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Is the statement $(∀x ∈ Q)(∃y ∈ Bbb Q)(x cdot y ∈ Bbb Z)$ true?


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0












$begingroup$


If $x=3/2$ and $y= 2/3$ this is true, but if, for example, $x=7/2$, this is false $(21/4 ∉ Z)$. So this predicate sentence is not correct.



Is this method of proof good?






discrete-mathematics predicate-logic







share|cite|improve this question









New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$












  • $begingroup$
    Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
    $endgroup$
    – Bernard
    yesterday












  • $begingroup$
    But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
    $endgroup$
    – Code xD
    yesterday






  • 1




    $begingroup$
    Yes, of course. So what?
    $endgroup$
    – Bernard
    yesterday










  • $begingroup$
    Yes, and... therefore...
    $endgroup$
    – Graham Kemp
    yesterday










  • $begingroup$
    My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
    $endgroup$
    – Code xD
    yesterday
















0












$begingroup$


If $x=3/2$ and $y= 2/3$ this is true, but if, for example, $x=7/2$, this is false $(21/4 ∉ Z)$. So this predicate sentence is not correct.



Is this method of proof good?






discrete-mathematics predicate-logic







share|cite|improve this question









New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
    $endgroup$
    – Bernard
    yesterday












  • $begingroup$
    But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
    $endgroup$
    – Code xD
    yesterday






  • 1




    $begingroup$
    Yes, of course. So what?
    $endgroup$
    – Bernard
    yesterday










  • $begingroup$
    Yes, and... therefore...
    $endgroup$
    – Graham Kemp
    yesterday










  • $begingroup$
    My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
    $endgroup$
    – Code xD
    yesterday














0












0








0





$begingroup$


If $x=3/2$ and $y= 2/3$ this is true, but if, for example, $x=7/2$, this is false $(21/4 ∉ Z)$. So this predicate sentence is not correct.



Is this method of proof good?






discrete-mathematics predicate-logic







share|cite|improve this question









New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




If $x=3/2$ and $y= 2/3$ this is true, but if, for example, $x=7/2$, this is false $(21/4 ∉ Z)$. So this predicate sentence is not correct.



Is this method of proof good?






discrete-mathematics predicate-logic




discrete-mathematics predicate-logic






share|cite|improve this question









New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Eevee Trainer

7,61621338




7,61621338






New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Code xDCode xD

33




33




New contributor




Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Code xD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
    $endgroup$
    – Bernard
    yesterday












  • $begingroup$
    But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
    $endgroup$
    – Code xD
    yesterday






  • 1




    $begingroup$
    Yes, of course. So what?
    $endgroup$
    – Bernard
    yesterday










  • $begingroup$
    Yes, and... therefore...
    $endgroup$
    – Graham Kemp
    yesterday










  • $begingroup$
    My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
    $endgroup$
    – Code xD
    yesterday


















  • $begingroup$
    Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
    $endgroup$
    – Bernard
    yesterday












  • $begingroup$
    But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
    $endgroup$
    – Code xD
    yesterday






  • 1




    $begingroup$
    Yes, of course. So what?
    $endgroup$
    – Bernard
    yesterday










  • $begingroup$
    Yes, and... therefore...
    $endgroup$
    – Graham Kemp
    yesterday










  • $begingroup$
    My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
    $endgroup$
    – Code xD
    yesterday
















$begingroup$
Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
$endgroup$
– Bernard
yesterday






$begingroup$
Your ‘proof’ is false. For $x=7/2$, just take $y=2/7$.
$endgroup$
– Bernard
yesterday














$begingroup$
But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
$endgroup$
– Code xD
yesterday




$begingroup$
But x * y = 7/2 * 2/7 = 1 and 1 ∈ Z, which is true
$endgroup$
– Code xD
yesterday




1




1




$begingroup$
Yes, of course. So what?
$endgroup$
– Bernard
yesterday




$begingroup$
Yes, of course. So what?
$endgroup$
– Bernard
yesterday












$begingroup$
Yes, and... therefore...
$endgroup$
– Graham Kemp
yesterday




$begingroup$
Yes, and... therefore...
$endgroup$
– Graham Kemp
yesterday












$begingroup$
My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
$endgroup$
– Code xD
yesterday




$begingroup$
My mistake, but if we take example x=8/2 and y= 2/7, then 16/14 ∉ Z . So this predicate sentence is not correct.
$endgroup$
– Code xD
yesterday










1 Answer
1






active

oldest

votes


















1












$begingroup$

The important thing about statements like $(forall x in mathbb{Q}) (exists y in mathbb{Q}): x times y in mathbb{Z}$ is that you can literally read them from left to right in order to understand what they say and how to prove/disprove them.




"For all $x$ in the rationals..." - so, we need to start with some arbitrary fraction $x$ which we have no control over - "... there exists $y$ in the rationals..." - next we get to pick any $y$ we choose (which may depend on $x$: we already have that available since it came earlier in the statement) - "... such that $x$ times $y$ is an integer." That is, finally we test the arbitrary $x$ we were given and the specific $y$ we chose to see if their product is an integer. If, for any possible $x$, we can pick a $y$ that makes it true, then the statement is true.




With that framework for the proof, can you show that the statement is true?






share|cite|improve this answer









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    active

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    1












    $begingroup$

    The important thing about statements like $(forall x in mathbb{Q}) (exists y in mathbb{Q}): x times y in mathbb{Z}$ is that you can literally read them from left to right in order to understand what they say and how to prove/disprove them.




    "For all $x$ in the rationals..." - so, we need to start with some arbitrary fraction $x$ which we have no control over - "... there exists $y$ in the rationals..." - next we get to pick any $y$ we choose (which may depend on $x$: we already have that available since it came earlier in the statement) - "... such that $x$ times $y$ is an integer." That is, finally we test the arbitrary $x$ we were given and the specific $y$ we chose to see if their product is an integer. If, for any possible $x$, we can pick a $y$ that makes it true, then the statement is true.




    With that framework for the proof, can you show that the statement is true?






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The important thing about statements like $(forall x in mathbb{Q}) (exists y in mathbb{Q}): x times y in mathbb{Z}$ is that you can literally read them from left to right in order to understand what they say and how to prove/disprove them.




      "For all $x$ in the rationals..." - so, we need to start with some arbitrary fraction $x$ which we have no control over - "... there exists $y$ in the rationals..." - next we get to pick any $y$ we choose (which may depend on $x$: we already have that available since it came earlier in the statement) - "... such that $x$ times $y$ is an integer." That is, finally we test the arbitrary $x$ we were given and the specific $y$ we chose to see if their product is an integer. If, for any possible $x$, we can pick a $y$ that makes it true, then the statement is true.




      With that framework for the proof, can you show that the statement is true?






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The important thing about statements like $(forall x in mathbb{Q}) (exists y in mathbb{Q}): x times y in mathbb{Z}$ is that you can literally read them from left to right in order to understand what they say and how to prove/disprove them.




        "For all $x$ in the rationals..." - so, we need to start with some arbitrary fraction $x$ which we have no control over - "... there exists $y$ in the rationals..." - next we get to pick any $y$ we choose (which may depend on $x$: we already have that available since it came earlier in the statement) - "... such that $x$ times $y$ is an integer." That is, finally we test the arbitrary $x$ we were given and the specific $y$ we chose to see if their product is an integer. If, for any possible $x$, we can pick a $y$ that makes it true, then the statement is true.




        With that framework for the proof, can you show that the statement is true?






        share|cite|improve this answer









        $endgroup$



        The important thing about statements like $(forall x in mathbb{Q}) (exists y in mathbb{Q}): x times y in mathbb{Z}$ is that you can literally read them from left to right in order to understand what they say and how to prove/disprove them.




        "For all $x$ in the rationals..." - so, we need to start with some arbitrary fraction $x$ which we have no control over - "... there exists $y$ in the rationals..." - next we get to pick any $y$ we choose (which may depend on $x$: we already have that available since it came earlier in the statement) - "... such that $x$ times $y$ is an integer." That is, finally we test the arbitrary $x$ we were given and the specific $y$ we chose to see if their product is an integer. If, for any possible $x$, we can pick a $y$ that makes it true, then the statement is true.




        With that framework for the proof, can you show that the statement is true?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        ChessanatorChessanator

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