# Divisibility 14

• Dec 20th 2008, 01:33 PM
Sea
Divisibility 14
Show that:

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

b) $\displaystyle a^{3}|b^{2}\Rightarrow a|b$
• Dec 20th 2008, 01:51 PM
Jhevon
Quote:

Originally Posted by Sea
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 ...
• Dec 20th 2008, 02:52 PM
Sea
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...
• Dec 20th 2008, 03:00 PM
Jhevon
Quote:

Originally Posted by Sea
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 $\displaystyle ka^3$ ?
• Dec 23rd 2008, 01:44 PM
Sea
Quote:

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...;):)...
• Dec 24th 2008, 12:22 PM
Sea
Quote:

Originally Posted by Sea
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...
• Dec 26th 2008, 10:38 AM
HallsofIvy
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 $\displaystyle a^3$?