I am currently reading Elementary Calculus: An Infinitesimal Approach by H. Jerome Keisler and was wondering if someone could help me with an aspect treated in the book.

On page 24 he says a number $\varepsilon$ is said to be infinitely small or infinitesimal if $$-a< \varepsilon < a$$ for every positive real number $a$. He then says the only real number that is infinitesimal is zero.

I really don't get that. What I understand is that in order for a number to be considered infinitely small it has to be bigger then $-a$ and smaller then $a$. Well if I take $a$ to be $-2$ that means that $-1$ would be infinitesimal since it is bigger than $-2$ but smaller then $2$. So then how can zero be the only real number that satisfies that condition?

  • $\begingroup$ Maybe some of this question is missing? If so, please edit to add the additional information. $\endgroup$ – Ken Jun 9 '15 at 3:53
  • $\begingroup$ Hi Ken, i typed the whole question however for some reason its only displaying a part of it. $\endgroup$ – samuel Jun 9 '15 at 3:55
  • $\begingroup$ I think it should work fine now $\endgroup$ – samuel Jun 9 '15 at 3:57
  • $\begingroup$ Suppose I have fixed the number $\epsilon$. Then $\epsilon$ is infinitesimal if and only if for every $a \in \mathbb{R}, a > 0$, we have that $- a < \epsilon < a$. For example, $-2 < -1 < + 2$, but NOT $-1/2 < -1 < + 1/2$. So $-1$ is not infinitesimal. $\endgroup$ – AJY Jun 9 '15 at 4:05
  • 1
    $\begingroup$ @FranciscoPresencia indeed. I never thought that it would be that fast and helpful glad i posted my question here. $\endgroup$ – samuel Jun 9 '15 at 19:07

Your example of taking $a$ to be $2$ and concluding that $1$ is infinitesimal since it is between $-2$ and $2$ is not a good example.

The reason for this is that the definition of an infinitesimal $\varepsilon$ is that $-a \leq \varepsilon \leq a$ for every positive real number $a$. You just picked some positive real number. This has to be true for every positive real number. That means $\varepsilon$ needs to be in $[-2, 2]$ and in $[-1, 1]$ and in $[-\frac{1}{2}, \frac{1}{2}]$ and in $[-\frac{1}{1000000}, \frac{1}{1000000}]$, and so on. That same $\varepsilon$ has to be in all of these at the same time to be an infinitesimal.

The only real number that satisfies that it is between $-a$ and $a$ for every real $a > 0$ is $\varepsilon = 0$.

So any number $\varepsilon$ other than $0$ that satisfies $-a \leq \varepsilon \leq a$ for every $a > 0$ real cannot itself be a real number, but there are plenty of infinitesimals that aren't real numbers. As we discussed, $0$ is the only number that's both real and infinitesimal.

  • 1
    $\begingroup$ Wow thank a lot.That was quick and very helpful especially the examples. $\endgroup$ – samuel Jun 9 '15 at 4:11
  • $\begingroup$ @samuel You're welcome! $\endgroup$ – layman Jun 9 '15 at 4:12

The point is that it is greater than $-a$ and less than $a$ for every $a$. So if you consider $e = -1$, you're correct that $-2 < -1 < 2$ but what about $-\frac{1}{2}$ and $\frac{1}{2}$? Clearly $-1$ does not lie between $-\frac{1}{2}$ and $\frac{1}{2}$. Likewise, if you had any non-zero real number $x$, $x$ does not lie between $-\frac{x}{2}$ and $\frac{x}{2}$. Thus the only real infinitesimal is $0$.

  • 2
    $\begingroup$ Cameron Williams thank you between your answer and the answer from user 46944 i understood what was meant. Thank a bunch $\endgroup$ – samuel Jun 9 '15 at 4:12
  • $\begingroup$ You're very welcome. Happy to help. Please don't hesitate to ask questions in the future! You're off to a great start. $\endgroup$ – Cameron Williams Jun 9 '15 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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