Use of the term “whenever” in Logic

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How does the word "whenever" translate into logic? For instance:




$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$







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    It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
    – Stan Tendijck
    1 hour ago







  • 1




    See my answer here.
    – Taroccoesbrocco
    56 mins ago














up vote
1
down vote

favorite












How does the word "whenever" translate into logic? For instance:




$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$







logic terminology logic-translation







share|cite|improve this question



















  • 1




    It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
    – Stan Tendijck
    1 hour ago







  • 1




    See my answer here.
    – Taroccoesbrocco
    56 mins ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











How does the word "whenever" translate into logic? For instance:




$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$







logic terminology logic-translation







share|cite|improve this question















How does the word "whenever" translate into logic? For instance:




$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$







logic terminology logic-translation




logic terminology logic-translation






share|cite|improve this question















share|cite|improve this question













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edited 19 mins ago









Taroccoesbrocco

4,35461535




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asked 1 hour ago









yoshi

1,018816




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  • 1




    It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
    – Stan Tendijck
    1 hour ago







  • 1




    See my answer here.
    – Taroccoesbrocco
    56 mins ago












  • 1




    It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
    – Stan Tendijck
    1 hour ago







  • 1




    See my answer here.
    – Taroccoesbrocco
    56 mins ago







1




1




It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago





It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago





1




1




See my answer here.
– Taroccoesbrocco
56 mins ago




See my answer here.
– Taroccoesbrocco
56 mins ago










3 Answers
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Whenever in this instant is used as the following :



$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$






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    "Whenever" can be a English representation of $Leftarrow$ or "is implied by".



    The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.



    so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$






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      $forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$






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        3 Answers
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        3 Answers
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        up vote
        4
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        Whenever in this instant is used as the following :



        $$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$






        share|cite|improve this answer








        New contributor




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





















          up vote
          4
          down vote













          Whenever in this instant is used as the following :



          $$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$






          share|cite|improve this answer








          New contributor




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



















            up vote
            4
            down vote










            up vote
            4
            down vote









            Whenever in this instant is used as the following :



            $$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$






            share|cite|improve this answer








            New contributor




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









            Whenever in this instant is used as the following :



            $$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$







            share|cite|improve this answer








            New contributor




            TheJedi 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 answer



            share|cite|improve this answer






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            answered 1 hour ago









            TheJedi

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                up vote
                3
                down vote













                "Whenever" can be a English representation of $Leftarrow$ or "is implied by".



                The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.



                so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$






                share|cite|improve this answer
























                  up vote
                  3
                  down vote













                  "Whenever" can be a English representation of $Leftarrow$ or "is implied by".



                  The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.



                  so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$






                  share|cite|improve this answer






















                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    "Whenever" can be a English representation of $Leftarrow$ or "is implied by".



                    The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.



                    so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$






                    share|cite|improve this answer












                    "Whenever" can be a English representation of $Leftarrow$ or "is implied by".



                    The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.



                    so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    James K

                    512212




                    512212




















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                        $forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$






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                          $forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$






                          share|cite|improve this answer






















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                            up vote
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                            $forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$






                            share|cite|improve this answer












                            $forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 58 mins ago









                            S.S.Danyal

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