For all integers of n, $n$ is divisible by $2 iff n^4$ is divisible by $2$












-4












$begingroup$



Provide the proof: $forall n in mathbb{Z}$, $n$ is divisible by $2
> iff n^4$ is divisible by $2$.




Just curious on how the proof of this statement would look like.






proof-writing







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$endgroup$








  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
    $endgroup$
    – manooooh
    Dec 12 '18 at 6:09






  • 1




    $begingroup$
    The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
    $endgroup$
    – Matti P.
    Dec 12 '18 at 6:14










  • $begingroup$
    @manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:22
















-4












$begingroup$



Provide the proof: $forall n in mathbb{Z}$, $n$ is divisible by $2
> iff n^4$ is divisible by $2$.




Just curious on how the proof of this statement would look like.






proof-writing







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
    $endgroup$
    – manooooh
    Dec 12 '18 at 6:09






  • 1




    $begingroup$
    The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
    $endgroup$
    – Matti P.
    Dec 12 '18 at 6:14










  • $begingroup$
    @manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:22














-4












-4








-4





$begingroup$



Provide the proof: $forall n in mathbb{Z}$, $n$ is divisible by $2
> iff n^4$ is divisible by $2$.




Just curious on how the proof of this statement would look like.






proof-writing







share|cite|improve this question











$endgroup$





Provide the proof: $forall n in mathbb{Z}$, $n$ is divisible by $2
> iff n^4$ is divisible by $2$.




Just curious on how the proof of this statement would look like.






proof-writing




proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 20:43









Bram28

61.8k44793




61.8k44793










asked Dec 12 '18 at 6:06









Dike DokeDike Doke

13




13








  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
    $endgroup$
    – manooooh
    Dec 12 '18 at 6:09






  • 1




    $begingroup$
    The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
    $endgroup$
    – Matti P.
    Dec 12 '18 at 6:14










  • $begingroup$
    @manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:22














  • 1




    $begingroup$
    Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
    $endgroup$
    – manooooh
    Dec 12 '18 at 6:09






  • 1




    $begingroup$
    The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
    $endgroup$
    – Matti P.
    Dec 12 '18 at 6:14










  • $begingroup$
    @manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:22








1




1




$begingroup$
Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
$endgroup$
– manooooh
Dec 12 '18 at 6:09




$begingroup$
Welcome to math.SE!! Please use MathJax to render the math correctly. Also, could you tell us what have you tried please?
$endgroup$
– manooooh
Dec 12 '18 at 6:09




1




1




$begingroup$
The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
$endgroup$
– Matti P.
Dec 12 '18 at 6:14




$begingroup$
The $Rightarrow$-side is quite easy to prove, since there are only two cases. So I would approach this by splitting the proof up to $Rightarrow$ and $Leftarrow$.
$endgroup$
– Matti P.
Dec 12 '18 at 6:14












$begingroup$
@manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
$endgroup$
– Dike Doke
Dec 12 '18 at 6:22




$begingroup$
@manooooh As far as the proof goes, I only showed n being an even integer, while the second part, where n^4 is divisible by 2, I did not understand if the second part is acting like a condition for the first part of the statement. Just confused on the "if and only if" i guess.
$endgroup$
– Dike Doke
Dec 12 '18 at 6:22










2 Answers
2






active

oldest

votes


















1












$begingroup$

A number $n$ is divisible by two if and only if $n$ is even $ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.



Now suppose that $n$ is even then $n=2k$ for some $k$ , so



$$n^4 = (2k)^4 = 2^4k^4 = 2cdot2^3cdot k^4 = 2cdot K$$



where $K=2^3cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that
$$n^4 = (2k+1)^4 =
16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$



where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).



Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:32












  • $begingroup$
    I've edited my answer to address this question.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:54










  • $begingroup$
    Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:55



















0












$begingroup$

One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:



$forall n in mathbb{Z}, forall p in mathbb{N}^+: n$ is divisible by $2
iff n^p$ is divisible by $2$



This we can prove by induction on $p$:



Base:



$p=1$.



We have to show that:



$$n text{ is divisible by } 2
iff n^1 text{ is divisible by }2$$



Well, that is trivially true, since $n^1=n$



Step:



Thye inductive hypothesis is that for some $k in mathbb{N}^+$, we have that:



$$n text{ is divisible by } 2
iff n^k text{ is divisible by }2$$



Now we need to show that:



$$n text{ is divisible by } 2
iff n^{k+1} text{ is divisible by }2$$



We would prove this if we can prove the following two things:



$$n text{ is divisible by } 2
Rightarrow n^{k+1} text{ is divisible by }2$$



and



$$n text{ is not divisible by } 2
Rightarrow n^{k+1} text{ is not divisible by }2$$



The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n cdot n^k$ is even as well



Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n cdot n^k$ is odd as well, and thus not divisible by $2$






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    2 Answers
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    2 Answers
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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    A number $n$ is divisible by two if and only if $n$ is even $ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.



    Now suppose that $n$ is even then $n=2k$ for some $k$ , so



    $$n^4 = (2k)^4 = 2^4k^4 = 2cdot2^3cdot k^4 = 2cdot K$$



    where $K=2^3cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that
    $$n^4 = (2k+1)^4 =
    16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$



    where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).



    Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
      $endgroup$
      – Dike Doke
      Dec 12 '18 at 6:32












    • $begingroup$
      I've edited my answer to address this question.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:54










    • $begingroup$
      Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:55
















    1












    $begingroup$

    A number $n$ is divisible by two if and only if $n$ is even $ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.



    Now suppose that $n$ is even then $n=2k$ for some $k$ , so



    $$n^4 = (2k)^4 = 2^4k^4 = 2cdot2^3cdot k^4 = 2cdot K$$



    where $K=2^3cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that
    $$n^4 = (2k+1)^4 =
    16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$



    where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).



    Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
      $endgroup$
      – Dike Doke
      Dec 12 '18 at 6:32












    • $begingroup$
      I've edited my answer to address this question.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:54










    • $begingroup$
      Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:55














    1












    1








    1





    $begingroup$

    A number $n$ is divisible by two if and only if $n$ is even $ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.



    Now suppose that $n$ is even then $n=2k$ for some $k$ , so



    $$n^4 = (2k)^4 = 2^4k^4 = 2cdot2^3cdot k^4 = 2cdot K$$



    where $K=2^3cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that
    $$n^4 = (2k+1)^4 =
    16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$



    where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).



    Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).






    share|cite|improve this answer











    $endgroup$



    A number $n$ is divisible by two if and only if $n$ is even $ldots$ equivalently, $n$ is even if and only if $n=2k$ for some $k$.



    Now suppose that $n$ is even then $n=2k$ for some $k$ , so



    $$n^4 = (2k)^4 = 2^4k^4 = 2cdot2^3cdot k^4 = 2cdot K$$



    where $K=2^3cdot k^4$. Therefore, $n^4$ is even and hence divisible by two. Conversely, suppose $n^4$ is divisible by $2$ but that $n$ were odd. Then repeating the previous we get that $n=2k+1$ for some $k$ and hence that
    $$n^4 = (2k+1)^4 =
    16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 = 2K+1$$



    where $K=8k^4 + 16k^3+24k^2+4k$, but then $n^4$ would be odd and hence not divisible by two (a contradiction).



    Notice: This second half of the proof is essentially proof by contraposition but I sort of artificially made it a proof by contradiction (because usually people new to proofs find these easier to follow).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 12 '18 at 17:53

























    answered Dec 12 '18 at 6:14









    SquirtleSquirtle

    4,2071641




    4,2071641












    • $begingroup$
      Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
      $endgroup$
      – Dike Doke
      Dec 12 '18 at 6:32












    • $begingroup$
      I've edited my answer to address this question.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:54










    • $begingroup$
      Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:55


















    • $begingroup$
      Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
      $endgroup$
      – Dike Doke
      Dec 12 '18 at 6:32












    • $begingroup$
      I've edited my answer to address this question.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:54










    • $begingroup$
      Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
      $endgroup$
      – Squirtle
      Dec 12 '18 at 17:55
















    $begingroup$
    Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:32






    $begingroup$
    Quick question: so the K for when n is odd will be (8k^3+16k^3+12k^2+4k) and to show 2K+1 for the second part of the proof?
    $endgroup$
    – Dike Doke
    Dec 12 '18 at 6:32














    $begingroup$
    I've edited my answer to address this question.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:54




    $begingroup$
    I've edited my answer to address this question.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:54












    $begingroup$
    Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:55




    $begingroup$
    Dike Doke.... also you should know that this question is not about "formal proofs" since it is not using formal logic. In the future, when you ask questions you ought to include your attempt at the problem.
    $endgroup$
    – Squirtle
    Dec 12 '18 at 17:55











    0












    $begingroup$

    One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:



    $forall n in mathbb{Z}, forall p in mathbb{N}^+: n$ is divisible by $2
    iff n^p$     is divisible by $2$



    This we can prove by induction on $p$:



    Base:



    $p=1$.



    We have to show that:



    $$n text{ is divisible by } 2
    iff n^1 text{ is divisible by }2$$



    Well, that is trivially true, since $n^1=n$



    Step:



    Thye inductive hypothesis is that for some $k in mathbb{N}^+$, we have that:



    $$n text{ is divisible by } 2
    iff n^k text{ is divisible by }2$$



    Now we need to show that:



    $$n text{ is divisible by } 2
    iff n^{k+1} text{ is divisible by }2$$



    We would prove this if we can prove the following two things:



    $$n text{ is divisible by } 2
    Rightarrow n^{k+1} text{ is divisible by }2$$



    and



    $$n text{ is not divisible by } 2
    Rightarrow n^{k+1} text{ is not divisible by }2$$



    The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n cdot n^k$ is even as well



    Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n cdot n^k$ is odd as well, and thus not divisible by $2$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:



      $forall n in mathbb{Z}, forall p in mathbb{N}^+: n$ is divisible by $2
      iff n^p$       is divisible by $2$



      This we can prove by induction on $p$:



      Base:



      $p=1$.



      We have to show that:



      $$n text{ is divisible by } 2
      iff n^1 text{ is divisible by }2$$



      Well, that is trivially true, since $n^1=n$



      Step:



      Thye inductive hypothesis is that for some $k in mathbb{N}^+$, we have that:



      $$n text{ is divisible by } 2
      iff n^k text{ is divisible by }2$$



      Now we need to show that:



      $$n text{ is divisible by } 2
      iff n^{k+1} text{ is divisible by }2$$



      We would prove this if we can prove the following two things:



      $$n text{ is divisible by } 2
      Rightarrow n^{k+1} text{ is divisible by }2$$



      and



      $$n text{ is not divisible by } 2
      Rightarrow n^{k+1} text{ is not divisible by }2$$



      The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n cdot n^k$ is even as well



      Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n cdot n^k$ is odd as well, and thus not divisible by $2$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:



        $forall n in mathbb{Z}, forall p in mathbb{N}^+: n$ is divisible by $2
        iff n^p$         is divisible by $2$



        This we can prove by induction on $p$:



        Base:



        $p=1$.



        We have to show that:



        $$n text{ is divisible by } 2
        iff n^1 text{ is divisible by }2$$



        Well, that is trivially true, since $n^1=n$



        Step:



        Thye inductive hypothesis is that for some $k in mathbb{N}^+$, we have that:



        $$n text{ is divisible by } 2
        iff n^k text{ is divisible by }2$$



        Now we need to show that:



        $$n text{ is divisible by } 2
        iff n^{k+1} text{ is divisible by }2$$



        We would prove this if we can prove the following two things:



        $$n text{ is divisible by } 2
        Rightarrow n^{k+1} text{ is divisible by }2$$



        and



        $$n text{ is not divisible by } 2
        Rightarrow n^{k+1} text{ is not divisible by }2$$



        The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n cdot n^k$ is even as well



        Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n cdot n^k$ is odd as well, and thus not divisible by $2$






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        $endgroup$



        One approach is to prove a more general result, of which your theorem is just a specific instance. That is, let's prove that in general:



        $forall n in mathbb{Z}, forall p in mathbb{N}^+: n$ is divisible by $2
        iff n^p$         is divisible by $2$



        This we can prove by induction on $p$:



        Base:



        $p=1$.



        We have to show that:



        $$n text{ is divisible by } 2
        iff n^1 text{ is divisible by }2$$



        Well, that is trivially true, since $n^1=n$



        Step:



        Thye inductive hypothesis is that for some $k in mathbb{N}^+$, we have that:



        $$n text{ is divisible by } 2
        iff n^k text{ is divisible by }2$$



        Now we need to show that:



        $$n text{ is divisible by } 2
        iff n^{k+1} text{ is divisible by }2$$



        We would prove this if we can prove the following two things:



        $$n text{ is divisible by } 2
        Rightarrow n^{k+1} text{ is divisible by }2$$



        and



        $$n text{ is not divisible by } 2
        Rightarrow n^{k+1} text{ is not divisible by }2$$



        The first half: If $n$ is divisible by $2$, then by inductive hypothesis $n^k$ is divisible by $2$ as well, and hence $n$ and $n^k$ are both even. But the product of two even numbers is even, and hence $n^{k+1}=n cdot n^k$ is even as well



        Second half: If $n$ is not divisible by $2$, then by inductive hypothesis $n^k$ is not divisible by $2$ either. Hence, they are both odd. But the product of two odd numbers is odd, and hence $n^{k+1}=n cdot n^k$ is odd as well, and thus not divisible by $2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 21:21









        Bram28Bram28

        61.8k44793




        61.8k44793






























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