Geometric Sequence Proof
$begingroup$
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
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
$endgroup$
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
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
asked Mar 18 '17 at 15:56
JenkinsMaJenkinsMa
1389
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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
$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
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?
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
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
$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
$endgroup$
add a comment |
$begingroup$
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
$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
$endgroup$
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
$$overbrace{ m^{r-1}, m^{r-2},m^{r-3},cdotscdots,m^2, m^1, 1}^{r text{ integer terms}}, overbrace{frac 1m, frac 1{m^2},cdotscdotscdots}^{text{non-integer terms}}$$
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
answered Mar 18 '17 at 16:45
hypergeometrichypergeometric
17.7k1758
17.7k1758
add a comment |
add a comment |
$begingroup$
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?
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
$endgroup$
add a comment |
$begingroup$
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?
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
$endgroup$
add a comment |
$begingroup$
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?
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
$endgroup$
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?
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
answered Mar 18 '17 at 17:48
AlephnullAlephnull
1,904621
1,904621
add a comment |
add a comment |
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$begingroup$
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