The answer is yes!

 

infinity*0= infinity (1-1)=infinity-infinity, which equals any number. because

 

infinity-infinity-3 is absorbed in infinity like a blackhole. and still equals infinity-infinity, likewise infinity-infinity-5 equals the same thing.

What?

The answer is yes!

 

infinity*0= infinity (1-1)=infinity-infinity, which equals any number. because

 

infinity-infinity-3 is absorbed in infinity like a blackhole. and still equals infinity-infinity, likewise infinity-infinity-5 equals the same thing.

Is Infinity - 99.9% of infinity still infinite?

Edited March 7, 201510 yr by Robittybob1

Keep in mind that my response here is in the context of the real numbers with addition and multiplication defined in the usual way.

The answer is yes!

infinity*0= infinity (1-1)=infinity-infinity, which equals any number. because

infinity-infinity-3 is absorbed in infinity like a blackhole. and still equals infinity-infinity, likewise infinity-infinity-5 equals the same thing.

The answer is yes, but not for the reason you claim.

Infinity is not a number, and thus arithmetic statements containing infinity, like [math]\infty - \infty[/math], aren't valid.

Rather, something like [math]0 \times \infty[/math] comes up in the context of some limiting process. The "indeterminate" aspect can be thought of as arising because we can take different "paths" towards [math]0 \times \infty[/math] depending on the limit in question, and arrive at different results.

Consider, for example, the following four limits, which all approach [math]0 \times \infty[/math] in the limit:

 

1. [math]\lim_{x \to \infty}0 \times x = 0[/math]

2. [math]\lim_{x \to \infty}\frac{1}{x} \times x = 1[/math]

3. [math]\lim_{x \to \infty}\frac{1}{2x} \times x = \frac{1}{2}[/math]

4. [math]\lim_{x \to \infty}\frac{1}{x} \times x^2 = \infty[/math]

Is Infinity - 99.9% of infinity still infinite?

"99.9% of infinity" isn't really valid, but if it were, then yes.