Sum/Product of two natural numbers is a natural number












1












$begingroup$


I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.



So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .



So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?



I am trying to prove this with the definition of a Natural numbers, but it's not working so far...



Thank you.






set-theory natural-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
    $endgroup$
    – Daniel Fischer♦
    Oct 9 '14 at 13:33










  • $begingroup$
    Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 13:57












  • $begingroup$
    inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 14:01










  • $begingroup$
    @DavidC You need to tag him, otherwise he won't get the notification of your comment.
    $endgroup$
    – Vincenzo Oliva
    Oct 11 '14 at 9:24






  • 1




    $begingroup$
    I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
    $endgroup$
    – Martin Sleziak
    Feb 1 '15 at 11:40
















1












$begingroup$


I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.



So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .



So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?



I am trying to prove this with the definition of a Natural numbers, but it's not working so far...



Thank you.






set-theory natural-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
    $endgroup$
    – Daniel Fischer♦
    Oct 9 '14 at 13:33










  • $begingroup$
    Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 13:57












  • $begingroup$
    inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 14:01










  • $begingroup$
    @DavidC You need to tag him, otherwise he won't get the notification of your comment.
    $endgroup$
    – Vincenzo Oliva
    Oct 11 '14 at 9:24






  • 1




    $begingroup$
    I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
    $endgroup$
    – Martin Sleziak
    Feb 1 '15 at 11:40














1












1








1


1



$begingroup$


I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.



So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .



So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?



I am trying to prove this with the definition of a Natural numbers, but it's not working so far...



Thank you.






set-theory natural-numbers







share|cite|improve this question











$endgroup$




I wanted to prove that the sum and the product of two natural numbers is a natural number. Intuitively it's clear to my why that is true, however I couldn't prove it.



So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , 1leq n$ .



So this is the information we have so far. But I am having difficulty proving that the sum/product of two natural numbers is a natural number. Can someone give me a clue how to prove this?



I am trying to prove this with the definition of a Natural numbers, but it's not working so far...



Thank you.






set-theory natural-numbers




set-theory natural-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 '15 at 11:38









Martin Sleziak

44.8k9118272




44.8k9118272










asked Oct 9 '14 at 13:29









Charles CarmichaelCharles Carmichael

30917




30917








  • 1




    $begingroup$
    First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
    $endgroup$
    – Daniel Fischer♦
    Oct 9 '14 at 13:33










  • $begingroup$
    Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 13:57












  • $begingroup$
    inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 14:01










  • $begingroup$
    @DavidC You need to tag him, otherwise he won't get the notification of your comment.
    $endgroup$
    – Vincenzo Oliva
    Oct 11 '14 at 9:24






  • 1




    $begingroup$
    I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
    $endgroup$
    – Martin Sleziak
    Feb 1 '15 at 11:40














  • 1




    $begingroup$
    First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
    $endgroup$
    – Daniel Fischer♦
    Oct 9 '14 at 13:33










  • $begingroup$
    Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 13:57












  • $begingroup$
    inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
    $endgroup$
    – Charles Carmichael
    Oct 9 '14 at 14:01










  • $begingroup$
    @DavidC You need to tag him, otherwise he won't get the notification of your comment.
    $endgroup$
    – Vincenzo Oliva
    Oct 11 '14 at 9:24






  • 1




    $begingroup$
    I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
    $endgroup$
    – Martin Sleziak
    Feb 1 '15 at 11:40








1




1




$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33




$begingroup$
First you need the definition of addition and multiplication. Then prove by induction for any fixed $minmathbb{N}$ that ${ninmathbb{N} : n+m inmathbb{N}} = mathbb{N}$ (and similarly for the product).
$endgroup$
– Daniel Fischer♦
Oct 9 '14 at 13:33












$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57






$begingroup$
Hi Daniel, can you verify this proof? Proof: Let there be $m in mathbb{N}$. We will define the set I={$n in mathbb{N} | n+m in mathbb{N}$ } So we want to prove that $ I = mathbb{N}$ , from the definition of I we derive $I subset mathbb{N}$ so we have to prove that $mathbb{N} subset I $ . We will prove this with induction. Basis for the induction: n=1 . Since $m in mathbb{N}$ and $mathbb{N}$ is inductive, $m+1 in mathbb{N}$ and therefore $1 in I$. Induction Hyp:$n in I$ , we will prove that $n+1 in I$ Induct Step: If $n in I$ then $n+m in mathbb{N}$ since $mathbb{N}$ is
$endgroup$
– Charles Carmichael
Oct 9 '14 at 13:57














$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01




$begingroup$
inductive $n+m+1 in mathbb{N}$ therefore $n+1 in I$ and therefore $I=mathbb{N}$, as desired. Is this correct? Thank you.
$endgroup$
– Charles Carmichael
Oct 9 '14 at 14:01












$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24




$begingroup$
@DavidC You need to tag him, otherwise he won't get the notification of your comment.
$endgroup$
– Vincenzo Oliva
Oct 11 '14 at 9:24




1




1




$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40




$begingroup$
I have added (set-theory) tag, since the question is about set-theoretical definition of natural numbers (as the smallest inductive set) based on Axiom of Infinity.
$endgroup$
– Martin Sleziak
Feb 1 '15 at 11:40










2 Answers
2






active

oldest

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0












$begingroup$

How are sum/product operations in $mathbb{N}$ defined in your context?



Let $n,minmathbb{N}$.



Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
$$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$



Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:



Addition (or "sum"):




  • If $m=0$, then $n+ m := m$

  • If $mneq 0$, then $n+ m := s(n+l)$


Multiplication (or "product"):




  • If $m=0$, then $ncdot m = 0$

  • If $mneq 0$, then $ncdot m = ncdot l + n$


Thus, by this definition and using induction, the result of both sum and product are also natural numbers.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The OP stated




    So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.




    I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.



    So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).



    With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.



    With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.



    Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).



    We conclude this by defining addition.



    If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:



    Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.



    The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
      $endgroup$
      – Asaf Karagila♦
      Jan 9 at 11:09











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    How are sum/product operations in $mathbb{N}$ defined in your context?



    Let $n,minmathbb{N}$.



    Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
    $$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$



    Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:



    Addition (or "sum"):




    • If $m=0$, then $n+ m := m$

    • If $mneq 0$, then $n+ m := s(n+l)$


    Multiplication (or "product"):




    • If $m=0$, then $ncdot m = 0$

    • If $mneq 0$, then $ncdot m = ncdot l + n$


    Thus, by this definition and using induction, the result of both sum and product are also natural numbers.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      How are sum/product operations in $mathbb{N}$ defined in your context?



      Let $n,minmathbb{N}$.



      Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
      $$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$



      Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:



      Addition (or "sum"):




      • If $m=0$, then $n+ m := m$

      • If $mneq 0$, then $n+ m := s(n+l)$


      Multiplication (or "product"):




      • If $m=0$, then $ncdot m = 0$

      • If $mneq 0$, then $ncdot m = ncdot l + n$


      Thus, by this definition and using induction, the result of both sum and product are also natural numbers.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        How are sum/product operations in $mathbb{N}$ defined in your context?



        Let $n,minmathbb{N}$.



        Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
        $$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$



        Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:



        Addition (or "sum"):




        • If $m=0$, then $n+ m := m$

        • If $mneq 0$, then $n+ m := s(n+l)$


        Multiplication (or "product"):




        • If $m=0$, then $ncdot m = 0$

        • If $mneq 0$, then $ncdot m = ncdot l + n$


        Thus, by this definition and using induction, the result of both sum and product are also natural numbers.






        share|cite|improve this answer











        $endgroup$



        How are sum/product operations in $mathbb{N}$ defined in your context?



        Let $n,minmathbb{N}$.



        Write $s(n)=ncup {n}$. Observe $s(n)inmathbb{N}$ by definition of inductive set, and recall
        $$mneq 0 implies exists linmathbb{N} big(m=s(l)big).$$



        Both addition and multiplication are binary operations in $mathbb{N}$ defined recursively as follows:



        Addition (or "sum"):




        • If $m=0$, then $n+ m := m$

        • If $mneq 0$, then $n+ m := s(n+l)$


        Multiplication (or "product"):




        • If $m=0$, then $ncdot m = 0$

        • If $mneq 0$, then $ncdot m = ncdot l + n$


        Thus, by this definition and using induction, the result of both sum and product are also natural numbers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 17:06

























        answered Dec 5 '18 at 18:27









        Dr PotatoDr Potato

        394




        394























            0












            $begingroup$

            The OP stated




            So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.




            I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.



            So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).



            With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.



            With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.



            Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).



            We conclude this by defining addition.



            If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:



            Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.



            The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
              $endgroup$
              – Asaf Karagila♦
              Jan 9 at 11:09
















            0












            $begingroup$

            The OP stated




            So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.




            I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.



            So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).



            With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.



            With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.



            Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).



            We conclude this by defining addition.



            If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:



            Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.



            The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
              $endgroup$
              – Asaf Karagila♦
              Jan 9 at 11:09














            0












            0








            0





            $begingroup$

            The OP stated




            So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.




            I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.



            So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).



            With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.



            With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.



            Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).



            We conclude this by defining addition.



            If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:



            Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.



            The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.






            share|cite|improve this answer









            $endgroup$



            The OP stated




            So our lecturer first defined what an inductive set is. Then he defined the Natural numbers as the intersection of all inductive sets. Then we proved the induction principle and $forall n in mathbb{N} , , 1 le n$.




            I think the OP meant to say $forall n in mathbb{N} , , 0 le n$, but in any case the following discussion will be instructive.



            So we are starting with the von Neumann ordinal $omega = mathbb N$ (see the wiki article Axiom of infinity).



            With this setup a binary operation called addition can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m + n$.



            With this setup a binary operation called mulitiplication can be defined mapping an ordered pair in $(m,n) in mathbb{N} times mathbb{N}$ to another natural number, $m * n$.



            Once you define these operations, you have to prove (abstractly) that these operations satisfy all the properties of addition and multiplication that you learned in elementary school (for example, that multiplication distributes over addition).



            We conclude this by defining addition.



            If $m,n in mathbb N$ they can be viewed as ordinals (well-ordered sets) and we can define addition, up to an isomorphism of well-ordered sets. But the following is also true:



            Theorem: Any two ordinals $X$, $Y$ are comparable, in the sense that either $X = Y$, or $X$ is an initial segment of $Y$ , or $Y$ is an initial segment of $X$.



            The OP has to show that if they add the ordinals $m$ and $n$ then $m + n$ is an initial segment of $mathbb N$. Since every (proper) initial segment of $mathbb N$ has a greatest element $s$, the mapping $m + n mapsto s$ is well-defined.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 11:05









            CopyPasteItCopyPasteIt

            4,1481628




            4,1481628












            • $begingroup$
              Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
              $endgroup$
              – Asaf Karagila♦
              Jan 9 at 11:09


















            • $begingroup$
              Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
              $endgroup$
              – Asaf Karagila♦
              Jan 9 at 11:09
















            $begingroup$
            Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
            $endgroup$
            – Asaf Karagila♦
            Jan 9 at 11:09




            $begingroup$
            Or maybe this takes place in a real analysis course, where commonly $Bbb N$ is just the intersection of all inductive subsets of $Bbb R$, where inductive means $1in A$ and $ain Ato a+1in A$. So maybe this has nothing to do with ordinals, and everything to do with induction. Not to mention that you completely sidestep the question and just say "Well, we can define addition and multiplication and then prove they satisfy what we want them to satisfy". Well, isn't this the whole point of this question?
            $endgroup$
            – Asaf Karagila♦
            Jan 9 at 11:09


















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