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?

  • Maybe some of this question is missing? If so, please edit to add the additional information. – Ken Jun 9 '15 at 3:53
  • Hi Ken, i typed the whole question however for some reason its only displaying a part of it. – samuel Jun 9 '15 at 3:55
  • I think it should work fine now – samuel Jun 9 '15 at 3:57
  • 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. – AJY Jun 9 '15 at 4:05
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    @FranciscoPresencia indeed. I never thought that it would be that fast and helpful glad i posted my question here. – samuel Jun 9 '15 at 19:07
up vote 9 down vote accepted

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.

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    Wow thank a lot.That was quick and very helpful especially the examples. – samuel Jun 9 '15 at 4:11
  • @samuel You're welcome! – 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 $0$.

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

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