0
$\begingroup$

How to prove $d>0$ is a divisor of $n$ iff $d=p_1^{b_1}p_2^{b_2}...p_r^{b_r}$ with $0< b_i<a_i$ for each $i$? here $n = p_1^{a_1}p_2^{a_2}...p_r^{a_r}$ with the $p_i$ distinct primes and the $a_i$ positive integers.

I am unsure of how to start this problem any solutions or hints are welcome

$\endgroup$
  • $\begingroup$ You want $0\le b_i\le a_i$. This is really just unique factorisation into primes. $\endgroup$ – Lord Shark the Unknown Oct 19 '17 at 15:54
  • $\begingroup$ Factor each $p_i^{a_i}= p_i^{b_i}p_i^{a_i- b_i}$.. Since "$0< b_i< a_i$" $a_i- b_i$ is positive. $\endgroup$ – user247327 Oct 19 '17 at 15:55
0
$\begingroup$

Hint: Try prove by contradiction.

Assume $d = p_1^{b_1}p_2^{b_2}...p_r^{b_r}p_{r+1}^{b_{r+1}}$ (i.e. $d$ has a prime factor that is not a factor of $n$). Divide $n$ by $d$.

Then, assume $d = p_1^{b_1+k}p_2^{b_2}...p_r^{b_r}$ (i.e., $d$ has one of the prime factors raised to a power higher than what it was raised in $n$). Again, divide $n$ by $d$.

Now, analyze the results. What can you say about them? Can you prove them to be integral?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.