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







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












$begingroup$



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

63.1k44793




63.1k44793










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 $2$






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036299%2ffor-all-integers-of-n-n-is-divisible-by-2-iff-n4-is-divisible-by-2%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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
















    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

    4,2091641




    4,2091641












    • $begingroup$
      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 $2$






        share|cite|improve this answer









        $endgroup$



        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

        63.1k44793




        63.1k44793






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036299%2ffor-all-integers-of-n-n-is-divisible-by-2-iff-n4-is-divisible-by-2%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more