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The set of all hyper-real numbers is denoted by $R^*$. Every real number is a member of $R^*$, but $R^*$ has other elements too. The infinitesimals in $R^*$ are of three kinds: positive, negative and zero.

I think zero is not an extension in the set of real numbers.

Question 1: Can we call any real number a hyper-real number, too? For example, $2$ is a real number, can we say that $2$ is a hyper-real number?

Question 2: Does the set of hyper-real numbers $R^*$ include such infinitesimals say $\epsilon$, such that $-a<\epsilon<a$ for every positive real number $a$?

Addition: Is it true that if $\epsilon$ is a positive infinitesimal, then $\epsilon>0$. However, $-\epsilon$ which is a negative infinitesimal is less than zero. But $0, \epsilon$ and $-\epsilon$ are greater than any negative real number and are less than any positive real number?

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Q1: yes (second statement), Q2: yes (zero is included and is real) – kaine Jun 12 '13 at 19:22
In the future, please try to make the title of your question more informative (I've done it for you now). E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Jun 12 '13 at 19:27
up vote 5 down vote accepted

Yes to both questions:

Note that in your definition (second statement) the reals are among ("members of" and so included in) the hyper-reals.

Every real number is a member of $R^∗$, but $R^∗$ has other elements too.

And since zero is a real number, then, it is also in the hyperreals. And $-a < 0 \lt a$ for all positive $a$.

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I've extended my question. Please can you answer what I've added. Thanks. – Samama Fahim Jun 12 '13 at 20:43
Hyper reals contain numbers that are greater than any real number, and less then any real number; the inverse of these numbers are infinesimal numbers, also hyperreals. And hence, there are many infinitesimal numbers $\epsilon$ such that for positive real $a$, $-a\lt \epsilon \lt a$ – amWhy Jun 12 '13 at 21:07
@amWhy: Nice way to handle a moving target posting ... :-) +1 – Amzoti Jun 13 '13 at 0:30

Just like every natural number is also an integer, every integer is also rational, every rational is also real, every real is also a hyperreal. So yes, $2$ is a hyperreal number.

The system of hyperreal numbers contains many infinitesimal numbers $\epsilon$ that satisfy $-a<\epsilon <a$ for all positive real $a$. For instance, the hyperreal represented by $(1,1/2,1/3,1/4,\cdots)$.

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