Talk:Real Plus Epsilon
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It seems that this must be true for ALL positive epsilon, but the statement of the theorem doesn't really seem to capture that. I can't think of a good way to rephrase it though; any thoughts? --Cynic (talk) 23:27, 28 May 2010 (UTC)
- Can't see the problem. As far as I can tell, it does. It says: let $a, b, \epsilon$ be real numbers such that $\epsilon$ is positive, i.e. $\epsilon > 0$, which means: for any $a, b, \epsilon$ such that ..., i.e. "any such that you care to choose" which is another way of saying "For all ..."
- Could be reworded (resymbolised) $\forall a, b \in \R, \epsilon \in \R^*_+$ I suppose, but I believe the way it's worded is neater and simpler.
- Have I missed the point? --Matt Westwood 07:18, 29 May 2010 (UTC)
- Well the criteria for $a \le b$ is that $\forall a, b \in \R , \epsilon \in \R_+^*: a < b + \epsilon$. The way the statement is put now, it seems that if $a < b + \epsilon$ for any particular set of $a, b, \epsilon$ then the second part must follow. But $a = 2, b = 1, \epsilon = 5$ makes the first statement true but the second statement false. Maybe it's just me misinterpreting it, but I think it needs reworking. I'll look at it at a more reasonable hour. --Cynic (talk) 05:32, 5 June 2010 (UTC)
- Good call! I understand what I've done wrong now. Fixing this up.--Matt Westwood 06:42, 5 June 2010 (UTC)
- Well the criteria for $a \le b$ is that $\forall a, b \in \R , \epsilon \in \R_+^*: a < b + \epsilon$. The way the statement is put now, it seems that if $a < b + \epsilon$ for any particular set of $a, b, \epsilon$ then the second part must follow. But $a = 2, b = 1, \epsilon = 5$ makes the first statement true but the second statement false. Maybe it's just me misinterpreting it, but I think it needs reworking. I'll look at it at a more reasonable hour. --Cynic (talk) 05:32, 5 June 2010 (UTC)
