About two sets of natural numbersNatural numbers not expressible as $x+s(x)$ nor $x+s(x)+l(x)$Circular list from the 2nd element of the result of repeatedly perfect shuffling a magnitude ordered list of natural numbers less than an even number.Looking for reading material on: Numbers, in whose internal decimal places appear all natural numbers as a sequence of digitsProperty for the natural numbers.How to find the highest [natural] radix base of a given number with a natural outputSeeking for an examples of non-trivial sets that can be used to generate all the natural numbersFind all natural numbers n such that…Universal Property to describe natural numbersLargest prime factors of two consecutive natural numbersModification of Euclids Proof of infinite primes to generate a sequence of distinct primes
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About two sets of natural numbers
Natural numbers not expressible as $x+s(x)$ nor $x+s(x)+l(x)$Circular list from the 2nd element of the result of repeatedly perfect shuffling a magnitude ordered list of natural numbers less than an even number.Looking for reading material on: Numbers, in whose internal decimal places appear all natural numbers as a sequence of digitsProperty for the natural numbers.How to find the highest [natural] radix base of a given number with a natural outputSeeking for an examples of non-trivial sets that can be used to generate all the natural numbersFind all natural numbers n such that…Universal Property to describe natural numbersLargest prime factors of two consecutive natural numbersModification of Euclids Proof of infinite primes to generate a sequence of distinct primes
$begingroup$
I would like to know if you know by which number should I multiply a list of natural numbers so that it never overlaps with a list of another set of natural numbers at some point in the future. The issue is that we have two different datasets with IDs that are natural numbers and we cannot modify any of the IDs unless we multiply them by a specific number, and we don't want that at some point in time the ID of one dataset finds another similar ID in the other dataset in the future. My limited math knowledge tells me that when taken to the infinite, there will be a clash of IDs at some point just because naturally there's no way to avoid the IDs in one dataset to happen at some point in the other dataset. But maybe I'm wrong, could you please help me?
Thank you!
elementary-number-theory
asked Mar 28 at 15:59
AriadnaAriadna
1
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Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
I would like to know if you know by which number should I multiply a list of natural numbers so that it never overlaps with a list of another set of natural numbers at some point in the future. The issue is that we have two different datasets with IDs that are natural numbers and we cannot modify any of the IDs unless we multiply them by a specific number, and we don't want that at some point in time the ID of one dataset finds another similar ID in the other dataset in the future. My limited math knowledge tells me that when taken to the infinite, there will be a clash of IDs at some point just because naturally there's no way to avoid the IDs in one dataset to happen at some point in the other dataset. But maybe I'm wrong, could you please help me?
Thank you!
elementary-number-theory
asked Mar 28 at 15:59
AriadnaAriadna
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35
add a comment |
$begingroup$
I would like to know if you know by which number should I multiply a list of natural numbers so that it never overlaps with a list of another set of natural numbers at some point in the future. The issue is that we have two different datasets with IDs that are natural numbers and we cannot modify any of the IDs unless we multiply them by a specific number, and we don't want that at some point in time the ID of one dataset finds another similar ID in the other dataset in the future. My limited math knowledge tells me that when taken to the infinite, there will be a clash of IDs at some point just because naturally there's no way to avoid the IDs in one dataset to happen at some point in the other dataset. But maybe I'm wrong, could you please help me?
Thank you!
elementary-number-theory
asked Mar 28 at 15:59
AriadnaAriadna
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I would like to know if you know by which number should I multiply a list of natural numbers so that it never overlaps with a list of another set of natural numbers at some point in the future. The issue is that we have two different datasets with IDs that are natural numbers and we cannot modify any of the IDs unless we multiply them by a specific number, and we don't want that at some point in time the ID of one dataset finds another similar ID in the other dataset in the future. My limited math knowledge tells me that when taken to the infinite, there will be a clash of IDs at some point just because naturally there's no way to avoid the IDs in one dataset to happen at some point in the other dataset. But maybe I'm wrong, could you please help me?
Thank you!
elementary-number-theory
elementary-number-theory
asked Mar 28 at 15:59
AriadnaAriadna
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:59
AriadnaAriadna
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:59
AriadnaAriadna
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 15:59
AriadnaAriadna
1
asked Mar 28 at 15:59
AriadnaAriadna
1
1
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ariadna is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35
add a comment |
$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35
$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35
$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can have two sets of numbers such that they never clash: For example, let both sets of numbers be $N_1 = N_2 = 1, 3, 5, dots $. Now, if you multiply $N_2$ by $2$, then the new set will be $2, 6, 10, dots $ which cannot clash with the set $N_1$ (since $N_1$ only contains odd numbers, while $N_2$ contains even numbers.
Of course, this depends on the set of numbers $N_1$ and $N_2$ in question.
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
$endgroup$
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
add a comment |
Your Answer
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1 Answer
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$begingroup$
You can have two sets of numbers such that they never clash: For example, let both sets of numbers be $N_1 = N_2 = 1, 3, 5, dots $. Now, if you multiply $N_2$ by $2$, then the new set will be $2, 6, 10, dots $ which cannot clash with the set $N_1$ (since $N_1$ only contains odd numbers, while $N_2$ contains even numbers.
Of course, this depends on the set of numbers $N_1$ and $N_2$ in question.
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
$endgroup$
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
add a comment |
$begingroup$
You can have two sets of numbers such that they never clash: For example, let both sets of numbers be $N_1 = N_2 = 1, 3, 5, dots $. Now, if you multiply $N_2$ by $2$, then the new set will be $2, 6, 10, dots $ which cannot clash with the set $N_1$ (since $N_1$ only contains odd numbers, while $N_2$ contains even numbers.
Of course, this depends on the set of numbers $N_1$ and $N_2$ in question.
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
$endgroup$
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
add a comment |
$begingroup$
You can have two sets of numbers such that they never clash: For example, let both sets of numbers be $N_1 = N_2 = 1, 3, 5, dots $. Now, if you multiply $N_2$ by $2$, then the new set will be $2, 6, 10, dots $ which cannot clash with the set $N_1$ (since $N_1$ only contains odd numbers, while $N_2$ contains even numbers.
Of course, this depends on the set of numbers $N_1$ and $N_2$ in question.
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
$endgroup$
You can have two sets of numbers such that they never clash: For example, let both sets of numbers be $N_1 = N_2 = 1, 3, 5, dots $. Now, if you multiply $N_2$ by $2$, then the new set will be $2, 6, 10, dots $ which cannot clash with the set $N_1$ (since $N_1$ only contains odd numbers, while $N_2$ contains even numbers.
Of course, this depends on the set of numbers $N_1$ and $N_2$ in question.
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
answered Mar 28 at 16:02
Siddharth BhatSiddharth Bhat
3,1471918
3,1471918
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
add a comment |
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
If you use prime numbers, this works over any multiplier
$endgroup$
– Don Thousand
Mar 28 at 16:11
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
$begingroup$
Thank you! although I think I didn't explain myself well. We need to dynamically merge two datasets, whose ID has the characteristics that I mentioned. Both sets of IDs start in 1, until infinity, and both of them grow by 1 to the next ID within their corresponding dataset. So, let's say that we multiply the first set of IDs by 2, maybe for some time we won't find, for example, the ID "2024" in the second dataset but we will find that ID at some point in the future in that second dataset, so we won't be able to uniquely identify the individual 2024 because another individual in dataset 2 exist
$endgroup$
– Ariadna
Mar 28 at 16:45
add a comment |
Ariadna is a new contributor. Be nice, and check out our Code of Conduct.
Ariadna is a new contributor. Be nice, and check out our Code of Conduct.
Ariadna is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It is impossible to do this unless you have some idea as to how the elements of the infinite lists are generated
$endgroup$
– Jorge Fernández Hidalgo
Mar 28 at 21:35