Prove that every hyperreal number between two infinitesimals is infinitesimal

(Very basic non-standard analysis: I wasn't sure where to post this)

My proof seems clumsy and probably wrong, so I was wondering if someone could help me out.

i. Let a and c be infinitesimal.

ii. Assume b is a hyperreal number that is a finite but not infinitesimal and lies in between a and c.

iii. Since c is infinitesimal but b is not, then b > c.

iv. But this is impossible since b lies in between a and c.

v. Therefore, by the above contradiction, if a and c are infinitesimal and b lies in between a and c, then b is infinitesimal.

Thanks for any help in clearing this up!

EDIT- here is a second attempt that I think is more concise and a bit more sound.

i. Assume a and c are infinitesimal.

ii. Let b be a hyperreal number such that a < b < c.

iii. Then b < c + a

iv. But c + a must be infinitesimal, since the addition of two infinitesimal numbers is infinitesimal.

v. Since b is less than an infinitesimal number, b is infinitesimal.

Is that better? Thanks again to anyone who reads this and tries to help.

Teeny, tiny proofs of teeny, tiny numbers or something like that...

I think in proof 1, you want to say "that is a finite number" instead of saying "that is a finite." iii. doesn't seem correct. Suppose b=-10,000. b is less than c, even if c is infinitesimal.

For proof 2, I don't see how you got iii. from ii.

I don't know if you have it, but I'd consider the "shadow" sh (equivalently standard part) of a, b, and c in your proof. Maybe something like this:

i. Assume a and c are infinitesimal.

ii. Let b indicate a non-infinitesimal hyperreal, such that c>= b >= a

iii. If b is not an infinitesimal, then its shadow sh(b) is also not an infinitesimal (it's shadows is a non-zero real number in this case).

iv. But, since c >= b >= a, and sh(c)=0, and sh(a)=0 (since both c and a are infinitesimal), we have 0 >= sh(b) >= 0.

v. But, of course, then sh(b)=0 an infinitesimal, which implies b as infinitesimal (contraposition on iii).

Problem comes as though, I don't know if you have the notion of a "shadow" or "standard part" to work with!