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Sorry for this dumb question, I know it's a very simple statement but my proof was different from the one given in the notes, and 9 out of 10 times when I try to write proofs myself they turn out to be incorrect, so I would just like to know if my proof given below is valid.

Proposition: If for every $b > 0$ we have $0 \le a < b$, then $a = 0$.

Proof
If $a > 0$, then let $b = a$, and we have $a \not < b$, a contradiction.
Therefore $a = 0$.

Vizuna
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  • It looks fine to me. – Timbuc Feb 25 '15 at 09:49
  • The proof is fine. I would start it with "Let's assume that $a\neq 0$. Then, because $0\leq a$, we have that $a>0$." – 5xum Feb 25 '15 at 09:50
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    Here $b$ is a "variable." By assuming $a>0,$ and choosing a particular value of $b$ (namely $b=a$) you have shown that the conditon $0 \leq a <b$ is not satisfied. This is fine. – Krish Feb 25 '15 at 09:54
  • It would be much better to formulate the proposition "If $a$ is a number such that for all $b>0$.. etc". The way you wrote it, it could be understood so that $a$ may depend on $b$. – Peter Franek Feb 25 '15 at 11:03

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