value $x_i = 1$ to each $x_i$. It works in this case, but generically, $x_i$'s may be different.
      $endgroup$
      – leastaction
      13 mins ago




      1




      1




      $begingroup$
      @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
      $endgroup$
      – Kuba
      10 mins ago




      $begingroup$
      @leastaction sure, which means there isn't one correct answer so every valid within constraints is correct?
      $endgroup$
      – Kuba
      10 mins ago











      2












      $begingroup$

      Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

      (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





      share|improve this answer









      $endgroup$

















        2












        $begingroup$

        Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

        (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





        share|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

          (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)





          share|improve this answer









          $endgroup$



          Sort[list, TrueQ@Simplify[#1 < #2, _Subscript > 0] &]

          (* Out: -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 13 mins ago









          MarcoBMarcoB

          37.6k556113




          37.6k556113





















              2












              $begingroup$

              In this case, we can use RankedMin and FullSimplify to get the answer you seek



              Assuming[
              Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
              FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
              (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


              This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






              share|improve this answer











              $endgroup$

















                2












                $begingroup$

                In this case, we can use RankedMin and FullSimplify to get the answer you seek



                Assuming[
                Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
                FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






                share|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  In this case, we can use RankedMin and FullSimplify to get the answer you seek



                  Assuming[
                  Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
                  FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                  (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                  This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.






                  share|improve this answer











                  $endgroup$



                  In this case, we can use RankedMin and FullSimplify to get the answer you seek



                  Assuming[
                  Subscript[x, 3] > 0 && Subscript[x, 7] > 0 && Subscript[x, 9] > 0,
                  FullSimplify[Table[RankedMin[list, i], i, 1, Length[list]]]]
                  (* -Subscript[x, 3] - Subscript[x, 9], -Subscript[x, 9], 0, Subscript[x, 7] *)


                  This has the advantage of not returning a (potentially) wrong answer if the sort order is uncertain.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 6 mins ago

























                  answered 13 mins ago









                  mikadomikado

                  6,6971929




                  6,6971929



























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Mathematica Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f193600%2fsort-with-assumptions%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Why is the 'in' operator throwing an error with a string literal instead of logging false?Why can't I use switch statement on a String?Python join: why is it string.join(list) instead of list.join(string)?Multiline String Literal in C#Why does comparing strings using either '==' or 'is' sometimes produce a different result?How to initialize an array's length in javascript?How can I print literal curly-brace characters in python string and also use .format on it?Why does ++[[]][+[]]+[+[]] return the string “10”?Why is char[] preferred over String for passwords?Why does this code using random strings print “hello world”?jQuery.inArray(), how to use it right?

                      बाताम इन्हें भी देखें सन्दर्भ दिक्चालन सूची1°05′00″N 104°02′0″E / 1.08333°N 104.03333°E / 1.08333; 104.033331°05′00″N 104°02′0″E / 1.08333°N 104.03333°E / 1.08333; 104.03333

                      How can we generalize the fact of finite dimensional vector space to an infinte dimensional case?$k[x]$-module and cyclic module over a finite dimensional vector spaceSubspace of a finite dimensional space is finite dimensionalIf V is an infinite-dimensional vector space, and S is an infinite-dimensional subspace of V, must the dimension of V/S be finite? ExplainWhy is an infinite dimensional space so different than a finite dimensional one?base for finite dimensional vector space is not infinite dimensional vector space?Any finite-dimensional vector space is the dual space of anotherHaving Trouble Understanding Meaning Of A Finite-Dimensional Vector SpaceProve that “Every subspaces of a finite-dimensional vector space is finite-dimensional”Ring as a finite dimensional Vector space over a field KQuestion regarding basis and dimension