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**: There exists ## a \in \mathbb{N} ## such that for all ## n,m \in \mathbb{Z} ## that satisfy ## n \cdot m = a ## then ## n > 0 \text{ or } m > 0 ##.**

__Prove/Disprove__

__My attempt (The statement's false, here's proof by contradiction ):__Suppose There exists ## a \in \mathbb{N} ## such that for all ## n,m \in \mathbb{Z} ## that satisfy ## n \cdot m = a ## then ## n > 0 \text{ or } m > 0 ##.

Let ## n,m \in \mathbb{Z} ## be arbitrary such that ## n \cdot m = a ##, then also ## n > 0 \text{ or } m > 0 ##.

## \underline{ \text{Case I , } } m < 0 ##:

In this case, since ## n \cdot m = a ## then ## n < 0 ## ( since ## a ## is a natural number )

Therefore ## n,m ## are negative, therefore it isn't true that ## n > 0 \text{ or } m > 0 ##, a contradiction.

Q.E.D

Is this disproof correct?

Also, My instructor said I should've had to "

**Choose**## a ##" and that therefore my disproof is not correct.

However I don't agree with him since I've assumed that

**there exists**## a ## then it is not correct to choose a value for ## a ## ( In my disproof ).

**:**

__Note__
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