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Proportionality without linear restriction?


Binary relation, reflexive, symmetric and transitiveWhat's $fracdydx$ of function $y=frac 12sin x$?Does type-theory have a concept of “relation”?Proving a change of variables is a variable over a ringThere exists a notation to “equations” where the unknown to solve is a kind of relation?does a univariate polynomial have a unique factorization?A property holds generically for polynomials - meaning?Can you turn a well-founded relation into a well-quasi-ordering?Laplace Transforms: How Do I Explain/Phrase What is Wrong with This Textbook Explanation?Can this notation be used to describe relations that are not functions?













0












$begingroup$


Is there a word to express the property $fracdydx>0$ of a relation?



I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Is there a word to express the property $fracdydx>0$ of a relation?



    I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Is there a word to express the property $fracdydx>0$ of a relation?



      I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.










      share|cite|improve this question









      $endgroup$




      Is there a word to express the property $fracdydx>0$ of a relation?



      I've heard the word "proportional" used to express this colloquially, although it is incorrect when $y=kx$ does not hold and it's another yet unknown polynomial over some restricted domain.







      functions derivatives polynomials relations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      Jorge BarriosJorge Barrios

      1054




      1054




















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          $begingroup$

          A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.






          share|cite|improve this answer









          $endgroup$












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            active

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            $begingroup$

            A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.






            share|cite|improve this answer









            $endgroup$

















              2












              $begingroup$

              A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.






              share|cite|improve this answer









              $endgroup$















                2












                2








                2





                $begingroup$

                A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.






                share|cite|improve this answer









                $endgroup$



                A differentiable functions $y : I to Bbb R$ (for some interval $I$) satisfying $fracdydx > 0$ is simply said to be (strictly) increasing.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                TravisTravis

                63k767150




                63k767150



























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