Prove at least one of the length, width, height of the box must be rational.












1












$begingroup$


Assume there is a big box be combined by finite small boxes and that the small boxes are not necessarily be the same.



The statement in my note is " If there is at least one of
length, width, height of each small box is rational, then at least one of the length, width, height of the big box must be rational. "



If all small boxes are the same, then there is nothing be proved.



I wonder this question can be proved by contradiction. That is, assume all of the length, width, height of the box are irrational.
But how to prove there is at least one of the small box having irrational length, width, and height？






geometry rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
    $endgroup$
    – Vasya
    Dec 19 '18 at 14:29










  • $begingroup$
    In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:33










  • $begingroup$
    here is an article containing $14$ proofs in the planar case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:34
















1












$begingroup$


Assume there is a big box be combined by finite small boxes and that the small boxes are not necessarily be the same.



The statement in my note is " If there is at least one of
length, width, height of each small box is rational, then at least one of the length, width, height of the big box must be rational. "



If all small boxes are the same, then there is nothing be proved.



I wonder this question can be proved by contradiction. That is, assume all of the length, width, height of the box are irrational.
But how to prove there is at least one of the small box having irrational length, width, and height？






geometry rational-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
    $endgroup$
    – Vasya
    Dec 19 '18 at 14:29










  • $begingroup$
    In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:33










  • $begingroup$
    here is an article containing $14$ proofs in the planar case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:34














1












1








1





$begingroup$


Assume there is a big box be combined by finite small boxes and that the small boxes are not necessarily be the same.



The statement in my note is " If there is at least one of
length, width, height of each small box is rational, then at least one of the length, width, height of the big box must be rational. "



If all small boxes are the same, then there is nothing be proved.



I wonder this question can be proved by contradiction. That is, assume all of the length, width, height of the box are irrational.
But how to prove there is at least one of the small box having irrational length, width, and height？






geometry rational-numbers







share|cite|improve this question











$endgroup$




Assume there is a big box be combined by finite small boxes and that the small boxes are not necessarily be the same.



The statement in my note is " If there is at least one of
length, width, height of each small box is rational, then at least one of the length, width, height of the big box must be rational. "



If all small boxes are the same, then there is nothing be proved.



I wonder this question can be proved by contradiction. That is, assume all of the length, width, height of the box are irrational.
But how to prove there is at least one of the small box having irrational length, width, and height？






geometry rational-numbers




geometry rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 14:35

























asked Dec 19 '18 at 14:10







user627221



















  • $begingroup$
    sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
    $endgroup$
    – Vasya
    Dec 19 '18 at 14:29










  • $begingroup$
    In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:33










  • $begingroup$
    here is an article containing $14$ proofs in the planar case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:34


















  • $begingroup$
    sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
    $endgroup$
    – Vasya
    Dec 19 '18 at 14:29










  • $begingroup$
    In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:33










  • $begingroup$
    here is an article containing $14$ proofs in the planar case.
    $endgroup$
    – lulu
    Dec 19 '18 at 14:34
















$begingroup$
sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
$endgroup$
– Vasya
Dec 19 '18 at 14:29




$begingroup$
sum of two positive irrational numbers cannot be a negative rational number to offset one rational dimension of a small box
$endgroup$
– Vasya
Dec 19 '18 at 14:29












$begingroup$
In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
$endgroup$
– lulu
Dec 19 '18 at 14:33




$begingroup$
In two dimensions the problem is well known. I suggest trying to adapt the arguments in that case.
$endgroup$
– lulu
Dec 19 '18 at 14:33












$begingroup$
here is an article containing $14$ proofs in the planar case.
$endgroup$
– lulu
Dec 19 '18 at 14:34




$begingroup$
here is an article containing $14$ proofs in the planar case.
$endgroup$
– lulu
Dec 19 '18 at 14:34










1 Answer
1






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oldest

votes


















1












$begingroup$

The key is to notice that an arbitrary box $B$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $q$ such that $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=0tag{1}$$



This equivalence is not difficult to prove. Indeed, if $B$ has one side of rational length $frac p q$ (say, along the $x$ axis), then that triple integral is the product of three integrals. One of these integrals is $$int_0^{frac p q}sin(2pi qx)dx=0$$
Conversely, suppose the triple integral is $0$, then at least one of the single integrals that forms its product must be $0$. Assume it's true for the one along the $x$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $sin(2pi qx)$, so it's a rational number.



With that equivalence proven, Let $B$ be the large box, and ${B_i}_{1leq i leq n}$ be the small boxes.
So all we have to do is to show that $B$ satisfies property $(1)$ for some integer $q$.



We know that each box $B_i$ has at least one dimension that's a rational number $frac {p_i}{q_i}$.



Let $$q=prod_{1leq i leq n}q_i$$ We can prove that property $(1)$ holds for that value of $q$.



The triple integral over $B$ can be decomposed as the sum of integrals over the small boxes
$$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=sum_i
intintint_{B_i} sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz$$



Since property $(1)$ is true with $q$ for each small box, each integral in the sum is $0$. This implies the integral over $B$ is also $0$, which implies the desired property.






share|cite|improve this answer









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    1 Answer
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    $begingroup$

    The key is to notice that an arbitrary box $B$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $q$ such that $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=0tag{1}$$



    This equivalence is not difficult to prove. Indeed, if $B$ has one side of rational length $frac p q$ (say, along the $x$ axis), then that triple integral is the product of three integrals. One of these integrals is $$int_0^{frac p q}sin(2pi qx)dx=0$$
    Conversely, suppose the triple integral is $0$, then at least one of the single integrals that forms its product must be $0$. Assume it's true for the one along the $x$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $sin(2pi qx)$, so it's a rational number.



    With that equivalence proven, Let $B$ be the large box, and ${B_i}_{1leq i leq n}$ be the small boxes.
    So all we have to do is to show that $B$ satisfies property $(1)$ for some integer $q$.



    We know that each box $B_i$ has at least one dimension that's a rational number $frac {p_i}{q_i}$.



    Let $$q=prod_{1leq i leq n}q_i$$ We can prove that property $(1)$ holds for that value of $q$.



    The triple integral over $B$ can be decomposed as the sum of integrals over the small boxes
    $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=sum_i
    intintint_{B_i} sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz$$



    Since property $(1)$ is true with $q$ for each small box, each integral in the sum is $0$. This implies the integral over $B$ is also $0$, which implies the desired property.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The key is to notice that an arbitrary box $B$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $q$ such that $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=0tag{1}$$



      This equivalence is not difficult to prove. Indeed, if $B$ has one side of rational length $frac p q$ (say, along the $x$ axis), then that triple integral is the product of three integrals. One of these integrals is $$int_0^{frac p q}sin(2pi qx)dx=0$$
      Conversely, suppose the triple integral is $0$, then at least one of the single integrals that forms its product must be $0$. Assume it's true for the one along the $x$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $sin(2pi qx)$, so it's a rational number.



      With that equivalence proven, Let $B$ be the large box, and ${B_i}_{1leq i leq n}$ be the small boxes.
      So all we have to do is to show that $B$ satisfies property $(1)$ for some integer $q$.



      We know that each box $B_i$ has at least one dimension that's a rational number $frac {p_i}{q_i}$.



      Let $$q=prod_{1leq i leq n}q_i$$ We can prove that property $(1)$ holds for that value of $q$.



      The triple integral over $B$ can be decomposed as the sum of integrals over the small boxes
      $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=sum_i
      intintint_{B_i} sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz$$



      Since property $(1)$ is true with $q$ for each small box, each integral in the sum is $0$. This implies the integral over $B$ is also $0$, which implies the desired property.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The key is to notice that an arbitrary box $B$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $q$ such that $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=0tag{1}$$



        This equivalence is not difficult to prove. Indeed, if $B$ has one side of rational length $frac p q$ (say, along the $x$ axis), then that triple integral is the product of three integrals. One of these integrals is $$int_0^{frac p q}sin(2pi qx)dx=0$$
        Conversely, suppose the triple integral is $0$, then at least one of the single integrals that forms its product must be $0$. Assume it's true for the one along the $x$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $sin(2pi qx)$, so it's a rational number.



        With that equivalence proven, Let $B$ be the large box, and ${B_i}_{1leq i leq n}$ be the small boxes.
        So all we have to do is to show that $B$ satisfies property $(1)$ for some integer $q$.



        We know that each box $B_i$ has at least one dimension that's a rational number $frac {p_i}{q_i}$.



        Let $$q=prod_{1leq i leq n}q_i$$ We can prove that property $(1)$ holds for that value of $q$.



        The triple integral over $B$ can be decomposed as the sum of integrals over the small boxes
        $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=sum_i
        intintint_{B_i} sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz$$



        Since property $(1)$ is true with $q$ for each small box, each integral in the sum is $0$. This implies the integral over $B$ is also $0$, which implies the desired property.






        share|cite|improve this answer









        $endgroup$



        The key is to notice that an arbitrary box $B$ aligned with the x-y-z axes in space has the property that one of its dimensions is rational if and only if there exists an integer $q$ such that $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=0tag{1}$$



        This equivalence is not difficult to prove. Indeed, if $B$ has one side of rational length $frac p q$ (say, along the $x$ axis), then that triple integral is the product of three integrals. One of these integrals is $$int_0^{frac p q}sin(2pi qx)dx=0$$
        Conversely, suppose the triple integral is $0$, then at least one of the single integrals that forms its product must be $0$. Assume it's true for the one along the $x$ axis. This implies that the domain is integration must be an interval whose length is a multiple of the period of $sin(2pi qx)$, so it's a rational number.



        With that equivalence proven, Let $B$ be the large box, and ${B_i}_{1leq i leq n}$ be the small boxes.
        So all we have to do is to show that $B$ satisfies property $(1)$ for some integer $q$.



        We know that each box $B_i$ has at least one dimension that's a rational number $frac {p_i}{q_i}$.



        Let $$q=prod_{1leq i leq n}q_i$$ We can prove that property $(1)$ holds for that value of $q$.



        The triple integral over $B$ can be decomposed as the sum of integrals over the small boxes
        $$intintint_B sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz=sum_i
        intintint_{B_i} sin(2pi qx)sin(2pi qy)sin(2pi qz)dxdydz$$



        Since property $(1)$ is true with $q$ for each small box, each integral in the sum is $0$. This implies the integral over $B$ is also $0$, which implies the desired property.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 14:47









        Stefan LafonStefan Lafon

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