Write a bijection to match the set of positive integers with the set of positive integers excluding the...
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0
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Write a bijection to match the set of positive integers with the set of positive
integers excluding the multiples of 4.
e.g. $$1mapsto1,2mapsto2, 3mapsto3, 4mapsto 5, 5mapsto , 6mapsto 7, 7mapsto9, dots $$
I don't have any idea where to start the process of defining such a bijection.
And is it possible to define more than one bijection for the particular question above?
sequences-and-series functions discrete-mathematics
edited Nov 16 at 18:19
Rustyn
6,79411743
add a comment |
up vote
0
down vote
favorite
Write a bijection to match the set of positive integers with the set of positive
integers excluding the multiples of 4.
e.g. $$1mapsto1,2mapsto2, 3mapsto3, 4mapsto 5, 5mapsto , 6mapsto 7, 7mapsto9, dots $$
I don't have any idea where to start the process of defining such a bijection.
And is it possible to define more than one bijection for the particular question above?
sequences-and-series functions discrete-mathematics
edited Nov 16 at 18:19
Rustyn
6,79411743
1
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Write a bijection to match the set of positive integers with the set of positive
integers excluding the multiples of 4.
e.g. $$1mapsto1,2mapsto2, 3mapsto3, 4mapsto 5, 5mapsto , 6mapsto 7, 7mapsto9, dots $$
I don't have any idea where to start the process of defining such a bijection.
And is it possible to define more than one bijection for the particular question above?
sequences-and-series functions discrete-mathematics
edited Nov 16 at 18:19
Rustyn
6,79411743
Write a bijection to match the set of positive integers with the set of positive
integers excluding the multiples of 4.
e.g. $$1mapsto1,2mapsto2, 3mapsto3, 4mapsto 5, 5mapsto , 6mapsto 7, 7mapsto9, dots $$
I don't have any idea where to start the process of defining such a bijection.
And is it possible to define more than one bijection for the particular question above?
sequences-and-series functions discrete-mathematics
sequences-and-series functions discrete-mathematics
edited Nov 16 at 18:19
Rustyn
6,79411743
edited Nov 16 at 18:19
Rustyn
6,79411743
edited Nov 16 at 18:19
Rustyn
6,79411743
edited Nov 16 at 18:19
Rustyn
6,79411743
edited Nov 16 at 18:19
Rustyn
6,79411743
6,79411743
asked Nov 16 at 18:05
user616589
1
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50
add a comment |
1
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50
1
1
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
For example you can try
$$f(n)=n + lfloor (n-1)/3rfloor.$$
Does it work? Are you able to find another bijection?
As a bonus question, find a bijection from the set of of positive integers with the set of positive integers excluding the multiples of a given integer $m>1$.
answered Nov 16 at 18:13
Robert Z
90k1056128
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
|
show 6 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
For example you can try
$$f(n)=n + lfloor (n-1)/3rfloor.$$
Does it work? Are you able to find another bijection?
As a bonus question, find a bijection from the set of of positive integers with the set of positive integers excluding the multiples of a given integer $m>1$.
answered Nov 16 at 18:13
Robert Z
90k1056128
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
|
show 6 more comments
up vote
1
down vote
For example you can try
$$f(n)=n + lfloor (n-1)/3rfloor.$$
Does it work? Are you able to find another bijection?
As a bonus question, find a bijection from the set of of positive integers with the set of positive integers excluding the multiples of a given integer $m>1$.
answered Nov 16 at 18:13
Robert Z
90k1056128
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
|
show 6 more comments
up vote
1
down vote
up vote
1
down vote
For example you can try
$$f(n)=n + lfloor (n-1)/3rfloor.$$
Does it work? Are you able to find another bijection?
As a bonus question, find a bijection from the set of of positive integers with the set of positive integers excluding the multiples of a given integer $m>1$.
answered Nov 16 at 18:13
Robert Z
90k1056128
For example you can try
$$f(n)=n + lfloor (n-1)/3rfloor.$$
Does it work? Are you able to find another bijection?
As a bonus question, find a bijection from the set of of positive integers with the set of positive integers excluding the multiples of a given integer $m>1$.
answered Nov 16 at 18:13
Robert Z
90k1056128
edited Nov 16 at 18:18
answered Nov 16 at 18:13
Robert Z
90k1056128
answered Nov 16 at 18:13
Robert Z
90k1056128
answered Nov 16 at 18:13
Robert Z
90k1056128
90k1056128
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
|
show 6 more comments
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
How did you find the bijection though? Is there a general process for defining a bijection for any given sets?
– user616589
Nov 16 at 18:24
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
I don't think there is a general rule. Why don't you try to find another bijection? Take a bijection from the set of of positive integers onto itself different from the identity and...
– Robert Z
Nov 16 at 18:28
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
The function works pretty well by the way, thanks.
– user616589
Nov 16 at 18:29
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
It's just that everytime I see a question like this, all I can do is try to match the values in the domain with the ones in the codomain and define a function out of it randomly, thought maybe there is an algorithm that I can stick through but there is no such a thing appearently.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
I did verify it, it works pretty well.
– user616589
Nov 16 at 18:32
|
show 6 more comments
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1
Hey Kaan! Nice question! One note though, when posting questions to Math.SE, I'd like to invite you to consider including your thoughts/attempts on solving the problem, (even if you have your doubts). For example, you wrote that you don't have any idea on where to start on this problem-- an explanation as to why you feel/think you're stuck would be sufficient! It helps show question answerers goodwill and makes for high quality questions. We appreciate you joining our community and good luck in the future!
– Rustyn
Nov 16 at 18:50