# Math Help - Prove that every hyperreal number between two infinitesimals is infinitesimal

1. ## 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.

2. ## 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!

3. No, I haven't come to shadow or standard part yet, at least not in any formal definition, but I believe I understand what they are (for 2x + ∆x, where ∆x is infinitesimal, and the result comes from the slope of y = x², the standard part is just 2x, right?).

As for where I got iii in my second attempt, I just kind of reasoned that if b is less than some number c, then b also has to be less than some number c plus some small positive number. Of course, I now see that I did not specify that a was positive, which is a problem.

Originally Posted by SunRiseAir
No, I haven't come to shadow or standard part yet, at least not in any formal definition, but I believe I understand what they are (for 2x + ∆x, where ∆x is infinitesimal, and the result comes from the slope of y = x², the standard part is just 2x, right?).
Yea, that's basically it. Looking at Henle's and Kleinberg's Infinitesimal Calculus p.40, we can define the shadow of a finite (my emphasis) hyperreal h, sh(h), as the unique real r infinitely close to h. So, for any finite hyperreal h there exists one and only one real number infinitely close to it. In other words, if h is some arbitrary finite hyperreal, then there exists a unique real number infinitely close to it. The converse does not hold though, since for any real number r there exist an uncountably infinity of finite hyperreals infinitely close to it. So, the shadow (or standard part) function sh does not have an inverse function. For two infinitesimals i_1, i_2, since sh(i_1)=sh(i_2) doesn't generally imply i_1=i_2, so sh is not injective (and thus we can't obtain a monomorphism). I'd hope it comes as clear that sh exhausts the co-domain R, or for all reals r in R, there exists at least one finite hyperreal i such that sh(i)=r, and thus sh qualifies as surjective (or onto). Once we join operations to the number systems, we can obtain an epimorphism, since we can set up a homomorphism from (i*, I_1, ..., I_n) to (r*, R_1, ..., R_n). Here, i* denotes the set of all finite hyperreal numbers, and I_1, ..., I_n indicate operations on i*, r* denotes the set of all real numbers, and R_1, ..., R_b indicate operations on r* such that arity of I_1 matches that of R_1, ..., the arity of I_n matches that of R_n. We have the homorphic equation sh(F(i_1, ..., i_n))=F'(sh(i_1), sh(i_2), ..., sh(i_n)) where F denotes any function on the finite hyperreals, while F' denotes any function on the reals. Or letting S to denote the shadow function we can more compactly write SFi_1,...,i_n=F'Si_1,...,Si_n.

More concretely this implies that we can define the shadow of a calculation using an operation in finite hyperreal arithmetic by real arithmetic (but NOT conversely since shadow doesn't have an inverse function... we only have a surjection here, not an injection also). For instance, suppose we have (i*, +') and (r*, +), or in other terms the hyperreals and reals under under addition respectively. So, let F=+', and F'=+ in the homomorphic equation sh(F(i_1, ..., i_n))=F'(sh(i_1), sh(i_2), ..., sh(i_n)). Since + and +' qualify as binary, we obtain sh(+'(i_1, i_2))=+(sh(i_1), sh(i_2)). So, for example, the shadow of the sum of say any infiniesimal 3' infinitely close to 3, any infinitesimal 4' infinitesimally close to 4, equals sh(3') plus sh(4')... 7. Of course, for non-standard calculus the catch comes as to realize that division of a finite hyperreal by any finite hyperreal isn't an operation on the finite hyperreals, since closure fails, but divison by 0 yields an *infinite* hyperreal, which implies division as closed on the hyperreals. Thus, computing a derivative can't get done by using the homomorphism shadow (or standard part) on a quotient directly, since it only got defined for finite hyperreals. Though, one can replace the numerator with equivalents and *eventually* use the homomorphism shadow. E. G. for
sh[((x+i)^2-x^2))/i] we can't compute it directly by using the above equality for the homophism sh, where i denotes an infinitesimal (sh(i)=0). But, since we can do arithmetic on infinitesimals, for the numerator we have ((x+i)^2-x^2))=(x^2+2xi+i^2-x^2)=2xi+i^2. So, given that i/i=1, we have sh([2xi+i^2/i]/i)=sh(2x+i). Now by the above homomorphism for shadow we have sh(2x+i)=sh(2x)+sh(i)=2x+0=2x, where x denotes any real number. So, sh[((x+i)^2-x^2))/i] =2x.

Hence I like the term "shadow" instead of the term "standard part" since the shadow of a finite hyperreal returns an image in the reals of a finite hyperreal.

I know I've digressed, but I have a feeling you might find some of that interesting. Sorry if I went to far away from your problem.

5. Originally Posted by SunRiseAir
(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.
The difficulty with this is that we do not know which of very many different approaches to non-standard analysis you are using.
But basically in each an infinitesimal is a hyper-real than is less that each positive real and greater than each negative real.
If b were not an infinitesimal then there is a positive real r such that $b\ge r$ which is a contradiction for a.
OR ELSE there is a negative real s such that $b\le s$ which is a contradiction for c.

So you had the right idea. And this way we avoid any but the most basic definitions.

Here is a free calculus textbook based on non-standard analysis.

6. Thanks Plato. The bold part is now obvious to me to be the key. I cannot forget that if b is not infinitesimal than there does exist a real number r that is less than or equal to it. Oddly, I did something similar on another proof (every positive hyperreal greater than some positive infinite number is positive infinite, and I assumed H is positive and infinite and H < K, then therefore any real number r has the relationship r < H < K, which means r < K, which means K is positive infinite).

Thanks for the online book.

Spoonwood, the information is definitely appreciated. I really want to try to understand this alternative approach to calculus. Maybe it's my bias, but it seems a lot more intuitive than the traditional "epsilon-delta" approach.

EDIT- so as to not bump the thread

Here's one final attempt (is this rigorous enough? I'm trying a more "standard" way):

Let a and c be two infinitesimal numbers, and b a hyperreal number such that a < b < c.

Since b is in between a and c, then |c - a| ≤ |c - b| + |b - a| by the triangle inequality. Then,

|c - a| ≤ |c - b - c + c| + |b - a - b + b|

|c - a| ≤ |c - b - c| +|c| + |b - a - b| +|b|

|c - a| ≤ |-b| +|c| + |-a| +|b|

|c - a| ≤ |b| +|c| + |a| +|b|

|c - a| ≤ 2|b| + |c| + |a|

|c - a| - 2|b| ≤ |c| + |a|

But since |c| + |a| is infinitesimal (because the sum of two infinitesimals is infinitesimal), then |c - a| - 2|b| is infinitesimal, since it is less than or equal to an infinitesimal number.

But it is possible for |c - a| - 2|b| to be infinitesimal only if either both |c - a| and 2|b| are infinite, both are finite but not infinitesimal, or both are infinitesimal.

Since |c - a| is infinitesimal, it therefore follows from above that 2|b| is infinitesimal, and it therefore follows that b is infinitesimal (since a real number times a infinitesimal number is infinitesimal).