Paradox inequality

Question

By doing exercises, I applied a reasoning in a demonstration, that looking at it well, it seems paradoxical.

Let's take a real number $\ a>0 $.

Then $b( \epsilon ) = a+ \epsilon > a $, for each $\ 0< \epsilon <1 $.

$\ b^2( \epsilon ) = (a+ \epsilon )^2 = (a^2 + 2 a \epsilon +\epsilon ^2) < a^2 + 2 a \epsilon + \epsilon $

$\ b^2( \epsilon ) < a^2 + \epsilon (2a+1)$

Since $\ \epsilon $ is as small as I want and it must always be that $\ b^2( \epsilon ) < a^2 + \epsilon (2a+1)$ then it means that:

$\ b^2( \epsilon ) \leq a^2 $

so $\ b( \epsilon ) \leq a $, that contradicts $\ b( \epsilon )> a $.

where am I wrong?

Thanks.

Answer 1

$\varepsilon$ is as small as you want, but it can never become negative.

Take $a=10$, and your process ends up with $b^2(\varepsilon) < 100 + 21 \varepsilon$.

This does not contradict $b^2(\varepsilon) \ge 100$.

Answer 2

You're taking the limit of $\epsilon \to 0$ on the left hand side, but on the right side $\epsilon$ is still fixed. What is actually true is that

$$ a^2 = b^2(0) = \lim_{\epsilon \to 0} b^2(\epsilon) \leq a^2 $$

Where the second equality is because $b^2$ is continous. Therefore, there is no such contradiction.

Answer 3

Why do you make it so complicated? Here is a much simpler way to get a contradiction using your idea. Define $$f(\epsilon)=\epsilon\text{ for }\epsilon\gt0.$$ So $$f(\epsilon)\gt0.$$ On the other hand, $$f(\epsilon)\le\epsilon.$$ Since I can make $\epsilon$ as small as I want this means that $$f(\epsilon)\le0,$$ which contradicts $f(\epsilon)\gt0.$

If you can see what I did wrong, maybe you can see that you made the same mistake in your argument.