Geometric Sequence Proof












0












$begingroup$


Problem:




(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.



(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.




My Solution for $a)$:




$a)$ First, we can analyze the case when $r geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.




I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?






sequences-and-series geometric-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can a geometric series be extended in both directions?
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:08










  • $begingroup$
    I guess so... Is this what I need to solve the problem?
    $endgroup$
    – JenkinsMa
    Mar 18 '17 at 16:19






  • 1




    $begingroup$
    Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:24










  • $begingroup$
    8,4,2,1,1/2,1/4,...
    $endgroup$
    – steven gregory
    Jan 3 '18 at 4:51










  • $begingroup$
    You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
    $endgroup$
    – BruceET
    13 hours ago
















0












$begingroup$


Problem:




(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.



(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.




My Solution for $a)$:




$a)$ First, we can analyze the case when $r geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.




I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?






sequences-and-series geometric-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can a geometric series be extended in both directions?
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:08










  • $begingroup$
    I guess so... Is this what I need to solve the problem?
    $endgroup$
    – JenkinsMa
    Mar 18 '17 at 16:19






  • 1




    $begingroup$
    Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:24










  • $begingroup$
    8,4,2,1,1/2,1/4,...
    $endgroup$
    – steven gregory
    Jan 3 '18 at 4:51










  • $begingroup$
    You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
    $endgroup$
    – BruceET
    13 hours ago














0












0








0


1



$begingroup$


Problem:




(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.



(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.




My Solution for $a)$:




$a)$ First, we can analyze the case when $r geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.




I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?






sequences-and-series geometric-series







share|cite|improve this question









$endgroup$




Problem:




(a) Determine all nonnegative integers $r$ such that it is possible for an infinite arithmetic sequence to contain exactly $r$ terms that are integers. Prove your answer.



(b) Determine all nonnegative integers $r$ such that it is possible for an infinite geometric sequence to contain exactly $r$ terms that are integers. Prove your answer.




My Solution for $a)$:




$a)$ First, we can analyze the case when $r geq 2$. In this case, there will be at least $2$ terms in the arithmetic sequence which are integers; $a+md$ and $a+nd$. WLOG, $m>n$. These terms are both integers, so $(m-n)d$ is also an integer. Adding $(m - n)d$ to $a + md$ gives $a + (2m - n)d$, which is also a member of the arithmetic sequence and an integer, so this process can continue, giving the arithmetic sequence an infinite number of terms. Therefore, $r$ can only be $r=1$.




I am currently stuck on $b)$, but I know that we could do a similar approach, but how would I do it?






sequences-and-series geometric-series




sequences-and-series geometric-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 18 '17 at 15:56









JenkinsMaJenkinsMa

1389




1389












  • $begingroup$
    Can a geometric series be extended in both directions?
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:08










  • $begingroup$
    I guess so... Is this what I need to solve the problem?
    $endgroup$
    – JenkinsMa
    Mar 18 '17 at 16:19






  • 1




    $begingroup$
    Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:24










  • $begingroup$
    8,4,2,1,1/2,1/4,...
    $endgroup$
    – steven gregory
    Jan 3 '18 at 4:51










  • $begingroup$
    You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
    $endgroup$
    – BruceET
    13 hours ago


















  • $begingroup$
    Can a geometric series be extended in both directions?
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:08










  • $begingroup$
    I guess so... Is this what I need to solve the problem?
    $endgroup$
    – JenkinsMa
    Mar 18 '17 at 16:19






  • 1




    $begingroup$
    Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
    $endgroup$
    – Parcly Taxel
    Mar 18 '17 at 16:24










  • $begingroup$
    8,4,2,1,1/2,1/4,...
    $endgroup$
    – steven gregory
    Jan 3 '18 at 4:51










  • $begingroup$
    You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
    $endgroup$
    – BruceET
    13 hours ago
















$begingroup$
Can a geometric series be extended in both directions?
$endgroup$
– Parcly Taxel
Mar 18 '17 at 16:08




$begingroup$
Can a geometric series be extended in both directions?
$endgroup$
– Parcly Taxel
Mar 18 '17 at 16:08












$begingroup$
I guess so... Is this what I need to solve the problem?
$endgroup$
– JenkinsMa
Mar 18 '17 at 16:19




$begingroup$
I guess so... Is this what I need to solve the problem?
$endgroup$
– JenkinsMa
Mar 18 '17 at 16:19




1




1




$begingroup$
Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
$endgroup$
– Parcly Taxel
Mar 18 '17 at 16:24




$begingroup$
Yes. Consider two integers in a geometric progression separated by some powers of the ratio $a$. Then it should be possible to extend the series in one direction infinitely to yield an infinite number of integers. Thus $r$ can only be one, which is achieved by (e.g.) a base value of 1 and a multiplier of $pi$.
$endgroup$
– Parcly Taxel
Mar 18 '17 at 16:24












$begingroup$
8,4,2,1,1/2,1/4,...
$endgroup$
– steven gregory
Jan 3 '18 at 4:51




$begingroup$
8,4,2,1,1/2,1/4,...
$endgroup$
– steven gregory
Jan 3 '18 at 4:51












$begingroup$
You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
$endgroup$
– BruceET
13 hours ago




$begingroup$
You have asked several questions here and gotten some nice answers, but I don't see where you have ever clicked to check mark to Accept an answer. Why is that?
$endgroup$
– BruceET
13 hours ago










2 Answers
2






active

oldest

votes


















1












$begingroup$

For part (b):



Take a positive integer $m$.



Construct a GP with first term $m^{r-1}$ and common ratio $frac 1m$. The GP is



$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Consider {a+nt} (where the sequence starts at n=0). r=0 is achieved when a,t are irrational. r=1 is achieved when a is an integer and t is irrational. r>=2 is impossible because given two integers n, m in the sequence (WLOG n


    Consider {a(t)^n} (where the sequence starts at n=0). r=0 is achieved when a is irrational and t is integral. r=1 is achieved when a is integral and t is irrational. r>=2 is impossible because given integers n, nt^k in the sequence t^k is integral and nt^(2k) is integral.



    What about the arithmetic and geometric sequences gives htem this property? Is there a more general class of sequences witht his property? What types of sequences have infinitely many r values?






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      For part (b):



      Take a positive integer $m$.



      Construct a GP with first term $m^{r-1}$ and common ratio $frac 1m$. The GP is



      $$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For part (b):



        Take a positive integer $m$.



        Construct a GP with first term $m^{r-1}$ and common ratio $frac 1m$. The GP is



        $$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For part (b):



          Take a positive integer $m$.



          Construct a GP with first term $m^{r-1}$ and common ratio $frac 1m$. The GP is



          $$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$






          share|cite|improve this answer









          $endgroup$



          For part (b):



          Take a positive integer $m$.



          Construct a GP with first term $m^{r-1}$ and common ratio $frac 1m$. The GP is



          $$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 18 '17 at 16:45









          hypergeometrichypergeometric

          17.7k1758




          17.7k1758























              1












              $begingroup$

              Consider {a+nt} (where the sequence starts at n=0). r=0 is achieved when a,t are irrational. r=1 is achieved when a is an integer and t is irrational. r>=2 is impossible because given two integers n, m in the sequence (WLOG n


              Consider {a(t)^n} (where the sequence starts at n=0). r=0 is achieved when a is irrational and t is integral. r=1 is achieved when a is integral and t is irrational. r>=2 is impossible because given integers n, nt^k in the sequence t^k is integral and nt^(2k) is integral.



              What about the arithmetic and geometric sequences gives htem this property? Is there a more general class of sequences witht his property? What types of sequences have infinitely many r values?






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Consider {a+nt} (where the sequence starts at n=0). r=0 is achieved when a,t are irrational. r=1 is achieved when a is an integer and t is irrational. r>=2 is impossible because given two integers n, m in the sequence (WLOG n


                Consider {a(t)^n} (where the sequence starts at n=0). r=0 is achieved when a is irrational and t is integral. r=1 is achieved when a is integral and t is irrational. r>=2 is impossible because given integers n, nt^k in the sequence t^k is integral and nt^(2k) is integral.



                What about the arithmetic and geometric sequences gives htem this property? Is there a more general class of sequences witht his property? What types of sequences have infinitely many r values?






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Consider {a+nt} (where the sequence starts at n=0). r=0 is achieved when a,t are irrational. r=1 is achieved when a is an integer and t is irrational. r>=2 is impossible because given two integers n, m in the sequence (WLOG n


                  Consider {a(t)^n} (where the sequence starts at n=0). r=0 is achieved when a is irrational and t is integral. r=1 is achieved when a is integral and t is irrational. r>=2 is impossible because given integers n, nt^k in the sequence t^k is integral and nt^(2k) is integral.



                  What about the arithmetic and geometric sequences gives htem this property? Is there a more general class of sequences witht his property? What types of sequences have infinitely many r values?






                  share|cite|improve this answer









                  $endgroup$



                  Consider {a+nt} (where the sequence starts at n=0). r=0 is achieved when a,t are irrational. r=1 is achieved when a is an integer and t is irrational. r>=2 is impossible because given two integers n, m in the sequence (WLOG n


                  Consider {a(t)^n} (where the sequence starts at n=0). r=0 is achieved when a is irrational and t is integral. r=1 is achieved when a is integral and t is irrational. r>=2 is impossible because given integers n, nt^k in the sequence t^k is integral and nt^(2k) is integral.



                  What about the arithmetic and geometric sequences gives htem this property? Is there a more general class of sequences witht his property? What types of sequences have infinitely many r values?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 18 '17 at 17:48









                  AlephnullAlephnull

                  1,904621




                  1,904621






























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