Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been assigned this problem for homework:

Show that, if $a < b + \epsilon$ for every $\epsilon \gt 0$, then $a\le b$.

I have tried to go about this using Induction, but I don't know what the base case would be. It is obvious to me in my mind, but I don't know how to put it into mathematical terms on paper. any hints?

share|cite|improve this question
Presumably $a$ and $b$ are arbitrary real numbers. If so, induction is not a suitable tool. – André Nicolas Jan 25 '13 at 21:00
(Standard) induction only works for statements that can be phrased as "for every $n$, P(n)" where $n$ is a natural number. Since $\epsilon$ is allowed to be any real number >0, induction won't work. – icurays1 Jan 25 '13 at 21:02
What strategy would you recommend. – Sam Jan 25 '13 at 21:08
@Sam - take a look at the answers below. – icurays1 Jan 25 '13 at 21:10

Hint: Use contradiction. If $a>b$, then show $a$ is not less than $b+\frac{a-b}{2}$. This contradiction implies $a\leq b$.

share|cite|improve this answer

I'm a snob, so I like using contrapositive instead of contradiction whenever possible. The contrapositive would go like this: "if $a>b$, then there exists an $\epsilon>0$ such that $a\geq b+\epsilon$." Finding this epsilon should be easy with a number line.

Edit: the original poster figured it out in the comments, so here's the full argument: if $a>b$, then certainly $a-b>0$. Hence set $\epsilon=a-b$; then, since $b+\epsilon=b+a-b=a$, we certainly have $a\leq b+\epsilon=a$ (since $a=a$).

share|cite|improve this answer
we could choose an actual number for ϵ or a representation using a and b? – Sam Jan 25 '13 at 21:12
A representation using $a$ and $b$, exactly - you can't pick say "$\epsilon=0.1$" because if $a$ and $b$ are closer, that $\epsilon$ won't work. – icurays1 Jan 25 '13 at 21:14
Should we let ϵ be equal to $a-b$? that way a would be greater than equal to $b+(a-b)$ therefore $a \geq a$ – Sam Jan 25 '13 at 21:16
Exactly - which is true, since $a=a$. I'll write the whole argument in the answer. – icurays1 Jan 25 '13 at 21:17
thank you so much ! I knew using induction was leading us down the wrong path – Sam Jan 25 '13 at 21:18

I'm assuming this is from a first proof-based course, so apologies if this comes off as condescending.

Induction is harder to use in this case for precisely the reason you stated: what is the base case? We don't have any equivalent of some smallest nonzero number 1 with which to start like we do in $\mathbb{N}$. Maybe we could start with 0.0001? No, 0.00001 is smaller. And 0.000001 is smaller still. This sort of idea is known as the Well-Ordering Principle and is maybe not immediately clear, but it's worth thinking about. In particular, $\mathbb{R}$ (the real numbers) is not well-ordered, since there is no smallest element in $(0,1)$.

But in statements involving inequalities it's often a good idea to try to do things by contradiction. In that case, we'd assume $a > b$ and try to show that this is impossible. Well, if $a > b$ then $a - b > 0$. But we are given that $a < b + \epsilon$ for any $\epsilon > 0$. So what value should we choose for $\epsilon$ to find our contradiction?

share|cite|improve this answer

assume a > b. Then a-b > 0. Then let ϵ = a - b. Then a < b + a-b = a , a contradiction so a <= b.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.