Construct a bijection from $Bbb N$ to $Bbb N times [3]$ ??
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Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!
combinatorics
edited Nov 27 at 11:23
Tianlalu
3,0101938
asked Nov 27 at 9:09
Caroline
61
add a comment |
up vote
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Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!
combinatorics
edited Nov 27 at 11:23
Tianlalu
3,0101938
asked Nov 27 at 9:09
Caroline
61
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
4
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!
combinatorics
edited Nov 27 at 11:23
Tianlalu
3,0101938
asked Nov 27 at 9:09
Caroline
61
Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!
combinatorics
combinatorics
edited Nov 27 at 11:23
Tianlalu
3,0101938
asked Nov 27 at 9:09
Caroline
61
edited Nov 27 at 11:23
Tianlalu
3,0101938
asked Nov 27 at 9:09
Caroline
61
edited Nov 27 at 11:23
Tianlalu
3,0101938
edited Nov 27 at 11:23
Tianlalu
3,0101938
edited Nov 27 at 11:23
Tianlalu
3,0101938
3,0101938
asked Nov 27 at 9:09
Caroline
61
asked Nov 27 at 9:09
Caroline
61
asked Nov 27 at 9:09
Caroline
61
61
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
4
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12
add a comment |
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
4
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12
1
1
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
4
4
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12
add a comment |
1 Answer
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Just divide the natural numbers by their residue modulo $3$, i.e. define the map $f colon mathbb{N} rightarrow mathbb{N} times {0,1,2 }$ as follows
begin{equation}
f(n)= begin{cases}
(n/3,0) quad text{if} 3|n \
(frac{n-1}{3},1) quad text{if} n=1 text{mod} 3 \
(frac{n-2}{3},2) quad text{if} n=2 text{mod} 3.
end{cases}
end{equation}
The basic idea is that since the cardinality of $mathbb{N}$ is infinite then $3 cdot |mathbb{N}|= |mathbb{N}|$. More concretely the set ${ 3n : n in mathbb{N} }$ is in bijection with $mathbb{N}$ just by the map $n mapsto 3n$, thus adding $0,1, 2$, i.e. shifting the multiples of $3$ by the possible residues of the division by $3$ we get all the natural numbers.
answered Nov 27 at 11:40
N.B.
689313
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1 Answer
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oldest
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up vote
0
down vote
Just divide the natural numbers by their residue modulo $3$, i.e. define the map $f colon mathbb{N} rightarrow mathbb{N} times {0,1,2 }$ as follows
begin{equation}
f(n)= begin{cases}
(n/3,0) quad text{if} 3|n \
(frac{n-1}{3},1) quad text{if} n=1 text{mod} 3 \
(frac{n-2}{3},2) quad text{if} n=2 text{mod} 3.
end{cases}
end{equation}
The basic idea is that since the cardinality of $mathbb{N}$ is infinite then $3 cdot |mathbb{N}|= |mathbb{N}|$. More concretely the set ${ 3n : n in mathbb{N} }$ is in bijection with $mathbb{N}$ just by the map $n mapsto 3n$, thus adding $0,1, 2$, i.e. shifting the multiples of $3$ by the possible residues of the division by $3$ we get all the natural numbers.
answered Nov 27 at 11:40
N.B.
689313
add a comment |
up vote
0
down vote
Just divide the natural numbers by their residue modulo $3$, i.e. define the map $f colon mathbb{N} rightarrow mathbb{N} times {0,1,2 }$ as follows
begin{equation}
f(n)= begin{cases}
(n/3,0) quad text{if} 3|n \
(frac{n-1}{3},1) quad text{if} n=1 text{mod} 3 \
(frac{n-2}{3},2) quad text{if} n=2 text{mod} 3.
end{cases}
end{equation}
The basic idea is that since the cardinality of $mathbb{N}$ is infinite then $3 cdot |mathbb{N}|= |mathbb{N}|$. More concretely the set ${ 3n : n in mathbb{N} }$ is in bijection with $mathbb{N}$ just by the map $n mapsto 3n$, thus adding $0,1, 2$, i.e. shifting the multiples of $3$ by the possible residues of the division by $3$ we get all the natural numbers.
answered Nov 27 at 11:40
N.B.
689313
add a comment |
up vote
0
down vote
up vote
0
down vote
Just divide the natural numbers by their residue modulo $3$, i.e. define the map $f colon mathbb{N} rightarrow mathbb{N} times {0,1,2 }$ as follows
begin{equation}
f(n)= begin{cases}
(n/3,0) quad text{if} 3|n \
(frac{n-1}{3},1) quad text{if} n=1 text{mod} 3 \
(frac{n-2}{3},2) quad text{if} n=2 text{mod} 3.
end{cases}
end{equation}
The basic idea is that since the cardinality of $mathbb{N}$ is infinite then $3 cdot |mathbb{N}|= |mathbb{N}|$. More concretely the set ${ 3n : n in mathbb{N} }$ is in bijection with $mathbb{N}$ just by the map $n mapsto 3n$, thus adding $0,1, 2$, i.e. shifting the multiples of $3$ by the possible residues of the division by $3$ we get all the natural numbers.
answered Nov 27 at 11:40
N.B.
689313
Just divide the natural numbers by their residue modulo $3$, i.e. define the map $f colon mathbb{N} rightarrow mathbb{N} times {0,1,2 }$ as follows
begin{equation}
f(n)= begin{cases}
(n/3,0) quad text{if} 3|n \
(frac{n-1}{3},1) quad text{if} n=1 text{mod} 3 \
(frac{n-2}{3},2) quad text{if} n=2 text{mod} 3.
end{cases}
end{equation}
The basic idea is that since the cardinality of $mathbb{N}$ is infinite then $3 cdot |mathbb{N}|= |mathbb{N}|$. More concretely the set ${ 3n : n in mathbb{N} }$ is in bijection with $mathbb{N}$ just by the map $n mapsto 3n$, thus adding $0,1, 2$, i.e. shifting the multiples of $3$ by the possible residues of the division by $3$ we get all the natural numbers.
answered Nov 27 at 11:40
N.B.
689313
answered Nov 27 at 11:40
N.B.
689313
answered Nov 27 at 11:40
N.B.
689313
answered Nov 27 at 11:40
N.B.
689313
689313
add a comment |
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Nov 27 at 9:12
4
Do you mean $[3] = {1,2,3}$?
– Wuestenfux
Nov 27 at 9:12
...or $[3] = {3}$?
– Vincent
Nov 27 at 11:32
I meant [3] as in {1,2,3}
– Caroline
Nov 29 at 1:12