Proof for $forall k >0 $, $xy subset y$

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I'm trying to show, $forall k >0 $, the following holds.



$$
(x,y) : subset > k
$$



I'm failing to figure out a contradiction.



Is there anyone to give a contradiction / proof for the statement?



How about when $k ge 1$?




real-analysis




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  • If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
    – fleablood
    Jul 21 at 0:44














up vote
0
down vote

favorite












I'm trying to show, $forall k >0 $, the following holds.



$$
(x,y) : subset > k
$$



I'm failing to figure out a contradiction.



Is there anyone to give a contradiction / proof for the statement?



How about when $k ge 1$?




real-analysis




share|cite|improve this question



















  • If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
    – fleablood
    Jul 21 at 0:44












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to show, $forall k >0 $, the following holds.



$$
(x,y) : subset > k
$$



I'm failing to figure out a contradiction.



Is there anyone to give a contradiction / proof for the statement?



How about when $k ge 1$?




real-analysis




share|cite|improve this question











I'm trying to show, $forall k >0 $, the following holds.



$$
(x,y) : subset > k
$$



I'm failing to figure out a contradiction.



Is there anyone to give a contradiction / proof for the statement?



How about when $k ge 1$?





real-analysis





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 21 at 0:34









moreblue

1738




1738











  • If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
    – fleablood
    Jul 21 at 0:44
















  • If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
    – fleablood
    Jul 21 at 0:44















If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
– fleablood
Jul 21 at 0:44




If $|xy| > k^2$ then $|x| > frac k^2y$. If $|y| le k$ then $|x| > frac k^2yge frac k^2k = k$. If $|y| > k$ then $|y| > k$.
– fleablood
Jul 21 at 0:44










2 Answers
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Hint: If$$(x,y), notsubset , ,$$then there would be a $(x,y)inmathbbR^2$ such that $|xy|>k^2$ and that both numbers $|x|$ and $|y|$ are smaller than or equal to $k$. Can you take it from here?






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    Just do the math. $|xy| > k^2 implies |x| > frac k^2y$ and $|y| > frac k^2$ and if $|x| le k$ then $|y| > frac k^2 ge frac k^2k = k$. And if $|y| le k$ then $|x| > frac k^2 ge frac k^2k = k$.



    Basically when ever you have $ab > M > 0$ then either $|a| > sqrt M$ or $|b| > sqrt M$.






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      2 Answers
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      Hint: If$$(x,y), notsubset , ,$$then there would be a $(x,y)inmathbbR^2$ such that $|xy|>k^2$ and that both numbers $|x|$ and $|y|$ are smaller than or equal to $k$. Can you take it from here?






      share|cite|improve this answer

























        up vote
        0
        down vote













        Hint: If$$(x,y), notsubset , ,$$then there would be a $(x,y)inmathbbR^2$ such that $|xy|>k^2$ and that both numbers $|x|$ and $|y|$ are smaller than or equal to $k$. Can you take it from here?






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          Hint: If$$(x,y), notsubset , ,$$then there would be a $(x,y)inmathbbR^2$ such that $|xy|>k^2$ and that both numbers $|x|$ and $|y|$ are smaller than or equal to $k$. Can you take it from here?






          share|cite|improve this answer













          Hint: If$$(x,y), notsubset , ,$$then there would be a $(x,y)inmathbbR^2$ such that $|xy|>k^2$ and that both numbers $|x|$ and $|y|$ are smaller than or equal to $k$. Can you take it from here?







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 21 at 0:39









          José Carlos Santos

          114k1698177




          114k1698177




















              up vote
              0
              down vote













              Just do the math. $|xy| > k^2 implies |x| > frac k^2y$ and $|y| > frac k^2$ and if $|x| le k$ then $|y| > frac k^2 ge frac k^2k = k$. And if $|y| le k$ then $|x| > frac k^2 ge frac k^2k = k$.



              Basically when ever you have $ab > M > 0$ then either $|a| > sqrt M$ or $|b| > sqrt M$.






              share|cite|improve this answer



























                up vote
                0
                down vote













                Just do the math. $|xy| > k^2 implies |x| > frac k^2y$ and $|y| > frac k^2$ and if $|x| le k$ then $|y| > frac k^2 ge frac k^2k = k$. And if $|y| le k$ then $|x| > frac k^2 ge frac k^2k = k$.



                Basically when ever you have $ab > M > 0$ then either $|a| > sqrt M$ or $|b| > sqrt M$.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Just do the math. $|xy| > k^2 implies |x| > frac k^2y$ and $|y| > frac k^2$ and if $|x| le k$ then $|y| > frac k^2 ge frac k^2k = k$. And if $|y| le k$ then $|x| > frac k^2 ge frac k^2k = k$.



                  Basically when ever you have $ab > M > 0$ then either $|a| > sqrt M$ or $|b| > sqrt M$.






                  share|cite|improve this answer















                  Just do the math. $|xy| > k^2 implies |x| > frac k^2y$ and $|y| > frac k^2$ and if $|x| le k$ then $|y| > frac k^2 ge frac k^2k = k$. And if $|y| le k$ then $|x| > frac k^2 ge frac k^2k = k$.



                  Basically when ever you have $ab > M > 0$ then either $|a| > sqrt M$ or $|b| > sqrt M$.







                  share|cite|improve this answer















                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jul 21 at 0:54


























                  answered Jul 21 at 0:46









                  fleablood

                  60.4k22575




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