Show that $n^3-n$ is divisible by $6$ using induction












6












$begingroup$


As homework, I have to prove that




$forall n in mathbb{N}: n^3-n$ is divisible by 6




I used induction



1) basis: $A(0): 0^3-0 = 6x$ , $x in mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?



2) requirement: $A(n):n^3-n=6x$



3) statement: $A(n+1): (n+1)^3-(n+1)=6x$



4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$



So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $forall n in mathbb{N}$



Did I do something wrong or is it that simple?






induction divisibility







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
    $endgroup$
    – Rick Decker
    Oct 11 '12 at 15:40
















6












$begingroup$


As homework, I have to prove that




$forall n in mathbb{N}: n^3-n$ is divisible by 6




I used induction



1) basis: $A(0): 0^3-0 = 6x$ , $x in mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?



2) requirement: $A(n):n^3-n=6x$



3) statement: $A(n+1): (n+1)^3-(n+1)=6x$



4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$



So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $forall n in mathbb{N}$



Did I do something wrong or is it that simple?






induction divisibility







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
    $endgroup$
    – Rick Decker
    Oct 11 '12 at 15:40














6












6








6





$begingroup$


As homework, I have to prove that




$forall n in mathbb{N}: n^3-n$ is divisible by 6




I used induction



1) basis: $A(0): 0^3-0 = 6x$ , $x in mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?



2) requirement: $A(n):n^3-n=6x$



3) statement: $A(n+1): (n+1)^3-(n+1)=6x$



4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$



So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $forall n in mathbb{N}$



Did I do something wrong or is it that simple?






induction divisibility







share|cite|improve this question











$endgroup$




As homework, I have to prove that




$forall n in mathbb{N}: n^3-n$ is divisible by 6




I used induction



1) basis: $A(0): 0^3-0 = 6x$ , $x in mathbb{N}_0$ // the 6x states that the result is a multiple of 6, right?



2) requirement: $A(n):n^3-n=6x$



3) statement: $A(n+1): (n+1)^3-(n+1)=6x$



4) step: $n^3-n+(n+1)^3-n=6x+(n+1)^3-n$



So when I resolve that I do get the equation: $n^3-n=6x$ so the statement is true for $forall n in mathbb{N}$



Did I do something wrong or is it that simple?






induction divisibility




induction divisibility






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 1 '16 at 13:59









Rodrigo de Azevedo

13.1k41959




13.1k41959










asked Oct 11 '12 at 14:49









Marco RauscherMarco Rauscher

36112




36112








  • 1




    $begingroup$
    I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
    $endgroup$
    – Rick Decker
    Oct 11 '12 at 15:40














  • 1




    $begingroup$
    I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
    $endgroup$
    – Rick Decker
    Oct 11 '12 at 15:40








1




1




$begingroup$
I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
$endgroup$
– Rick Decker
Oct 11 '12 at 15:40




$begingroup$
I know what you meant, but it's confusing when in both of steps (2) and (3) you use $6x$. It would be clearer if you had said in (2) $n^3-n=6x$ for some integer $x$ and in (3) you had said $(n+1)^3-(n+1)=6y$ for some integer $y$.
$endgroup$
– Rick Decker
Oct 11 '12 at 15:40










3 Answers
3






active

oldest

votes


















5












$begingroup$

No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then
$$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$
is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
    $endgroup$
    – mathguy
    Aug 1 '16 at 13:33





















2












$begingroup$

No need for induction.
$$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
which are three consecutive integers. So one must be divisible by 3.




Check for $n=1$: $1^3-1=0=3cdot 0$



Assume it's true for $n=k$.
If you let $n=k+1$ you get
$$begin{align*}
(k+1)^3-(k+1)&=k^3+3k^2+2 \
&=k^3+3k^2+2k\
&=3cdot (k^2+k)+(k^3-k)end{align*}$$
which is divisible by 3






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
    $endgroup$
    – Marco Rauscher
    Oct 11 '12 at 14:54










  • $begingroup$
    @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
    $endgroup$
    – yroc
    Nov 23 '15 at 19:59



















1












$begingroup$

If $n^3-n=6m$,



$$(n+1)^3-(n+1)=n^3+3n^2+2n=6m+3(n^2+n).$$



Then, by an auxiliary induction $n^2+n$ is even.



Indeed, $0^2+0$ is even and if $n^2+n=2k$,



$$(n+1)^2+(n+1)=n^2+3n+2=2k+2(n+1)$$ which is even.



Finally, $3$ times an even number is a multiple of $6$.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then
    $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$
    is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
      $endgroup$
      – mathguy
      Aug 1 '16 at 13:33


















    5












    $begingroup$

    No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then
    $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$
    is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
      $endgroup$
      – mathguy
      Aug 1 '16 at 13:33
















    5












    5








    5





    $begingroup$

    No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then
    $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$
    is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.






    share|cite|improve this answer









    $endgroup$



    No, your argument is not quite right (or at least not clear to me). You must show that if $A(n)$ is true, then $A(n+1)$ follows. $A(n)$ here is the statement "$n^3-n$ is divisible by $6$". Assuming $A(n)$ is true, then
    $$(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1=(n^3-n)+3n(n+1)$$
    is divisible by $6$ because (1) $n^3-n$ is a multiple of $6$ by assumption, and (2) $3n(n+1)$ is divisible by $6$ because one of $n$ or $n+1$ must be even (this is related to what Alex was pointing out). Therefore $A(n)$ implies $A(n+1)$ and, if $A(n)$ is true for some value of $n$, then all higher integer values of $n$ follow. You correctly showed that $A(0)$ is true.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Oct 11 '12 at 14:58









    JonathanJonathan

    6,78511533




    6,78511533








    • 2




      $begingroup$
      Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
      $endgroup$
      – mathguy
      Aug 1 '16 at 13:33
















    • 2




      $begingroup$
      Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
      $endgroup$
      – mathguy
      Aug 1 '16 at 13:33










    2




    2




    $begingroup$
    Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
    $endgroup$
    – mathguy
    Aug 1 '16 at 13:33






    $begingroup$
    Of course, the argument that "one of $n$ or $n+1$ must be even" is no different from "one of $n, n-1, n+1$ must be divisible by 3". To stay within the spirit of the problem, the fact that $3n(n+1)$ is divisible by 6 should also be proved by induction.
    $endgroup$
    – mathguy
    Aug 1 '16 at 13:33













    2












    $begingroup$

    No need for induction.
    $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
    which are three consecutive integers. So one must be divisible by 3.




    Check for $n=1$: $1^3-1=0=3cdot 0$



    Assume it's true for $n=k$.
    If you let $n=k+1$ you get
    $$begin{align*}
    (k+1)^3-(k+1)&=k^3+3k^2+2 \
    &=k^3+3k^2+2k\
    &=3cdot (k^2+k)+(k^3-k)end{align*}$$
    which is divisible by 3






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
      $endgroup$
      – Marco Rauscher
      Oct 11 '12 at 14:54










    • $begingroup$
      @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
      $endgroup$
      – yroc
      Nov 23 '15 at 19:59
















    2












    $begingroup$

    No need for induction.
    $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
    which are three consecutive integers. So one must be divisible by 3.




    Check for $n=1$: $1^3-1=0=3cdot 0$



    Assume it's true for $n=k$.
    If you let $n=k+1$ you get
    $$begin{align*}
    (k+1)^3-(k+1)&=k^3+3k^2+2 \
    &=k^3+3k^2+2k\
    &=3cdot (k^2+k)+(k^3-k)end{align*}$$
    which is divisible by 3






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
      $endgroup$
      – Marco Rauscher
      Oct 11 '12 at 14:54










    • $begingroup$
      @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
      $endgroup$
      – yroc
      Nov 23 '15 at 19:59














    2












    2








    2





    $begingroup$

    No need for induction.
    $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
    which are three consecutive integers. So one must be divisible by 3.




    Check for $n=1$: $1^3-1=0=3cdot 0$



    Assume it's true for $n=k$.
    If you let $n=k+1$ you get
    $$begin{align*}
    (k+1)^3-(k+1)&=k^3+3k^2+2 \
    &=k^3+3k^2+2k\
    &=3cdot (k^2+k)+(k^3-k)end{align*}$$
    which is divisible by 3






    share|cite|improve this answer











    $endgroup$



    No need for induction.
    $$n^3-n=n(n^2-1)=n(n-1)(n+1)$$
    which are three consecutive integers. So one must be divisible by 3.




    Check for $n=1$: $1^3-1=0=3cdot 0$



    Assume it's true for $n=k$.
    If you let $n=k+1$ you get
    $$begin{align*}
    (k+1)^3-(k+1)&=k^3+3k^2+2 \
    &=k^3+3k^2+2k\
    &=3cdot (k^2+k)+(k^3-k)end{align*}$$
    which is divisible by 3







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Oct 11 '12 at 15:00

























    answered Oct 11 '12 at 14:53









    AlexAlex

    2,131718




    2,131718












    • $begingroup$
      Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
      $endgroup$
      – Marco Rauscher
      Oct 11 '12 at 14:54










    • $begingroup$
      @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
      $endgroup$
      – yroc
      Nov 23 '15 at 19:59


















    • $begingroup$
      Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
      $endgroup$
      – Marco Rauscher
      Oct 11 '12 at 14:54










    • $begingroup$
      @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
      $endgroup$
      – yroc
      Nov 23 '15 at 19:59
















    $begingroup$
    Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
    $endgroup$
    – Marco Rauscher
    Oct 11 '12 at 14:54




    $begingroup$
    Thank you for your answer. I know that way, but I wanted to know if I used the principle of induction in the correct way.
    $endgroup$
    – Marco Rauscher
    Oct 11 '12 at 14:54












    $begingroup$
    @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
    $endgroup$
    – yroc
    Nov 23 '15 at 19:59




    $begingroup$
    @Alex There is a need for induction if the question says "prove by induction" (as has been known to happen on a test).
    $endgroup$
    – yroc
    Nov 23 '15 at 19:59











    1












    $begingroup$

    If $n^3-n=6m$,



    $$(n+1)^3-(n+1)=n^3+3n^2+2n=6m+3(n^2+n).$$



    Then, by an auxiliary induction $n^2+n$ is even.



    Indeed, $0^2+0$ is even and if $n^2+n=2k$,



    $$(n+1)^2+(n+1)=n^2+3n+2=2k+2(n+1)$$ which is even.



    Finally, $3$ times an even number is a multiple of $6$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $n^3-n=6m$,



      $$(n+1)^3-(n+1)=n^3+3n^2+2n=6m+3(n^2+n).$$



      Then, by an auxiliary induction $n^2+n$ is even.



      Indeed, $0^2+0$ is even and if $n^2+n=2k$,



      $$(n+1)^2+(n+1)=n^2+3n+2=2k+2(n+1)$$ which is even.



      Finally, $3$ times an even number is a multiple of $6$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $n^3-n=6m$,



        $$(n+1)^3-(n+1)=n^3+3n^2+2n=6m+3(n^2+n).$$



        Then, by an auxiliary induction $n^2+n$ is even.



        Indeed, $0^2+0$ is even and if $n^2+n=2k$,



        $$(n+1)^2+(n+1)=n^2+3n+2=2k+2(n+1)$$ which is even.



        Finally, $3$ times an even number is a multiple of $6$.






        share|cite|improve this answer









        $endgroup$



        If $n^3-n=6m$,



        $$(n+1)^3-(n+1)=n^3+3n^2+2n=6m+3(n^2+n).$$



        Then, by an auxiliary induction $n^2+n$ is even.



        Indeed, $0^2+0$ is even and if $n^2+n=2k$,



        $$(n+1)^2+(n+1)=n^2+3n+2=2k+2(n+1)$$ which is even.



        Finally, $3$ times an even number is a multiple of $6$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 1 '16 at 13:57









        Yves DaoustYves Daoust

        129k675227




        129k675227






























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