Basic Properties Explanation

Question

In regards to divisibility I am having trouble wrapping my head around some of the concepts, more specifically some of the general properties of divisibility.

for example, why is it possible for x|y, x|z => x|(y-z)?

In order to attempt to work this out to myself I tried to run through it logically. I understand that x|y = yc

GJ

What does $b$ stand for in $x|y \wedge x|z \implies x|(y-b)$? — barak manos, Sep 04 '14 at 12:45
There are many more things to revise in this question. For example, how can $x|y$ be equal to a numerical value other than $1$ and $0$??? (let alone the fact that those numerical value appear to be non-integer). — barak manos, Sep 04 '14 at 12:48
$x|y$ is a Boolean expression. Either it is true or it is false. So you may consider it as $1$ or $0$ if you want, but you can't really refer to that as a numerical value. — barak manos, Sep 04 '14 at 12:51
Well, I suppose they can be equal to any integer so long as y is a multiple of x — Confused_Teacher, Sep 04 '14 at 12:52
are you fine with the notation below $$x | y \implies y = cx$$ ? — AgentS, Sep 04 '14 at 12:52
$x|y$ is true if and only if $x$ divides $y$ to an integer multiple. You cannot say that $x|y$ equals some numerical value. It just doesn't make much sense. — barak manos, Sep 04 '14 at 12:53
$x|y $ is read as "x divides y" so as you have said earlier "y is a multiple of x"; all this means is that you can find some integer $c$ such that $$y = cx$$ — AgentS, Sep 04 '14 at 12:55
Okay, I am comfortable with y = cx, but again how does this lead to the final conclusion? — Confused_Teacher, Sep 04 '14 at 12:56
May be an example makes it clear.
Say : $y = 15$ and $z = 9$. It is easy to see that $3 | y$ and $3 | z$ . According to the property in question, $3$ must divide $y-z$ also; check if it really divides or not — AgentS, Sep 04 '14 at 13:05

score 1 · Answer 1 · answered Sep 04 '14 at 13:02

Let $\mathbb{Z}$ be the set of integers.

Assume that $x \in \mathbb{Z}$, $y \in \mathbb{Z}$ and $z \in \mathbb{Z}$.

Then

$$x \mid y \Longrightarrow y = cx,$$

for some $c \in \mathbb{Z}$.

Similarly,

$$x \mid z \Longrightarrow z = dx,$$

for some $d \in \mathbb{Z}$.

Subtracting the two equations involving $x, y, z$ we get:

$$y - z = (c - d)x,$$

for some $e = c - d \in \mathbb{Z}$.

Therefore, we get $x \mid (y - z)$, and we are done.

Note the following:

The difference of two integers is again an integer.

Definitions are if-and-only-if statements.

score 0 · Answer 2 · answered Sep 04 '14 at 13:26

In writing “$\,x|y = yc\,$”, you have made a serious error. The point is that $x|y$ is not a number, but a sentence. This sentence says that there is some other number $m$ for which $y=mx$. Similarly, $x|z$ says that there is some other number $n$ for which $z=nx$. I think that once you see this, and are holding the (otherwise unspecified) numbers $m$ and $n$ in your hand, you will see how to prove that $x\,|\,(y\pm z)$

Basic Properties Explanation

2 Answers2