Let m be an element of the Natural Numbers and n be an element of the Integers. If m*n is an element of the Natural Numbers, then n is an element of the Natural Numbers.

Any help with the above proof would be appreciated!

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- Feb 17th 2011, 02:10 PMjstarks44444Multiplication of Natural Numbers proof
Let m be an element of the Natural Numbers and n be an element of the Integers. If m*n is an element of the Natural Numbers, then n is an element of the Natural Numbers.

Any help with the above proof would be appreciated! - Feb 17th 2011, 02:32 PMPlato
One again, if you would post a list of axioms, definitions, and theorems you would receive better help.

Not having a such here a guess based upon the definition of $\displaystyle \mathbb{N}$ you posted before.

Suppose that $\displaystyle n\notin\mathbb{N}$

Based on the given we know that $\displaystyle n\in \mathbb{Z}$ so from the axiom $\displaystyle n=0\text{ or }-n\in \mathbb{N}$.

You have proven that $\displaystyle j>0$ for all $\displaystyle j\in\mathbb{N} $.

What would that say about $\displaystyle mn~?$