Element Set Theory Problem

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I have this set problem I need to find an answer for, and the last time I did element set theory was awhile back. It goes:



Solve for $n$ where $n$ = $min (B' cap N)$



$n in mathbbN$



$x in A$



$A = a: 1 ≤ a ≤ 5, a in mathbbN$



$B = b: b = x^2$



Thanks in advance!




elementary-set-theory




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  • What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
    – Mauro ALLEGRANZA
    Aug 2 at 13:55










  • Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • What is the role of $A$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
    – Mauro ALLEGRANZA
    Aug 2 at 14:00






  • 1




    So, the conclusion is therefore n=2, as it is the lowest not square number?
    – Stevie Martin
    Aug 2 at 14:05















up vote
0
down vote

favorite












I have this set problem I need to find an answer for, and the last time I did element set theory was awhile back. It goes:



Solve for $n$ where $n$ = $min (B' cap N)$



$n in mathbbN$



$x in A$



$A = a: 1 ≤ a ≤ 5, a in mathbbN$



$B = b: b = x^2$



Thanks in advance!




elementary-set-theory




share|cite|improve this question





















  • What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
    – Mauro ALLEGRANZA
    Aug 2 at 13:55










  • Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • What is the role of $A$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
    – Mauro ALLEGRANZA
    Aug 2 at 14:00






  • 1




    So, the conclusion is therefore n=2, as it is the lowest not square number?
    – Stevie Martin
    Aug 2 at 14:05













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have this set problem I need to find an answer for, and the last time I did element set theory was awhile back. It goes:



Solve for $n$ where $n$ = $min (B' cap N)$



$n in mathbbN$



$x in A$



$A = a: 1 ≤ a ≤ 5, a in mathbbN$



$B = b: b = x^2$



Thanks in advance!




elementary-set-theory




share|cite|improve this question













I have this set problem I need to find an answer for, and the last time I did element set theory was awhile back. It goes:



Solve for $n$ where $n$ = $min (B' cap N)$



$n in mathbbN$



$x in A$



$A = a: 1 ≤ a ≤ 5, a in mathbbN$



$B = b: b = x^2$



Thanks in advance!





elementary-set-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 2 at 15:13









packetpacket

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asked Aug 2 at 13:50









Stevie Martin

32




32











  • What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
    – Mauro ALLEGRANZA
    Aug 2 at 13:55










  • Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • What is the role of $A$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
    – Mauro ALLEGRANZA
    Aug 2 at 14:00






  • 1




    So, the conclusion is therefore n=2, as it is the lowest not square number?
    – Stevie Martin
    Aug 2 at 14:05

















  • What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
    – Mauro ALLEGRANZA
    Aug 2 at 13:55










  • Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • What is the role of $A$ ?
    – Mauro ALLEGRANZA
    Aug 2 at 13:56










  • If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
    – Mauro ALLEGRANZA
    Aug 2 at 14:00






  • 1




    So, the conclusion is therefore n=2, as it is the lowest not square number?
    – Stevie Martin
    Aug 2 at 14:05
















What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
– Mauro ALLEGRANZA
Aug 2 at 13:55




What is $B$ ? Are $b$ and $x$ naturals ? that means that $b$ is a "square" number : $4,9$ etc
– Mauro ALLEGRANZA
Aug 2 at 13:55












Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
– Mauro ALLEGRANZA
Aug 2 at 13:56




Is $B'$ the complement (with respect to $mathbb N$) of $B$ ?
– Mauro ALLEGRANZA
Aug 2 at 13:56












What is the role of $A$ ?
– Mauro ALLEGRANZA
Aug 2 at 13:56




What is the role of $A$ ?
– Mauro ALLEGRANZA
Aug 2 at 13:56












If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
– Mauro ALLEGRANZA
Aug 2 at 14:00




If so, $n$ will be the least natural number in $B'$, i.e. the least natural number that is not a square. We have $1^2=1$ and $2^2=4$. Thus, the conclusion is ...
– Mauro ALLEGRANZA
Aug 2 at 14:00




1




1




So, the conclusion is therefore n=2, as it is the lowest not square number?
– Stevie Martin
Aug 2 at 14:05





So, the conclusion is therefore n=2, as it is the lowest not square number?
– Stevie Martin
Aug 2 at 14:05











1 Answer
1






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oldest

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0
down vote



accepted










We can try to address the problem by steps.



Let $mathbb N$ the set of natural numbers, i.e. $mathbb N = 1,2,3,ldots $.



Let $B = b in mathbb N mid b=x^2, text for some x in mathbb N $.



Thus, $B = 1,4,9,16,ldots $.



$B'$ is the complement of $B$.



Both $B$ and $B'$ are subsets of $mathbb N$; thus :




$B' cap mathbb N = B'$.




Conclusion : the least $n in (B' cap mathbb N)$ will be the least element in $B'$, i.e. the least natural number that is not a square.



Obviously, $B' = 2,3,5,ldots $, and thus :





$n=2$.








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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    We can try to address the problem by steps.



    Let $mathbb N$ the set of natural numbers, i.e. $mathbb N = 1,2,3,ldots $.



    Let $B = b in mathbb N mid b=x^2, text for some x in mathbb N $.



    Thus, $B = 1,4,9,16,ldots $.



    $B'$ is the complement of $B$.



    Both $B$ and $B'$ are subsets of $mathbb N$; thus :




    $B' cap mathbb N = B'$.




    Conclusion : the least $n in (B' cap mathbb N)$ will be the least element in $B'$, i.e. the least natural number that is not a square.



    Obviously, $B' = 2,3,5,ldots $, and thus :





    $n=2$.








    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      We can try to address the problem by steps.



      Let $mathbb N$ the set of natural numbers, i.e. $mathbb N = 1,2,3,ldots $.



      Let $B = b in mathbb N mid b=x^2, text for some x in mathbb N $.



      Thus, $B = 1,4,9,16,ldots $.



      $B'$ is the complement of $B$.



      Both $B$ and $B'$ are subsets of $mathbb N$; thus :




      $B' cap mathbb N = B'$.




      Conclusion : the least $n in (B' cap mathbb N)$ will be the least element in $B'$, i.e. the least natural number that is not a square.



      Obviously, $B' = 2,3,5,ldots $, and thus :





      $n=2$.








      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        We can try to address the problem by steps.



        Let $mathbb N$ the set of natural numbers, i.e. $mathbb N = 1,2,3,ldots $.



        Let $B = b in mathbb N mid b=x^2, text for some x in mathbb N $.



        Thus, $B = 1,4,9,16,ldots $.



        $B'$ is the complement of $B$.



        Both $B$ and $B'$ are subsets of $mathbb N$; thus :




        $B' cap mathbb N = B'$.




        Conclusion : the least $n in (B' cap mathbb N)$ will be the least element in $B'$, i.e. the least natural number that is not a square.



        Obviously, $B' = 2,3,5,ldots $, and thus :





        $n=2$.








        share|cite|improve this answer













        We can try to address the problem by steps.



        Let $mathbb N$ the set of natural numbers, i.e. $mathbb N = 1,2,3,ldots $.



        Let $B = b in mathbb N mid b=x^2, text for some x in mathbb N $.



        Thus, $B = 1,4,9,16,ldots $.



        $B'$ is the complement of $B$.



        Both $B$ and $B'$ are subsets of $mathbb N$; thus :




        $B' cap mathbb N = B'$.




        Conclusion : the least $n in (B' cap mathbb N)$ will be the least element in $B'$, i.e. the least natural number that is not a square.



        Obviously, $B' = 2,3,5,ldots $, and thus :





        $n=2$.









        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 2 at 14:08









        Mauro ALLEGRANZA

        60.6k346105




        60.6k346105






















             

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