# Thread: Prove/disprove the following statement

1. ## Prove/disprove the following statement

For all x, there exists a y such that, .

My attempt at the solution:

Assuming x=/= y,
x|z can be written as xk=z
y|z can be written as yk=z
xk=yk
x=y
Therefore, the statement is false.

I have been told this solution is wrong. Can someone please tell me what I did wrong or give me a suggestion how to go about proving this?

Thanks

2. ## Re: Prove/disprove the following statement

Originally Posted by nubshat
For all x, there exists a y such that, .
My attempt at the solution:
Assuming x=/= y,
x|z can be written as xk=z
y|z can be written as yk=z
xk=yk
x=y
Therefore, the statement is false.
Your mistake is in using the variable $k$ twice.
Moreover, we must assume that all variables are positive integers?

You are correct to say that the statement is false.
The only way to show a statement is false is to provide a counter-example.

Think about $x=2$

3. ## Re: Prove/disprove the following statement

Originally Posted by Plato
Your mistake is in using the variable $k$ twice.
Moreover, we must assume that all variables are positive integers?

You are correct to say that the statement is false.
The only way to show a statement is false is to provide a counter-example.

Think about $x=2$
Sorry I forgot to mention the universe of discourse is Z\{0}.

Why would x=2 not work?

If x=2, when z is not divisible by x (example: z=7), there will always exist a y that is also not divisible by 7(example: y=25).
Also, when z is divisible by x (example: z=8), there will always be a y that is a multiple of z (example: y=1).

4. ## Re: Prove/disprove the following statement

Originally Posted by nubshat
Why would x=2 not work?
If x=2, when z is not divisible by x (example: z=7), there will always exist a y that is also not divisible by 7(example: y=25).
Also, when z is divisible by x (example: z=8), there will always be a y that is a multiple of z (example: y=1).
You do not understand how to read the statement. We are in $\mathbb{Z}\setminus\{0\}$ so now the statement is true.
.It says "For every number $x$ there is a number $y\ne x$ such that for every number $z$ it is the case that $x|z~{\color{blue}\iff}~y|z$

That is an if an if and only if which means if $2|6$ then $y|6$ must be true.
Then what is $y$? $y\ne 3$ because $2|8\text{ but}3\not | 8$.

So what is $y$ it has to work for very even $z$ ? Can you find one?