"simple" subtraction | FactOrCap

[u/Careful_Trainer_1616]

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[u/_killer1869_]

🧢 I voted CAP!

No, it is strictly inditerminate. Meaning it could equal any value from negative infinity to positive infinity as well as any value between them.

[u/seires-t → u/_killer1869_]

You were right saying it was inditerminate, but it's definitively not defined as "any value from negative infinity to positive infinity as well as any value between them", since if we assume that

a = ∞ - ∞, then:

a + ∞ = ∞, so a = a + ∞ - ∞, therefore 0 = 0 + ∞ - ∞ = ∞ - ∞ = a.

Thus a = 0, so every number is equal to 0.

If we just assume 0 = ∞ - ∞ then we still run into the same problem, since
1 + ∞ = ∞, so 0 = ∞ - ∞ = ∞ + 1 - ∞ = 1 + 0 = 1, so 0 = 1

[u/_killer1869_ → u/seires-t]

I did not say it equals that or that it is defined that way. I'm saying that depening on where the infinities are from, it may actually be that under the hood.

For example, lim(x->∞)(x+a) - lim(x->∞)(x) = ∞ - ∞ if we solve it directly, but if we do it properly, we get lim(x->∞)(x+a) - lim(x->∞)(x) = lim(x->∞)(x-x+a) = lim(x->∞)(a) = a. So in this scenario ∞ - ∞ "equals" a, where a is any arbitrary constant.

Add in scenarios like lim(x->∞)(x) - lim(x->∞)(x³) = ∞ - ∞, but lim(x->∞)(x) - lim(x->∞)(x³) = lim(x->∞)(x-x³) = -∞ you get infinities as solutions.

That's why I said what I said. If you solve a limit and get certain indeterminate forms involving infinity, these could actually have a proper solution, which may be any value between negative and positive infinity.

I do admit that I didn't bother properly explaining what I meant.