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Does the sign matter for proportionality?


What does this formula do?what is the constant of proportionality?I did not understand the approach needed to solve this problem on inverse proportionalityCalculating Volume for the Perfect CompostWhy does proportionality yield multiplicitiveness and not addition?Why does 1:2 = 1/2?Why does 100% represent the whole of somethingSimple Ratios Shortcut - Why does it work?How do I figure out the proportionality between two aspect ratiosHow is the joint proportionality (joint variation) equation (i.e. $y = kxz$) logically correct?













3












$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?










share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago















3












$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?










share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago













3












3








3





$begingroup$


This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?










share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




This question arose from Physics, where the force on an object attached on a spring is proportional to the displacement to the equilibrium (that is, the rest position). Also, if the displacement to the equilibrium is positive, the force will be negative, as it tries to pull the object back (i.e. if you pull a string, the force is opposite to your direction of pull).



Therefore, it can be said that:



$$F propto -x$$
Where $F$ is the force and $x$ is the displacement from equilibrium



Is this the same as:
$$F propto x$$
According to the relation of proportionality, it should be, but my friend says that putting the second one is not correct.



Both are equivalent right?







ratio






share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 46 mins ago









Brian

1,529516




1,529516






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asked 3 hours ago









Sebi SSebi S

182




182




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New contributor





Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Sebi S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago
















  • $begingroup$
    It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
    $endgroup$
    – David Peterson
    3 hours ago















$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago




$begingroup$
It depends on definitions/context. Using the definition on wikipedia, yes they are equivalent.
$endgroup$
– David Peterson
3 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






share|cite|improve this answer









$endgroup$




















    3












    $begingroup$

    Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.






          share|cite|improve this answer









          $endgroup$



          Yes, proportionality does not tell anything about the sign of the proportionality constant. This is probably done in physics so that the proportionality constant $k$ can be considered to be always positive. This has a physical interpretation , so it is convenient for it to be positive.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          John DoeJohn Doe

          12.3k11341




          12.3k11341





















              3












              $begingroup$

              Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

                Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.






                  share|cite|improve this answer









                  $endgroup$



                  Mathematically they are equivalent, but physycist may want to differentiate between a force in the same direction as the displacement, and a force opposite to the displacement, as they lead to physically different behavior of the system. Thus the minus sign matters.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Adam LatosińskiAdam Latosiński

                  9809




                  9809




















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