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Mathematics
OpenStudy (anonymous):

Let \(x\) and \(a\) be real numbers. Suppose that \(x

OpenStudy (zarkon):

assume x>a let d=x-a d>0 ...

OpenStudy (zarkon):

I assume x<a+e is true for all e>0

OpenStudy (anonymous):

Yes, Zarkon, that's correct. Thank you for pointing that out. I'll try what you proposed.

OpenStudy (anonymous):

Does this seem legit? Proof: Let \(x\) and \(a\) be real numbers and suppose that \(x<a+\epsilon\) for every positive number \(\epsilon\). Assume that \(x>a\), and let \(d=x-a\). Then \(d>0\). It follows that \(x<a+d\), \(x<a+x-a\), \(x<x\). The last statement is a contradiction. Therefore, \(x\leq a\). \(\blacksquare\)\[\]

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