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I am asked to prove: If $a-\epsilon< b$ for every $\epsilon> 0$, then $a \le b$.

I have worked out that a proof by contradiction, assume $a>b$, would be best in this case but I am not too sure where to proceed from there. Should I rearrange the first part I am given? Must I pick a specific epsilon?

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    If $a > b$, then $a - \frac{a-b}{2} > b$, and $\frac{a-b}{2}>0$. – 2017-01-20
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    "Must I pick a specific epsilon" You must pick an epsilon where this fails. So pick an epsilon where $a - \epsilon \ge b$. To find such an epsilon solve $a \ge b + \epsilon$ and $a - b \ge \epsilon$. If $a > b$ any epsilon equal or less than $a -b$ will do. For simplicity use $\epsilon = a-b$. – 2017-01-20
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    @fleablood I thought to use e=a-b but I am not to sure where to go from there. Would I then have to plug it into my assumption of a-e – 2017-01-20
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    If e = a-b then a - e = b, but that contradicts our hypothesis that a- e < b for ***all*** e. – 2017-01-20

3 Answers 3

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How to get inspiration in how to do a proof by contradiction:

We know we want:

Step 1: $a \not \le b$

Step 2: $a > b$

Step 3: ?????????

Step 4: $a - \epsilon \ge b$ (Profit!)

So what's step 3?

Work back from step 4:

$a - \epsilon \ge b$

$a - b \ge \epsilon$

$a- b \ge \epsilon > 0$

$a - b > 0$ so

$a > b$

Step 2:

So proof is:

Suppose $a \not \le b$

$a > b$

$a-b > 0$. Select any $\epsilon$ so that $a-b \ge \epsilon > 0$

Then $a-\epsilon \ge b$. A contradiction.

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To do it mostly directly with a hail-mary proof by contradiction at the end.

$a - \epsilon < b$ for all $\epsilon$ means

$a - b < \epsilon$ for all $\epsilon$.

$\epsilon$ can be any positive number.

If $a-b > 0$ then if $\epsilon = a-b$ we have

$a-b < \epsilon = a -b$. Impossible.

So $a - b \le 0$ and $a \le b$.

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$a - \epsilon < b$ means $a -b < \epsilon$ means $a -b$ is less then all positive numbers.

The proof hinges on the observation that only 0 and negative numbers are smaller than all positive numbers. This is because no positive number is less than itself.

So here is proof number umpteen:

$0 < a - b < a-b = \epsilon$ is impossible.

So it can't be that $a - \epsilon < a- \epsilon =b$ for $\epsilon = a-b$.

But $a - \epsilon < b$ for all $\epsilon > 0$.

So that must mean $\epsilon = a-b > 0$ is impossible.

The only way that is impossible, as $\epsilon $ can be any positive number, is that $a-b$ is not positive.

$a - b \le 0$.

And $a \le b$.

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Proof (by contradiction). Assume that $a-\epsilon0$ and suppose $a>b$. Take $\epsilon_0=a-b$. Then $\epsilon_0>0$. Using the assumption (as $\epsilon_0$ is a specific value of $\epsilon$), we get $$a-\epsilon_0

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Assume that a>b.

You are given that $\forall\epsilon>0,$ $a-\epsilon b,$ $$\epsilon = (a-b) >0$$ and it fits the bill since $a-\epsilon = b \ge b.$