Are real numbers also hyperreal? Are there hyperreal $\epsilon$ between $-a$ and $a$ for any positive real $a$?

Question

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?

Answer 1

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 $a$.

Answer 2

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 $(1,1/2,1/3,1/4,\cdots)$.