# Definition of Divisibility

1. Sep 26, 2015

### S.R

Question:
If x | y, (is true), then x ≤ y and x ≠ 0.

For instance, if x > y, then there are no integer solutions to equation kx = y and thus, x does not divide y.
Is this a correct proposition?

2. Sep 26, 2015

### Staff: Mentor

What do you think?

3. Sep 26, 2015

### WWGD

Up to sign, if you extend divisibility to the whole of $\mathbb Z$.

4. Sep 26, 2015

### S.R

I suppose if x and y are negative, then the converse is true.
For instance, -2 | -4 is true, but -4 | -2 is false.

Last edited: Sep 26, 2015
5. Sep 27, 2015

### Staff: Mentor

Why aren't you using the usual definition of divisibility; i.e., For integers x and y, x | y iff y = kx for some integer k.

6. Sep 27, 2015

### nuuskur

Happens to be correct, but merely a proposition unless you also prove it.
If $x|y$ and $x>y$ then there exists $c\in\mathbb{Z}\setminus\{0\}$ such that $cx = y$. Figure out how this contradicts and you are home free :) Also, $\mathbb{Z}$ does not form a group under multiplication, so you can't "divide" both sides by some non zero scalar.

7. Sep 27, 2015

### S.R

Assuming x, y ∈ N:
If x | y and x > y, then there exists an integer k such that kx = y or k = y/x.
However, since x > y, the expression y/x is not an integer.
Therefore, we can conclude x does not divide y, since no integer k exists such that kx = y.

Last edited: Sep 27, 2015
8. Sep 27, 2015

### nuuskur

Were you operating with real numbers (in a complete ordered field) the statement $k = \frac{y}{x}$ is meaningful, but you can't do that if you are strictly operating within $\mathbb{Z}$.
The conclusion is correct that there is no such integer $k$ that produces the desired result.
Alternatively we could use the given that $x>y$. Let's look for a candidate $k\in\mathbb{Z}$ such that $kx=y$
Since $k$ is an integer it can be expressed as a sum $k :=\pm ( 1+1+...+1)$ (there is actually a reason why this is true, but I don't want to burst the bubble)

Looking at positive integers first
Then $kx = y$ becomes $(1+1+...+1)x = y$. Due to the distributive property the previous can also be written as $x + x + ... + x = y$
But we already have that $x>y$ therefore $x+x+...+x = y$ can never be true.

And if we look in the negative integers then $-x -x ..-x = y$. Under addition, $\mathbb{Z}$ does form a group, so we can say there exists $-(-x) = x$ such that $x+(-x) = 0$ and what would follow is $y+x+x+...+x = 0$, which can only be true if $y>x$, violating one of our established criteria.
The initial assumption that such an integer exists is therefore incorrect. And $x|y$ only if $0\neq |x|\leq |y|$

In actuality, such divisibility is defined in a ring with no zero terms. A ring is an algebraic structure that forms an abelian group under addition, multiplication and addition are connected via the distributive property. A ring with no zero terms (can't be too sure of the English terminology here) is a ring where no two ring elements' product is 0.

We don't know anything about "dividing both sides by.." for this problem. There is no group under multiplication.

EDIT: oop, You assumed x,y to be in N, then the negative integer part is unnecessary.
Actually, this proof contains inaccuracies, as well :/

Last edited: Sep 27, 2015
9. Sep 27, 2015

### Staff: Mentor

If x | y, then it's not possible for x to be larger than y. "x divides y" means that x is a factor of y.
I don't get what you're trying to do here.

10. Sep 27, 2015

### nuuskur

Challenge the definition :D I agree, it's fruitless.