Prove the following: Let $n$ be an integer. If $2|n^2$, then $4|n^2$.












0












$begingroup$


I came up with the following:



$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$.
So, $n^2|(2k)^2$,
$n^2 = 4k^2$, where $k^2 ∈ Z$.



So, if $2|n^2$, then $4|n^2$ as desired.



What do you all think?






proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 4:24










  • $begingroup$
    Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
    $endgroup$
    – Dave
    Dec 6 '18 at 4:24










  • $begingroup$
    See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:31










  • $begingroup$
    Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
    $endgroup$
    – user334732
    Dec 6 '18 at 18:57












  • $begingroup$
    Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
    $endgroup$
    – user334732
    Dec 6 '18 at 19:46
















0












$begingroup$


I came up with the following:



$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$.
So, $n^2|(2k)^2$,
$n^2 = 4k^2$, where $k^2 ∈ Z$.



So, if $2|n^2$, then $4|n^2$ as desired.



What do you all think?






proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 4:24










  • $begingroup$
    Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
    $endgroup$
    – Dave
    Dec 6 '18 at 4:24










  • $begingroup$
    See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:31










  • $begingroup$
    Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
    $endgroup$
    – user334732
    Dec 6 '18 at 18:57












  • $begingroup$
    Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
    $endgroup$
    – user334732
    Dec 6 '18 at 19:46














0












0








0


0



$begingroup$


I came up with the following:



$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$.
So, $n^2|(2k)^2$,
$n^2 = 4k^2$, where $k^2 ∈ Z$.



So, if $2|n^2$, then $4|n^2$ as desired.



What do you all think?






proof-writing







share|cite|improve this question











$endgroup$




I came up with the following:



$2|n^2$ implies that $2|n*n$. We proved in class that if $q|b*p$, then $q|b$ or $q|p$. Therefore, if $2|n^2$, then $2|n$ or $2|n$.
So, $2|n$ implies $n=2k$, for some $k ∈ Z$.
So, $n^2|(2k)^2$,
$n^2 = 4k^2$, where $k^2 ∈ Z$.



So, if $2|n^2$, then $4|n^2$ as desired.



What do you all think?






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 8 '18 at 1:22







Mettal

















asked Dec 6 '18 at 4:21









MettalMettal

257




257












  • $begingroup$
    Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 4:24










  • $begingroup$
    Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
    $endgroup$
    – Dave
    Dec 6 '18 at 4:24










  • $begingroup$
    See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:31










  • $begingroup$
    Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
    $endgroup$
    – user334732
    Dec 6 '18 at 18:57












  • $begingroup$
    Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
    $endgroup$
    – user334732
    Dec 6 '18 at 19:46


















  • $begingroup$
    Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
    $endgroup$
    – saulspatz
    Dec 6 '18 at 4:24










  • $begingroup$
    Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
    $endgroup$
    – Dave
    Dec 6 '18 at 4:24










  • $begingroup$
    See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:31










  • $begingroup$
    Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
    $endgroup$
    – user334732
    Dec 6 '18 at 18:57












  • $begingroup$
    Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
    $endgroup$
    – user334732
    Dec 6 '18 at 19:46
















$begingroup$
Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
$endgroup$
– saulspatz
Dec 6 '18 at 4:24




$begingroup$
Well if $n$ is odd, then $2nmid n^2$ so you're done with that case.
$endgroup$
– saulspatz
Dec 6 '18 at 4:24












$begingroup$
Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
$endgroup$
– Dave
Dec 6 '18 at 4:24




$begingroup$
Should it be "if $2mid color{red}{n}$ then $4mid n^2$"?
$endgroup$
– Dave
Dec 6 '18 at 4:24












$begingroup$
See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
$endgroup$
– tarit goswami
Dec 6 '18 at 4:31




$begingroup$
See the statement is - if $2|n^2$ then $4|n^2$. You don't need to think about odd $n$.
$endgroup$
– tarit goswami
Dec 6 '18 at 4:31












$begingroup$
Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
$endgroup$
– user334732
Dec 6 '18 at 18:57






$begingroup$
Hi and welcome to MSE. Have you done prime factorisation? Assuming so, what are the properties of the exponents of the prime factors in the every factorisation of a square number?
$endgroup$
– user334732
Dec 6 '18 at 18:57














$begingroup$
Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
$endgroup$
– user334732
Dec 6 '18 at 19:46




$begingroup$
Try adding the answer to this to your question: Is this a problem from a text, or from a class? If from a class, which one, and what theorems have they seen recently? If a text, which one? For example, are they in a context of abstract algebra, or number theory, or intro to proofs, or something else? See this discussion: chat.stackexchange.com/transcript/message/47930322#47930322
$endgroup$
– user334732
Dec 6 '18 at 19:46










4 Answers
4






active

oldest

votes


















0












$begingroup$

Suppose not. Then $n^2=2times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.




Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2implies 4|n^2$ is $4nmid n^2implies 2nmid n$. So, if we can prove the contrapositive statement is true then we are done.



$$4nmid n^2implies 2nmid nimplies 2nmid ncdot n $$
So we are sone!






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there any problem to understand this?
    $endgroup$
    – tarit goswami
    Dec 6 '18 at 4:56










  • $begingroup$
    No. I think I got it. Thank you!
    $endgroup$
    – Mettal
    Dec 6 '18 at 4:58



















0












$begingroup$

If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $Bbb Z$). So $n=2m $ for some $m in Bbb Z$. Hence $n^2=4m^2$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.






    share|cite|improve this answer











    $endgroup$





















      -1












      $begingroup$

      An import ant part of writing any proof is to determine from the outset what you can assume to be true.



      If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}cdot p_2^{i_2}cdot p_3^{i_3}ldots$



      Then the prime factorisation of $n^2=p_1^{2i_2}cdot p_2^{2i_2}cdot p_3^{2i_3}ldots$



      Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.



      (In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).






      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%2f3028028%2fprove-the-following-let-n-be-an-integer-if-2n2-then-4n2%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        Suppose not. Then $n^2=2times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.




        Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2implies 4|n^2$ is $4nmid n^2implies 2nmid n$. So, if we can prove the contrapositive statement is true then we are done.



        $$4nmid n^2implies 2nmid nimplies 2nmid ncdot n $$
        So we are sone!






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Is there any problem to understand this?
          $endgroup$
          – tarit goswami
          Dec 6 '18 at 4:56










        • $begingroup$
          No. I think I got it. Thank you!
          $endgroup$
          – Mettal
          Dec 6 '18 at 4:58
















        0












        $begingroup$

        Suppose not. Then $n^2=2times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.




        Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2implies 4|n^2$ is $4nmid n^2implies 2nmid n$. So, if we can prove the contrapositive statement is true then we are done.



        $$4nmid n^2implies 2nmid nimplies 2nmid ncdot n $$
        So we are sone!






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Is there any problem to understand this?
          $endgroup$
          – tarit goswami
          Dec 6 '18 at 4:56










        • $begingroup$
          No. I think I got it. Thank you!
          $endgroup$
          – Mettal
          Dec 6 '18 at 4:58














        0












        0








        0





        $begingroup$

        Suppose not. Then $n^2=2times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.




        Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2implies 4|n^2$ is $4nmid n^2implies 2nmid n$. So, if we can prove the contrapositive statement is true then we are done.



        $$4nmid n^2implies 2nmid nimplies 2nmid ncdot n $$
        So we are sone!






        share|cite|improve this answer











        $endgroup$



        Suppose not. Then $n^2=2times (2m+1)$ for some positive integer $m$. But that means, $n^2=4m+2$. Which is a contradiction, as, any perfect square can leave remainder $0,1$ after division by $4$.




        Here is a more satisfactory arument. A statement and it's contrapositive statement has same truth value. Here, the contrapositive statement of $2|n^2implies 4|n^2$ is $4nmid n^2implies 2nmid n$. So, if we can prove the contrapositive statement is true then we are done.



        $$4nmid n^2implies 2nmid nimplies 2nmid ncdot n $$
        So we are sone!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 4:48

























        answered Dec 6 '18 at 4:35









        tarit goswamitarit goswami

        1,7591421




        1,7591421












        • $begingroup$
          Is there any problem to understand this?
          $endgroup$
          – tarit goswami
          Dec 6 '18 at 4:56










        • $begingroup$
          No. I think I got it. Thank you!
          $endgroup$
          – Mettal
          Dec 6 '18 at 4:58


















        • $begingroup$
          Is there any problem to understand this?
          $endgroup$
          – tarit goswami
          Dec 6 '18 at 4:56










        • $begingroup$
          No. I think I got it. Thank you!
          $endgroup$
          – Mettal
          Dec 6 '18 at 4:58
















        $begingroup$
        Is there any problem to understand this?
        $endgroup$
        – tarit goswami
        Dec 6 '18 at 4:56




        $begingroup$
        Is there any problem to understand this?
        $endgroup$
        – tarit goswami
        Dec 6 '18 at 4:56












        $begingroup$
        No. I think I got it. Thank you!
        $endgroup$
        – Mettal
        Dec 6 '18 at 4:58




        $begingroup$
        No. I think I got it. Thank you!
        $endgroup$
        – Mettal
        Dec 6 '18 at 4:58











        0












        $begingroup$

        If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $Bbb Z$). So $n=2m $ for some $m in Bbb Z$. Hence $n^2=4m^2$.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $Bbb Z$). So $n=2m $ for some $m in Bbb Z$. Hence $n^2=4m^2$.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $Bbb Z$). So $n=2m $ for some $m in Bbb Z$. Hence $n^2=4m^2$.






            share|cite|improve this answer









            $endgroup$



            If $2|n^2$, $n^2$ is even. Note that $n^2$ is even if and only if $n$ is even (in $Bbb Z$). So $n=2m $ for some $m in Bbb Z$. Hence $n^2=4m^2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 4:28









            Thomas ShelbyThomas Shelby

            2,525321




            2,525321























                0












                $begingroup$

                If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.






                share|cite|improve this answer











                $endgroup$


















                  0












                  $begingroup$

                  If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.






                  share|cite|improve this answer











                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.






                    share|cite|improve this answer











                    $endgroup$



                    If $2|m^2$ then it means its square root $m$ must be even, or else $m$ would be irrational. If it were odd then there would be no $2$ as part of its prime factors, so multiplying it by itself would not create an even number. We can define $n$ such that $2n=m$. Since $m^2=(2n)^2=4n^2$, then we can easily show $4|4n^2$ and therefore $4|m^2$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 6 '18 at 4:45

























                    answered Dec 6 '18 at 4:38









                    KykyKyky

                    450213




                    450213























                        -1












                        $begingroup$

                        An import ant part of writing any proof is to determine from the outset what you can assume to be true.



                        If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}cdot p_2^{i_2}cdot p_3^{i_3}ldots$



                        Then the prime factorisation of $n^2=p_1^{2i_2}cdot p_2^{2i_2}cdot p_3^{2i_3}ldots$



                        Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.



                        (In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).






                        share|cite|improve this answer









                        $endgroup$


















                          -1












                          $begingroup$

                          An import ant part of writing any proof is to determine from the outset what you can assume to be true.



                          If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}cdot p_2^{i_2}cdot p_3^{i_3}ldots$



                          Then the prime factorisation of $n^2=p_1^{2i_2}cdot p_2^{2i_2}cdot p_3^{2i_3}ldots$



                          Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.



                          (In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).






                          share|cite|improve this answer









                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            An import ant part of writing any proof is to determine from the outset what you can assume to be true.



                            If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}cdot p_2^{i_2}cdot p_3^{i_3}ldots$



                            Then the prime factorisation of $n^2=p_1^{2i_2}cdot p_2^{2i_2}cdot p_3^{2i_3}ldots$



                            Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.



                            (In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).






                            share|cite|improve this answer









                            $endgroup$



                            An import ant part of writing any proof is to determine from the outset what you can assume to be true.



                            If you can assume the fundamental theorem of arithmetic then you have the unique prime factorisation of $n=p_1^{i_2}cdot p_2^{i_2}cdot p_3^{i_3}ldots$



                            Then the prime factorisation of $n^2=p_1^{2i_2}cdot p_2^{2i_2}cdot p_3^{2i_3}ldots$



                            Then since every exponent is even, the power of every prime factor of $n^2$ is either $1$ or a power of $p^2$, which generalises the result you need to every prime.



                            (In $2^3=8$, 2 is the prime, 3 is the exponent and 8 is the power).







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 7 '18 at 10:37









                            user334732user334732

                            4,26311240




                            4,26311240






























                                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%2f3028028%2fprove-the-following-let-n-be-an-integer-if-2n2-then-4n2%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

                                Probability when a professor distributes a quiz and homework assignment to a class of n students.

                                Image
                                0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
                                Read more

                                Aardman Animations

                                Image
                                Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
                                Read more

                                Are they similar matrix

                                Image
                                0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
                                Read more