# Is infinity times 0 indeterminate?

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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?

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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 by Robittybob1
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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 $\infty - \infty$, aren't valid.

Rather, something like $0 \times \infty$ 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 $0 \times \infty$ depending on the limit in question, and arrive at different results.

Consider, for example, the following four limits, which all approach $0 \times \infty$ in the limit:

1. $\lim_{x \to \infty}0 \times x = 0$

2. $\lim_{x \to \infty}\frac{1}{x} \times x = 1$

3. $\lim_{x \to \infty}\frac{1}{2x} \times x = \frac{1}{2}$

4. $\lim_{x \to \infty}\frac{1}{x} \times x^2 = \infty$

Is Infinity - 99.9% of infinity still infinite?

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

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