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Math Help - Divisibility 14

  1. #1
    Sea
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    Divisibility 14

    Show that:



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



    b) a^{3}|b^{2}\Rightarrow a|b
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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
    Show that:



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



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

    a \nmid b and b = aq + r, for q,r<br />
\in \mathbb{Z}, 0 < r < a

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

    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...

    a \nmid b and b = aq + r, for q,r<br />
\in \mathbb{Z}

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

    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 a \nmid b \implies a^3 \nmid b^3.

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

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

    a) a^{3}|b^{3}\Rightarrow a|b
    a^3|b^3  \Rightarrow  b^3=k\times a^3  \Rightarrow b^3-(k^{1/3})^3\times a^3 = 0.<br />


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


    we see that

    (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, b = a\times k^{1/3}.


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


    and


    Thanks Jerry......
    Last edited by Sea; December 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) a^{3}|b^{2}\Rightarrow a|b



    I think..


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

    <br />
b=q_1^{f_1} \times q_2^{f_2}......

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


     (\Rightarrow ):


     a^3=p_1^{3e_1}\times p_2^{3e_2}......<br />


    <br /> <br />
b^2=q_1^{2f_1} \times q_2^{2f_2}......




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


    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; December 24th 2008 at 12:44 PM.
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    Jhevon has suggest twice now that you prove the contrapositive: tha tif a does NOT divide b, then a^3 does not divide 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. b^3= (ma+k)^3= m^3a^3+ 3ka^2+ 3k^2a+ k^3. Can you prove that 3ka^2+ 3k^2a+ k^3 is not a multiple of a^3?
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