How can one prove that if x is an integer greater than 2, then x/(x-1) is a not an integer?












0














How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?



Intuitively I can see this is true but how to prove it?






elementary-number-theory







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  • 2




    Presumably you mean "not an integer"
    – Henry
    Nov 27 at 8:33










  • oops! Yes, "not an integer".
    – pirsquare
    Nov 27 at 8:39


















0














How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?



Intuitively I can see this is true but how to prove it?






elementary-number-theory







share|cite|improve this question




















  • 2




    Presumably you mean "not an integer"
    – Henry
    Nov 27 at 8:33










  • oops! Yes, "not an integer".
    – pirsquare
    Nov 27 at 8:39
















0












0








0







How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?



Intuitively I can see this is true but how to prove it?






elementary-number-theory







share|cite|improve this question















How can one prove that if x is an integer greater than 2, then x/(x-1) is a not integer?



Intuitively I can see this is true but how to prove it?






elementary-number-theory




elementary-number-theory






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share|cite|improve this question













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edited Nov 27 at 8:40

























asked Nov 27 at 8:31









pirsquare

379319




379319








  • 2




    Presumably you mean "not an integer"
    – Henry
    Nov 27 at 8:33










  • oops! Yes, "not an integer".
    – pirsquare
    Nov 27 at 8:39
















  • 2




    Presumably you mean "not an integer"
    – Henry
    Nov 27 at 8:33










  • oops! Yes, "not an integer".
    – pirsquare
    Nov 27 at 8:39










2




2




Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33




Presumably you mean "not an integer"
– Henry
Nov 27 at 8:33












oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39






oops! Yes, "not an integer".
– pirsquare
Nov 27 at 8:39












3 Answers
3






active

oldest

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5














$$dfrac x{x-1}=1+dfrac1{x-1} $$



So, $x-1(ne0)$ must divide $1implies x-1=pm1$






share|cite|improve this answer





























    2














    Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)



    $$frac x{x-1}-1=frac 1{x-1}gt0$$



    $$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$



    So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.






    share|cite|improve this answer





























      2














      The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.



      Indeed, for $2<x_1<x_2$:
      $$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$






      share|cite|improve this answer





















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        3 Answers
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        3 Answers
        3






        active

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        active

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        active

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        5














        $$dfrac x{x-1}=1+dfrac1{x-1} $$



        So, $x-1(ne0)$ must divide $1implies x-1=pm1$






        share|cite|improve this answer


























          5














          $$dfrac x{x-1}=1+dfrac1{x-1} $$



          So, $x-1(ne0)$ must divide $1implies x-1=pm1$






          share|cite|improve this answer
























            5












            5








            5






            $$dfrac x{x-1}=1+dfrac1{x-1} $$



            So, $x-1(ne0)$ must divide $1implies x-1=pm1$






            share|cite|improve this answer












            $$dfrac x{x-1}=1+dfrac1{x-1} $$



            So, $x-1(ne0)$ must divide $1implies x-1=pm1$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 8:34









            lab bhattacharjee

            223k15156274




            223k15156274























                2














                Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)



                $$frac x{x-1}-1=frac 1{x-1}gt0$$



                $$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$



                So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.






                share|cite|improve this answer


























                  2














                  Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)



                  $$frac x{x-1}-1=frac 1{x-1}gt0$$



                  $$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$



                  So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)



                    $$frac x{x-1}-1=frac 1{x-1}gt0$$



                    $$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$



                    So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.






                    share|cite|improve this answer












                    Since one answer has already been given, here is a slightly different way of proceeding (there are tweaks which can be made to fill out the proof, if necessary)



                    $$frac x{x-1}-1=frac 1{x-1}gt0$$



                    $$2-frac x{x-1}=frac {x-2}{x-1}gt 0$$



                    So $frac x{x-1}$ lies strictly between the consecutive integers $1$and $2$, and the second of the inequalities shows explicitly why the condition $xgt 2$ comes in.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 at 8:47









                    Mark Bennet

                    80.4k981179




                    80.4k981179























                        2














                        The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.



                        Indeed, for $2<x_1<x_2$:
                        $$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$






                        share|cite|improve this answer


























                          2














                          The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.



                          Indeed, for $2<x_1<x_2$:
                          $$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$






                          share|cite|improve this answer
























                            2












                            2








                            2






                            The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.



                            Indeed, for $2<x_1<x_2$:
                            $$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$






                            share|cite|improve this answer












                            The function $f(x)=frac{x}{x-1}$ is decreasing for $xin (2,+infty)$ and bounded $1<f(x)<2$.



                            Indeed, for $2<x_1<x_2$:
                            $$1<frac{x_2}{x_2-1}<frac{x_1}{x_1-1}<2.$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 27 at 9:06









                            farruhota

                            19.2k2736




                            19.2k2736






























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