# Operations with infinity

1. May 3, 2014

### Jhenrique

Is correct to state that:

$x \infty = \text{sgn}(x) \infty$

$\infty^x = \infty^{\text{sgn}(x)}$

?

2. May 3, 2014

### HallsofIvy

If you are talking about the "usual" real number system, the are NO "operations with infinity" because "infinity" is not a real number. And there are several different ways to create number systems which include "infinity" as a number. Which are you talking about?

3. May 4, 2014

### Jhenrique

Your answer complicated more the things... look, exp(-∞) = 0, exp(0) = 1, exp(∞) = ∞... appears make sense make operation with ∞...

4. May 4, 2014

### HallsofIvy

Not with the usual real number system, they don't- except a short hand for limits.
(Not "exp(0)= 1". I have no problem with that and I don't know why you included it here.)

5. May 4, 2014

### Jhenrique

For shows that infinity acts like a number...

6. May 4, 2014

### micromass

They make sense in some number systems equipped with infinity, but not in others. For example, on the affine real line, the above operations are true, see http://en.wikipedia.org/wiki/Extended_real_number_line
But on the projective real line, they are false: http://en.wikipedia.org/wiki/Real_projective_line
There are many other systems which allow an infinity and where the above might make sense or not, so you need to specify.

I don't see what $e^0 = 1$ has to do with infinity.

And infinity is certainly not a real number. It might act like on in some ways, but not in others.

7. May 4, 2014