If $a+\epsilon > b$ for each $\epsilon >0$, can we conclude that $a>b$?
Please help me to clarify the above. Thanks in advance.
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Martin Sleziak
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Peter
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Well, $a \geq b.$ – Will Jagy Sep 29 '13 at 04:10
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8Be careful: "for every $\epsilon > 0$, if $a+\epsilon > b$, then $a > b$" is a different statement than "if, for every $\epsilon > 0$, $a + \epsilon > b$, then $a > b$." One of them is a lot closer to being true than the other. – Micah Sep 29 '13 at 04:25
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1Related: Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$?. Some of the posts linked there might be of interest, too. – Martin Sleziak May 03 '17 at 12:30
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Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? – user1110341 May 27 '23 at 15:19
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No, you cannot: for every $\epsilon>0$ it’s true that $1+\epsilon>1$, but it’s not true that $1>1$. What you can conclude is that $a\ge b$. To see this, suppose that $a<b$, and let $\epsilon=b-a$; then $\epsilon>0$, but $a+\epsilon=a+(b-a)=b\not>b$, contradicting the hypothesis. Thus, it must be the case that $a\ge b$.
Brian M. Scott
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4@Don: Then obviously it’s not the case that $a+\epsilon>b$ for each $\epsilon>0$, so the example is irrelevant. – Brian M. Scott Sep 29 '13 at 05:19
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1Yeah Don with the example your using you go beyond the hpyothesis imagaine in ur example if u chose epsilon to = 15 then it would work since 5 + 15 = 20 > 10 ie 5+ϵ > 10, but here we had to use a particular epsilon choice. The whole point is we wanted it to work for ANY epsilon that a person want to choose, no matter how small or how big as long as it is greater than zero. to techniquely u wnat ur choice to be as small as possible her to make sure its holds for any epislon, if it works for the smallest number(dont exist so we use ϵ) it obvisly works for larger numbers – Dec 08 '13 at 18:56
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Hello Sir, this is a beautiful answer, but I can't get it how proving $a+\epsilon = b$ from $a<b$ would help us derive $a \ge b$. I have been trying to figure this out for couple of hours which no results. – alu Jan 15 '21 at 18:02
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1@alu: It’s a proof by contradiction. We are given that $a+\epsilon>a$ for each positive $\epsilon$. To get a contradiction, we suppose that $a<b$. Then for $\epsilon=b-a$, which is positive, we have a+\epsilon=b$. This contradicts what we were given. Thus, the supposition that $a<b$ must be false, since it is what led us to the contradiction, and therefore $a\ge b$. – Brian M. Scott Jan 15 '21 at 18:57
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Okay, thank you for explaining, I understand it now :). I understood it's a proof by contradiction, I didn't understand how starting from $1+ \epsilon > 1$ would help us find $a \ge b$. Now I know, we start from $a + \epsilon > b$, which is given in the question and find its contradiction. So I suppose you meant to type $a + \epsilon > b$ in the comment above, clarify me if I'm wrong. – alu Jan 16 '21 at 10:21
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@BrianM.Scott I just want to make sure, since you haven't replied. It will be $a+\epsilon>b$ in your comment at the beginning, isn't it? And I remembered today, I thought you were trying to prove $1+\epsilon>1$, that's why I got so confused. But now it's all clear. – alu Jan 17 '21 at 12:42
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1@alu: It was supposed to say: To get a contradiction, we suppose that $a<b$. then for $\epsilon=b-a$, which is positive, we have $a+\epsilon=b$. This contradicts what we were given. That is, we were told that $a+\epsilon$ is always greater than $b$ when $\epsilon>0$, and now we’ve found a counterexample. – Brian M. Scott Jan 18 '21 at 07:11