Is this a typo in Section 1.8.1 Mathematics for Computer Science?












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enter image description here



Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?



This is from MIT's Mathematics for Computer Science






proof-explanation proof-theory







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  • $begingroup$
    Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
    $endgroup$
    – DanielWainfleet
    4 hours ago
















3












$begingroup$


enter image description here



Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?



This is from MIT's Mathematics for Computer Science






proof-explanation proof-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
    $endgroup$
    – DanielWainfleet
    4 hours ago














3












3








3





$begingroup$


enter image description here



Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?



This is from MIT's Mathematics for Computer Science






proof-explanation proof-theory







share|cite|improve this question









$endgroup$




enter image description here



Am I completely mistake or is it suppose to say $n^2$ is a multiple of 2 and therefore $n$ must be a multiple of 4?



This is from MIT's Mathematics for Computer Science






proof-explanation proof-theory




proof-explanation proof-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









doctopusdoctopus

1413




1413












  • $begingroup$
    Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
    $endgroup$
    – DanielWainfleet
    4 hours ago


















  • $begingroup$
    Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
    $endgroup$
    – DanielWainfleet
    4 hours ago
















$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago




$begingroup$
Generally: If $m,n$ are positive integers and $m^{1/n}$ is not an integer then $m^{1/n}$ is irrational. Proved by a similar method.
$endgroup$
– DanielWainfleet
4 hours ago










2 Answers
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No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.






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    3












    $begingroup$

    No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
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      $begingroup$

      No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.






          share|cite|improve this answer









          $endgroup$



          No. $6^2$ is a multiple of $2$ but $6$ is not a multiple of $4$. If $n=2k$ then for sure $n^2=4k^2$. So, MIT is right.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          RandallRandall

          10.9k11431




          10.9k11431























              3












              $begingroup$

              No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.






                  share|cite|improve this answer









                  $endgroup$



                  No, it is correct. The point is that $2d^2=n^2$ implies $n^2$ is even, and only even numbers square to give an even number, so $n$ much be even, so $n^2$ is then actually a multiple of $4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 4 hours ago









                  John DoeJohn Doe

                  12.2k11341




                  12.2k11341






























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