1
$\begingroup$

Note this problem is about the investigation of the E-Zone.

Part A: Describe all E-primes.

Answer: E-primes are greater than and equal to 1 and its only E-zone factors are 1 and itself (similar to any primes). Also it is not divisible by 4, which is the same as saying 2 times an odd number.

Part B: Show that every even number can be factored as a product of E-primes.

So my book, which is Joseph Silverman's A Friendly Introduction to Number Theory, states that this proof is just a mimic of the proof of this fact for ordinary numbers, however I was wondering if there was another way of proving this statement?

$\endgroup$
  • 3
    $\begingroup$ Mind telling us what the E-zone is? $\endgroup$ – anon Dec 31 '13 at 5:45
  • 2
    $\begingroup$ A world where only even numbers exists. $\endgroup$ – Username Unknown Dec 31 '13 at 5:46
  • $\begingroup$ Dear god, this was the first text I ever read on number theory. I got the book when I was 13 years old. Watch out for the chapter on perfect numbers, he starts spewing poetry. $\endgroup$ – Ethan Dec 31 '13 at 5:49
  • $\begingroup$ @Ethan I was enjoying the problems in this book until I go to this part. $\endgroup$ – Username Unknown Dec 31 '13 at 5:52
  • $\begingroup$ @DramaFreak Haha my 7th grade math teacher gave me the book. I stopped reading after the chapters on quadratic reciprocity lol $\endgroup$ – Ethan Dec 31 '13 at 5:53
4
$\begingroup$

Each even number $n$ is of the form $n=2^ku$ where $u$ is odd and $k\ge 1$. If $k=1$ then $n$ is already an E-prime. Otherwise let the first E-prime be $2u$ and all others be $2$ as many as needed.

Note this method relies on a small part of unique factorization, and it is not the same as the way suggested to solve it in the Silverman text.

Note: For part A you may want to be more explicit and say an E-prime is any number of the form $2u$ with $u$ odd. Also it may or may not need a sign, depending on whether the E-zone includes negative integers.

Elaboration of part B argument: First, the statement that

Every even number $n$ is of the form $2^ku$ with $u$ odd and $k\ge 1$

can be shown by induction. Base case $n=2=2\cdot 1$ is of the desired form with $k=1,u=1$. Now suppose $n>2$ is even so that $n=2k$ for some $k>1$. If $k$ is odd then we have the desired form $n=2^1u$ where $u=k$. If $k$ is even, then $k=2k'$ and then $n=2\cdot(2k').$ At this point $2k'$ is an even number less than $n$ so by the inductive hypotheses $2k'=2^tu$ with $t\ge 1$ and $u$ odd. Putting this back into $n=2 \cdot 2k'$ gives $n=2^{t+1}u$ which is of the desired form, finishing the inductive step.

Now that's done, and you want to show every even number $n$ is a product of E-primes, where these are all numbers of the form $2u$ with $u$ odd (from part A). Note first that $2=2\cdot 1$ qualifies as an E-prime using $u=1$. Since $n$ is even we can use the above shown fact and write $n=2^ku$ with $k\ge 1$ and $u$ odd. There are two cases:

case 1: $k=1$. In this case $n=2^1u=2u$ with $u$ odd, so $n$ is itself an E-prime. It is thus the product of a single E-prime and the result holds in this case.

case 2: $k>1$. Here we can write $k=1+r$ and then $$n=2^ku=(2u)\cdot 2 \cdot 2 \cdots \cdot 2,$$ where there are $r$ copies of the E-prime $2$ being multiplied together after the initial E-prime $(2u)$. So in case 2 also, we have $n$ as a product of E-primes.

I hope this helps explain part B...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I understand what you are saying but I am not sure how to use it in the proof or how to start it for that matter. $\endgroup$ – Username Unknown Jan 1 '14 at 2:34
  • $\begingroup$ @DramaFreak Do you already understand part A? I could help more if I knew what step(s) in the logic are not clear to you. $\endgroup$ – coffeemath Jan 1 '14 at 2:58
  • $\begingroup$ I understand part A. If you could that would help $\endgroup$ – Username Unknown Jan 1 '14 at 22:10
  • $\begingroup$ @DramaFreak I just added a (somewhat lengthy) part in my answer hopefully clearing up part B. $\endgroup$ – coffeemath Jan 1 '14 at 22:35

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.