# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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