# I Can I do this for hyperreals?

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1. Dec 10, 2016

### Buffu

I need to prove whether the given expression is finite but not infinitesimal/infinite/infinitesimal.$$H-K\over H^2 + K^2$$ Where $H, K$ are +ve infinite numbers.

I did the following,

Let $H = \alpha K, \alpha \in R$
then,
$${\alpha K-K\over \alpha^2K^2 + K^2} \implies {\alpha - 1\over K(\alpha^2 + 1)}$$
Since ${\alpha - 1\over (\alpha^2 + 1)}$ is finite and $K$, infinite. Thus the given expression is infinitesimal.

I am not sure if this is correct.
Am I correct ?

2. Dec 10, 2016

### Staff: Mentor

It all depends on the assumption $H=\alpha K$ with $\alpha \in R$.
So a) does it mean $\alpha \in \mathbb{R}$ or $\alpha \in {}^*\mathbb{R}$ and b) this is meant by "+ve infinite"?

I'm not used to hyperreals, so maybe the question is silly. But I try to rule out cases like $H=K^L$ or $H=\varepsilon + K$ in which it becomes intricate. If you may assume $H= \text{ real }\cdot K$, then your argument seems to be correct.

3. Dec 10, 2016

### Buffu

I meant $\alpha \in \mathbb{R}$ ie set of reals.
If I am right then for a infinitesimal $\varepsilon$, there are two infinities $1/\varepsilon$ and $-1/\varepsilon$.
Former being larger than every real and latter being smaller than every real.

My main problem is that I don't know if my method proves that the given expression is infinitesimal for all the possible relations between $H,K$, some which you mentioned,$H=K^L, H=\varepsilon + K...$

Do you know any other method to prove this ?

4. Dec 10, 2016

### Staff: Mentor

As far as I can see, the infinite hyperreals are basically defined by inverse infinitesimals.
So I would write $\varepsilon = \frac{1}{K}$ and $\eta = \frac{1}{H}$.
This leads to an expression $(\varepsilon - \eta)\frac{\varepsilon \cdot \eta}{\varepsilon^2 + \eta^2}$ which boils it down to the difference of two infinitesimals around zero: $\varepsilon - \eta$. Maybe one can formally work with $st$ from here on. It is in principle the same argument as yours, I simply found more sources dealing with infinitesimals than with infinites.

5. Dec 10, 2016

### Buffu

Sum/product of two infinitesimal is infinitesimal. But the quotient is indeterminate form.

So, $(\varepsilon - \eta)\frac{\varepsilon \cdot \eta}{\varepsilon^2 + \eta^2}$ =infinitesimal(infinitesimal/infinitesimal) = indeterminate.
How should I proceed now ?
I don't get this part ??

6. Dec 10, 2016

### Staff: Mentor

I'm not sure, whether the above is a valid transformation or not and you should go on. $H+K$ or $\frac{\varepsilon}{\eta}$ are already indeterminate, i.e. the question itself has indeterminates. My thought has been, that $\frac{\varepsilon \eta}{\varepsilon^2 + \eta^2}=1$ (I think) and $\varepsilon - \eta$ is left. Then I would have proceeded like this:
$$\varepsilon - \eta = \varepsilon + (-\eta) = \varepsilon + \delta \text{ all infinitesimal }$$

7. Dec 10, 2016

### Buffu

$\frac{\varepsilon \eta}{\varepsilon^2 + \eta^2}=1$
Why do you think that ? I want to know. Most likely you are right, I don't know much non-standard analysis.

Most questions in the exercise were in indeterminate form, we need to transform them to determinate form using algebra, and then find there nature.

8. Dec 12, 2016

### Buffu

9. Dec 12, 2016

### haruspex

Not an area I've ever indulged in before, but from a bit of reading...
Seems to me we have to allow for H and K (or ε and η) being of different orders. So assuming a ratio which is real is not valid.
The sum of squares term will necessarily be dominated by the higher infinite order (lower infinitesimal order) if they are different. So $\frac{\epsilon\eta}{\epsilon^2+\eta^2}$ is either real or infinitesimal. it cannot be infinite. You can arrive at the same using H and K.

10. Dec 12, 2016

### Staff: Mentor

How can it be real? If they are of the same order, we have the right approach above. If one is of higher order, it should dominate, so we divide (lower order)/(higher order) - the result should still be infinitesimal.

11. Dec 12, 2016

### haruspex

Yes, my fault, I was just looking at the $\frac{\eta\epsilon}{\eta^2+\epsilon^2}$ term being discussed. Now I look back at the oroginal post I agree with you - must be infinitesimal.

12. Dec 12, 2016

### Buffu

I will give a try to complete solution :-

Case 1: - $H > K$

$${H - K \over H^2 + K^2 } = {1/H - K/H^2 \over 1 + K^2/H^2 } = {\text{infinitesimal} - \text{infinitesimal} \over \text{finite} +\text{infinitesimal} } = {\text{infinitesimal} \over \text{finite}} = \text{infinitesimal}$$

Case 2:- $K > H$
$${H - K \over H^2 + K^2 } = -\left( { K - H \over H^2 + K^2 }\right)$$
Then we proceed as above to get negative infinitesimal.

Case 3:- $H = K$
$${H - K \over H^2 + K^2 } = 0$$
Which is a real infinitesimal.

@haruspex , @mfb , @fresh_42

Is this correct ?

13. Dec 13, 2016

### haruspex

Your argument does not work in case 3. H-K need not be zero. It could as much as the same infinity as H and K.
Use a method similar to the other cases.

14. Dec 13, 2016

### Staff: Mentor

H=K could be meant literally. But then K2/H2 is "infinitesimal or finite".

15. Dec 13, 2016

### Buffu

Finite but not infinitesimal, because then $K^2/H^2 = 1$, am I correct ?

Last edited: Dec 13, 2016
16. Dec 13, 2016

### Buffu

Why ? $H - K = H - H = 0$.

17. Dec 13, 2016

### Staff: Mentor

That's not what I meant, but my post was a bit short:
If K=H means literally identical in case 3, then H>K includes every case where H>K, including (for example) H=w+1, K=w, in this case K2/H2 is finite.

From H>K, you cannot conclude that K2/H2 is infinitesimal. It could be, but you don't know, it could also be finite.

18. Dec 13, 2016

### Buffu

$H = w + 1, K = w$
$w = 1/\varepsilon$

$${H\over K} = { w + 1\over w } = {{ 1 + \varepsilon \over \varepsilon } \over 1/\varepsilon} = { 1 + \varepsilon} = \text{Finite}$$

So how should I solve the question now.

19. Dec 13, 2016

### Staff: Mentor

The way you did, just replace "infinitesimal" by "finite or infinitesimal" at the place we discussed.

20. Dec 13, 2016

### Buffu

I did not get what you said , Should I replace "Infinitesimal" everywhere with "Infinitesimal or finite".