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!










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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
















1














Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!










share|cite|improve this question




















  • 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








1







Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!










share|cite|improve this question















Can anyone help with this question? I’m having a lot of trouble understanding how to do this problem. Any help is appreciated!







combinatorics






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edited Nov 27 at 11:23









Tianlalu

3,09621038




3,09621038










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














  • 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










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.






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    1 Answer
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    0














    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.






    share|cite|improve this answer


























      0














      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.






      share|cite|improve this answer
























        0












        0








        0






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 11:40









        N.B.

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        689313






























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