How Does the Archimedean Property Imply b^2 ≤ 0?

[Abraham]

This isn't really hw. I need someone to explain a certain line in a proof:

" b² $\leq$ $\frac{1}{n}$ for all n in the natural numbers. This implies that b² $\leq$ 0 (a consequence of the Archimedean property). "

I don't see how the Archimedean is applied in this context. This is my understanding of the property: For real numbers x,y, x>0, there exists a natural number n such that nx > y.

I don't see how this proves b² $\leq$ $\frac{1}{n}$ $\Rightarrow$ b² $\leq$ 0.

Thanks

 

[Dick]

Apply the Archimedean property with b^2=x and y=1. You know b^2>=0. If b is not zero then b^2>0. If nb^2>1 then b^2>1/n. It's a proof by contradiction.