1

Given a statement like $b$ divides $a$ means there exists some integer $k$ such that $a = b\cdot k$, does that mean if $b$ does not divide $a$, then there are integers $q$ and $r$ such that $a = q\cdot b +r$ for $0<r<b$ according to the division algorithm?

Am I assuming too much here?

EDIT: Let me also clarify that I am assuming $a,b\in\mathbb{Z}$ and both are positive.

MrStormy83
  • 639
  • 4
  • 14
  • 1
    Yes, what you say is a correct interpretation of "$b$ does not divide $a$" for integers $a$ and $b$. You could also have phrased it other ways such as $a\not\equiv 0\pmod{b}$, $a$ is not a multiple of $b$, there does not exist any $k$ such that $a=bk$, for all integers $k$ one has $a\neq bk$, etc... – JMoravitz Oct 04 '17 at 00:11
  • Depends. Are $a$ and $b$ integers and is $b$ positive? – Hagen von Eitzen Oct 04 '17 at 00:12
  • Yes, $a$ and $b$ are integers and $b$ is positive. – MrStormy83 Oct 04 '17 at 00:16

3 Answers3

2

Yes. There's a simpler representation.

To say $b$ does not divide $a$ means that $b$ is not a factor of $a$. That is, if you try to do the division, you'll have leftovers - the remainder $r$.

Sean Roberson
  • 9,110
1

What you've said is true. That said, depending on what class you're taking (I assume this is for a class), you might need to prove that $r$ can be chosen such that $0 < r < b$, and that the existence of such an $r$ implies that $b\nmid a$ (it depends how detailed you're expected to be).

Carl Schildkraut
  • 32,931
  • I am using this fact to prove a different statement. What I would like to show is that if the order of an element b, does not divide the order of an element a, then the order of a*b is larger than the order of a. I just wanted to get the context behind "does not divide" correct. – MrStormy83 Oct 04 '17 at 00:14
0

No.

Consider the case $b=-5$ and $a=9$. Then we say $b\nmid a$ because there is no integer $k$ with $a=bk$. But you cannot find $q,r$ with $a=qb+r$ and $0<r<b$, simply because the second condition cannot be fulfilled.

Hagen von Eitzen
  • 374,180
  • Because this is actually about orders of elements within an abelian group G, it is assumed that $b$ and $a$ are both positive integers. – MrStormy83 Oct 04 '17 at 00:17