Mathematical Induction & Sequencing
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Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$
Must there be an index k for which $q_k > 5^{50}$?
sequences-and-series combinatorics
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
$endgroup$
add a comment |
$begingroup$
Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$
Must there be an index k for which $q_k > 5^{50}$?
sequences-and-series combinatorics
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
$endgroup$
add a comment |
$begingroup$
Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$
Must there be an index k for which $q_k > 5^{50}$?
sequences-and-series combinatorics
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
$endgroup$
Let $q_1$ be an arbitrary positive real number and define the sequence {$ q_{n};$} by $ q_{n+1};$ = $ q_{n};$ + $frac{1}{q_n}$
Must there be an index k for which $q_k > 5^{50}$?
sequences-and-series combinatorics
sequences-and-series combinatorics
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
edited Nov 23 '18 at 22:32
Scientifica
6,37641335
6,37641335
asked Apr 2 '17 at 14:24
user431696
add a comment |
add a comment |
3 Answers
3
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oldest
votes
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It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$
But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
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add a comment |
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Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.
Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.
Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
$endgroup$
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
add a comment |
$begingroup$
Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.
If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.
Therefore the series is not bounded and will eventually reach any arbitrarily large number.
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
$endgroup$
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$
But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
$endgroup$
add a comment |
$begingroup$
It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$
But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
$endgroup$
add a comment |
$begingroup$
It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$
But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
$endgroup$
It is easy to show $q_n$ is a strictly increasing sequence of positive numbers. Thus if it is bounded, it converges. Let us assume $q_n$ is bounded. If $q_n$ converges, then $lim_{nrightarrowinfty}(q_n) =lim_{nrightarrowinfty}(q_{n+1}) =lim_{nrightarrowinfty}(q_n + frac{1}{q_n}) iff lim_{nrightarrowinfty} (frac{1}{q_n} =0)$ $ iff lim_{nrightarrowinfty}{q_n} = infty$
But then $q_n$ is not bounded, which is a contradiction.
Thus $q_n$ is unbounded, and for all M we can find an n for which $q_n >M$
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
edited Apr 2 '17 at 15:25
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
answered Apr 2 '17 at 15:12
Daphna KeidarDaphna Keidar
1996
1996
add a comment |
add a comment |
$begingroup$
Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.
Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.
Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
$endgroup$
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
add a comment |
$begingroup$
Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.
Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.
Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
$endgroup$
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
add a comment |
$begingroup$
Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.
Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.
Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
$endgroup$
Let's assume the sequence were bounded. Since it is a strictly increasing sequence, it follows that is must converge to some finite limit. Let's call that limit $q$. Note that since for $q_nle 1$ we have $q_{n+1} = q_n + 1/q_n > 1/q_n ge 1$, therefore $q>1$.
Now since $q_n$ converges to $q$, for every $epsilon>0$ there's an $N$ so that $q_n>q-epsilon$ for all $nge N$. Let's specifically take $epsilon=1/q$. Note that $epsilon>0$ because of $q>1$. Then you've got some $N$ so that $q-epsilon<q_N<q$. But then, $q_{N+1}=q_N+1/q_N > q-epsilon +1/q_N = q - 1/q + 1/q_N > q -1/q + 1/q = q$, in contradiction to the assumption that $q$ is an upper bound. Therefore the assumption that the sequence is bounded must be wrong.
Since the sequence is unbounded and strictly monotonous growing, it will grow above any bound, including the bound $5^{50}$.
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
edited Apr 2 '17 at 15:30
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
answered Apr 2 '17 at 15:21
celtschkceltschk
29.8k75599
29.8k75599
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
add a comment |
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
$1/q$ isn't better?
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:25
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
@RafaBudría: Yes; I originally named the limit $a$ and then renamed it to $q$, but then erroneously continued to use $a$ in the text added after doing the change.
$endgroup$
– celtschk
Apr 2 '17 at 15:28
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
I supposed that. Still remain some a's :)
$endgroup$
– Rafa Budría
Apr 2 '17 at 15:29
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
$begingroup$
@RafaBudría: I hope now I missed no further one; thank you.
$endgroup$
– celtschk
Apr 2 '17 at 15:30
add a comment |
$begingroup$
Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.
If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.
Therefore the series is not bounded and will eventually reach any arbitrarily large number.
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
$endgroup$
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
add a comment |
$begingroup$
Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.
If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.
Therefore the series is not bounded and will eventually reach any arbitrarily large number.
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
$endgroup$
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
add a comment |
$begingroup$
Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.
If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.
Therefore the series is not bounded and will eventually reach any arbitrarily large number.
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
$endgroup$
Since $q_1$ is positive and greater $0$ and $frac{1}{q_1}$ is positive and greater $0$, $q_2=q_1+frac{1}{q_1}$ will be positive as well. Furthermore since $frac{1}{q_1}$ is greater 0, $q_2=q_1+frac{1}{q_1}$ will be greater than $q_1$ and by induction this means that the sequence is strictly increasing.
If the sequence were bounded, that would mean it would converge to a certain value. If that were the case, the difference between two consecutive numbers in the sequence would need to tend to $0$ as $n$ tends to infinity, $limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=0$. Substituting $q_{n+1}=q_n+frac{1}{q_n}$, we get $0=limlimits_{nrightarrowinfty}(q_{n+1}-q_n)=limlimits_{nrightarrowinfty}(q_n+frac{1}{q_n}-q_n)=limlimits_{nrightarrowinfty}(frac{1}{q_n})$, yet $limlimits_{nrightarrowinfty}(frac{1}{q_n})=0Leftrightarrow limlimits_{nrightarrowinfty}q_n=infty$, which would contradicts the series being bounded.
Therefore the series is not bounded and will eventually reach any arbitrarily large number.
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
edited Apr 2 '17 at 15:35
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
answered Apr 2 '17 at 14:29
ThorgottThorgott
566314
566314
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
add a comment |
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
1
1
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
$begingroup$
Doesn't follow. You proved it's increasing, not that it's unbounded. I think in fact it's unbounded, but proof is needed.
$endgroup$
– Rafa Budría
Apr 2 '17 at 14:34
add a comment |
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