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





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        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

            2,765421




            2,765421























                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

                    462213




                    462213























                        -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).






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                            $endgroup$



                            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

                            4,25911240




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