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CGandC
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Prove/Disprove: 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 ##.
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: I understand I can disprove the statement directly by explicitly choosing a value for ## a ## , but such disproof is different from what I did and I want to know if what I did is correct.
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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