Divisibility 14

Sea · #1 $Old$ December 20th, 2008, 03:33 PM

Show that:

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

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

Jhevon · #2 $Old$ December 20th, 2008, 03:51 PM

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 ...

Sea · #3 $Old$ December 20th, 2008, 04:52 PM

Thanks...but I cannot see the proves...

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

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

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

I didn't know...

Jhevon · #4 $Old$ December 20th, 2008, 05:00 PM

Quote:

Originally Posted by Sea $View Post$

Thanks...but I cannot see the proves...

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

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

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

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 $ka^3$ ?

Sea · #5 $Old$ December 23rd, 2008, 03:44 PM

Quote:

Show that:

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

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

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...

...

Sea · #6 $Old$ December 24th, 2008, 02:22 PM

Quote:

Originally Posted by Sea $View Post$

Show that:

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

I think..

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

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

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

$(\Rightarrow ):$

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

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

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

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

and

Thanks Jerry...

HallsofIvy · #7 $Old$ December 26th, 2008, 12:38 PM

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 $a^3$ ?