Proof that an odd integer multiplied by $3$ and squared is always odd












4














I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:



"For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."



So far I know I can build this problem by stating the two following things:




  1. Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)

  2. Prove all odd numbers square also remain odd.


What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)



I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.



Thanks ahead of time guys!






elementary-number-theory proof-writing







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    I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:



    "For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."



    So far I know I can build this problem by stating the two following things:




    1. Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)

    2. Prove all odd numbers square also remain odd.


    What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)



    I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.



    Thanks ahead of time guys!






    elementary-number-theory proof-writing







    share|cite|improve this question



























      4












      4








      4







      I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:



      "For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."



      So far I know I can build this problem by stating the two following things:




      1. Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)

      2. Prove all odd numbers square also remain odd.


      What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)



      I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.



      Thanks ahead of time guys!






      elementary-number-theory proof-writing







      share|cite|improve this question















      I'm working with a proof in a discrete structures CS course, and I am a little confused by how to build up some logic for the argument. Currently we're working with symbolic logic, the problem statement is:



      "For all integers n, if n is an odd integer, then $ 3n^2$ is also an odd integer."



      So far I know I can build this problem by stating the two following things:




      1. Prove all odd numbers, multiplied by $3$ remain odd. (how can I go about proving this?)

      2. Prove all odd numbers square also remain odd.


      What would be the best way about proving this? Would it be simpler to prove all numbers multiplied by an odd number remain odd (since an odd number squared is still being multiplied by an odd number)



      I'm not really looking for a solution to be handed to me, I just feel a little stuck on where to go from here, and would like some direction on what to do next.



      Thanks ahead of time guys!






      elementary-number-theory proof-writing




      elementary-number-theory proof-writing






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      edited Jan 4 at 8:49









      Eevee Trainer

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      asked Nov 1 '14 at 20:31









      dustybladedustyblade

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          4 Answers
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          Consider that any odd number $x$ can be written as, for some integer $n$
          $$x=2n+1.$$
          So, if you had, for instance two odd $x_1$ and $x_2$, you could say
          $$x_1=2n_1+1$$
          $$x_2=2n_2+1.$$
          When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
          and hence show the product is odd?






          share|cite|improve this answer





























            2














            The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.






            share|cite|improve this answer





















            • It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
              – Milo Brandt
              Nov 1 '14 at 20:40










            • A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
              – Matt Samuel
              Nov 1 '14 at 20:42





















            2














            Any odd number $n$ can be written as $2p+1$.



            $3n^2$ = $3(2p+1)^2$



            =$3(4p^2+4p+1)$



            The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.



            $3$ times an odd number is odd.






            share|cite|improve this answer





























              0














              Probably the simplest proof you can make for this:




              • Let $n in mathbb{N}$ be odd.

              • Thus, $2 not mid n$.

              • Thus, $2 not mid n^2$.

              • We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.


              This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.



              (I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)






              share|cite|improve this answer





















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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

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                active

                oldest

                votes






                active

                oldest

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                3














                Consider that any odd number $x$ can be written as, for some integer $n$
                $$x=2n+1.$$
                So, if you had, for instance two odd $x_1$ and $x_2$, you could say
                $$x_1=2n_1+1$$
                $$x_2=2n_2+1.$$
                When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
                and hence show the product is odd?






                share|cite|improve this answer


























                  3














                  Consider that any odd number $x$ can be written as, for some integer $n$
                  $$x=2n+1.$$
                  So, if you had, for instance two odd $x_1$ and $x_2$, you could say
                  $$x_1=2n_1+1$$
                  $$x_2=2n_2+1.$$
                  When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
                  and hence show the product is odd?






                  share|cite|improve this answer
























                    3












                    3








                    3






                    Consider that any odd number $x$ can be written as, for some integer $n$
                    $$x=2n+1.$$
                    So, if you had, for instance two odd $x_1$ and $x_2$, you could say
                    $$x_1=2n_1+1$$
                    $$x_2=2n_2+1.$$
                    When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
                    and hence show the product is odd?






                    share|cite|improve this answer












                    Consider that any odd number $x$ can be written as, for some integer $n$
                    $$x=2n+1.$$
                    So, if you had, for instance two odd $x_1$ and $x_2$, you could say
                    $$x_1=2n_1+1$$
                    $$x_2=2n_2+1.$$
                    When you multiply these expressions, can you see how we would express the result in following form $$x_1x_2=2(text{some integer})+1$$
                    and hence show the product is odd?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 1 '14 at 20:37









                    Milo BrandtMilo Brandt

                    39.4k475139




                    39.4k475139























                        2














                        The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.






                        share|cite|improve this answer





















                        • It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                          – Milo Brandt
                          Nov 1 '14 at 20:40










                        • A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                          – Matt Samuel
                          Nov 1 '14 at 20:42


















                        2














                        The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.






                        share|cite|improve this answer





















                        • It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                          – Milo Brandt
                          Nov 1 '14 at 20:40










                        • A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                          – Matt Samuel
                          Nov 1 '14 at 20:42
















                        2












                        2








                        2






                        The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.






                        share|cite|improve this answer












                        The definition of "odd" is "not divisible by 2". You can see that multiplying an odd number by 3 and squaring preserves oddness by manipulating the prime factorization, which will still contain no powers of 2.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 1 '14 at 20:37









                        Matt SamuelMatt Samuel

                        37.4k63665




                        37.4k63665












                        • It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                          – Milo Brandt
                          Nov 1 '14 at 20:40










                        • A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                          – Matt Samuel
                          Nov 1 '14 at 20:42




















                        • It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                          – Milo Brandt
                          Nov 1 '14 at 20:40










                        • A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                          – Matt Samuel
                          Nov 1 '14 at 20:42


















                        It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                        – Milo Brandt
                        Nov 1 '14 at 20:40




                        It's worth noting that this only applies to integers; by this definition, $sqrt{2}$ is odd, meaning the product of two "odd" numbers $sqrt{2}cdot sqrt{2}=2$ is even!
                        – Milo Brandt
                        Nov 1 '14 at 20:40












                        A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                        – Matt Samuel
                        Nov 1 '14 at 20:42






                        A positive real number has a prime factorization if and only if it is an integer (at least if we assume the exponents of the primes are nonnegative integers).
                        – Matt Samuel
                        Nov 1 '14 at 20:42













                        2














                        Any odd number $n$ can be written as $2p+1$.



                        $3n^2$ = $3(2p+1)^2$



                        =$3(4p^2+4p+1)$



                        The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.



                        $3$ times an odd number is odd.






                        share|cite|improve this answer


























                          2














                          Any odd number $n$ can be written as $2p+1$.



                          $3n^2$ = $3(2p+1)^2$



                          =$3(4p^2+4p+1)$



                          The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.



                          $3$ times an odd number is odd.






                          share|cite|improve this answer
























                            2












                            2








                            2






                            Any odd number $n$ can be written as $2p+1$.



                            $3n^2$ = $3(2p+1)^2$



                            =$3(4p^2+4p+1)$



                            The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.



                            $3$ times an odd number is odd.






                            share|cite|improve this answer












                            Any odd number $n$ can be written as $2p+1$.



                            $3n^2$ = $3(2p+1)^2$



                            =$3(4p^2+4p+1)$



                            The term in parenthesis is odd, since $4p^2$ and $4p$ are both even for any value of $p$.



                            $3$ times an odd number is odd.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 1 '14 at 21:34









                            Rao A.Rao A.

                            1673




                            1673























                                0














                                Probably the simplest proof you can make for this:




                                • Let $n in mathbb{N}$ be odd.

                                • Thus, $2 not mid n$.

                                • Thus, $2 not mid n^2$.

                                • We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.


                                This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.



                                (I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)






                                share|cite|improve this answer


























                                  0














                                  Probably the simplest proof you can make for this:




                                  • Let $n in mathbb{N}$ be odd.

                                  • Thus, $2 not mid n$.

                                  • Thus, $2 not mid n^2$.

                                  • We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.


                                  This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.



                                  (I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)






                                  share|cite|improve this answer
























                                    0












                                    0








                                    0






                                    Probably the simplest proof you can make for this:




                                    • Let $n in mathbb{N}$ be odd.

                                    • Thus, $2 not mid n$.

                                    • Thus, $2 not mid n^2$.

                                    • We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.


                                    This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.



                                    (I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)






                                    share|cite|improve this answer












                                    Probably the simplest proof you can make for this:




                                    • Let $n in mathbb{N}$ be odd.

                                    • Thus, $2 not mid n$.

                                    • Thus, $2 not mid n^2$.

                                    • We also note $2 not mid 3$, and thus, $2 not mid 3n^2$.


                                    This proof essentially makes heavy use of the fact that if $d mid ab$, then $d mid a$ or $d mid b$. The contrapositive of that claim is if $d not mid a$ and $d not mid b$, then $d not mid ab$.



                                    (I recognize this post is old, but I felt like this proof is simpler and more intuitive, so hopefully people will appreciate this alternate proof as well.)







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Jan 4 at 8:48









                                    Eevee TrainerEevee Trainer

                                    5,0211734




                                    5,0211734






























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