I have a very, very elementar question.
If the assumption is that $a>0$ and then in a proof it is shown that $a\geq 0$, is that a contradiction?
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1No, $a > 0$ and $a \geq 0$ are compatible. It would be a contradiction if $a\leq 0$ were shown. – Daniel Fischer Nov 04 '14 at 19:50
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$\geq$: Greater than or equal to; if one of those is true, the statement is true. – Robert Cardona Nov 04 '14 at 20:00
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What if I told you the answer to your question was "yes" ? Would you believe me? Why or why not? – DanielV Nov 04 '14 at 20:51
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It's only a contradiction if $a=0$. In other words, if your proof can deduce $a=0$, it's a contradiction. Afterall, $1>0$ and $1\geq 0$ is ok. But if in the course of a proof you get $1=0$, then something has gone wrong.
Alex R.
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$a\geqslant0$ means $a=0\:\vee\:a\gt0$. If you are given that $a\gt0$ as being true then the statement $$a\gt0\:\wedge\:(a=0\:\vee\:a\gt0),$$ is also true since $a=0\:\vee\:a\gt0$ is true. (Why? Remember when $p\vee q$ is true.)
Hakim
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