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Thread: Divisibility 14

  1. #1
    Sea
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    Divisibility 14

    Show that:



    a) $\displaystyle a^{3}|b^{3}\Rightarrow a|b$



    b) $\displaystyle a^{3}|b^{2}\Rightarrow a|b$
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sea View Post
    Show that:



    a) $\displaystyle a^{3}|b^{3}\Rightarrow a|b$



    b) $\displaystyle a^{3}|b^{2}\Rightarrow a|b$
    In both cases, use the contrapositive.

    To start you off, both proofs would begin in the following way:

    Assume $\displaystyle a \nmid b$. Then, by the division algorithm, we can write $\displaystyle b = aq + r$, for $\displaystyle q,r \in \mathbb{Z}$, $\displaystyle 0 < r < a$. So ...
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    Sea
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    Thanks...but I cannot see the proves...

    $\displaystyle a \nmid b$ and $\displaystyle b = aq + r$, for $\displaystyle q,r
    \in \mathbb{Z}, 0 < r < a$

    $\displaystyle b^{3}=(aq+r)^{3}=a^{3}q^{3}+3a^{2}qr+3aqr^{2}+r^{3 }$

    $\displaystyle a^{3}\mid a^{3}q^{3}+3a^{2}q^{2}r+3aqr^{2}+r^{3}$



    I didn't know...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sea View Post
    Thanks...but I cannot see the proves...

    $\displaystyle a \nmid b$ and $\displaystyle b = aq + r$, for $\displaystyle q,r
    \in \mathbb{Z}$

    $\displaystyle b^{3}=(aq+r)^{3}=a^{3}q^{3}+3a^{2}qr+3aqr^{2}+r^{3 }$

    $\displaystyle a^{3}\mid a^{3}q^{3}+3a^{2}q^{2}r+3aqr^{2}+r^{3}$



    I didn't know...
    keep in mind what you want to prove here, you want to show that $\displaystyle a \nmid b \implies a^3 \nmid b^3$.

    that means you have to show $\displaystyle b^3 = k a^3$ is NOT true for any integer $\displaystyle k$, given that $\displaystyle a \nmid b$.

    now, is it possible to write $\displaystyle a^3q^3 + 3a^2q^2r + 3aqr^2 + r^3$ in the form $\displaystyle ka^3$ ?
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  5. #5
    Sea
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    Show that:

    a) $\displaystyle a^{3}|b^{3}\Rightarrow a|b$
    $\displaystyle a^3|b^3 \Rightarrow b^3=k\times a^3 \Rightarrow b^3-(k^{1/3})^3\times a^3 = 0.
    $


    Using the identity $\displaystyle p^3-q^3 = (p-q)(p^2+p\times q+q^2)$


    we see that

    $\displaystyle (b-a\times k^{1/3})(b^2+b \times a\times k^{1/3}+k^{2/3}\times a^2) = 0.$


    The second factor cannot be zero. Hence, $\displaystyle b = a\times k^{1/3}$.


    Because $\displaystyle b$ and $\displaystyle a $are integers,$\displaystyle k^{1/3}$ must be an integer.


    and


    Thanks Jerry......
    Last edited by Sea; Dec 23rd 2008 at 01:55 PM.
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  6. #6
    Sea
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    Quote Originally Posted by Sea View Post
    Show that:

    b) $\displaystyle a^{3}|b^{2}\Rightarrow a|b$



    I think..


    $\displaystyle a=p_1^{e_1}\times p_2^{e_2}......$

    $\displaystyle
    b=q_1^{f_1} \times q_2^{f_2}......$

    where p1,... and q1,...are primes and e1,... and f1,... are powers, then


    $\displaystyle (\Rightarrow ):$


    $\displaystyle a^3=p_1^{3e_1}\times p_2^{3e_2}......
    $


    $\displaystyle

    b^2=q_1^{2f_1} \times q_2^{2f_2}......$




    $\displaystyle

    a^3|b^2 \Rightarrow p_1^{3e_1}\times p_2^{3e_2}...... | q_1^{2f_1} \times q_2^{2f_2}......
    $


    is a whole number. So, there can't be more p's than q's, every q is a p, etc.



    and


    Thanks Jerry...
    Last edited by Sea; Dec 24th 2008 at 12:44 PM.
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  7. #7
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    Jhevon has suggest twice now that you prove the contrapositive: tha tif a does NOT divide b, then $\displaystyle a^3$ does not divide $\displaystyle b^3$. Why have you not even tried that?

    If a does not divide b then b= ma+ k for some k greater than 0 and less than a. $\displaystyle b^3= (ma+k)^3= m^3a^3+ 3ka^2+ 3k^2a+ k^3$. Can you prove that $\displaystyle 3ka^2+ 3k^2a+ k^3$ is not a multiple of $\displaystyle a^3$?
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