I was reading through the proof of this and I'm having trouble grasping what the contradiction is. Since $a\leq b+\varepsilon$ this is equivalent to $a-\varepsilon\leq b$. Suppose $b<a$ and taking $\varepsilon=\frac{1}{2}(a-b)$ we have $a-\varepsilon=a-\frac{1}{2}(a-b)=\frac{1}{2}a+\frac{1}{2}b>b$. At this point they state that this is a contradiction to the hypothesis but I don't see how it contradicts that. The contradiction or error I see is that then we have $b<a-\varepsilon\leq b\Rightarrow b<b$ which makes no sense. Although this doesn't appear to be the hypothesis. Also I was curious as to the intuition behind setting $\varepsilon=\frac{1}{2}(a-b)$.
Asked
Active
Viewed 690 times
-1
-
for any two number $x,y$ , $x>y$ and $x\leq y$ can not hold together – Black-horse Dec 04 '15 at 06:45
-
See also: http://math.stackexchange.com/questions/679038/intuition-if-a-leq-b-epsilon-for-all-epsilon0-then-a-leq-b – Martin Sleziak Aug 29 '16 at 04:48
-
How is that +9 when my question is -1... After looking back at it the last line implies of the proof implies $1/2a > 1/2b$ and there's the contradiction. – Craig Oct 08 '16 at 03:32
-
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? – user1110341 May 27 '23 at 15:21