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What on earth was the reason for choosing 23i? It's neither largest in cardinality nor magnitude

While 14+9 = 23, |23i| > |14+9i| for the tie breaker (-98i is what I consider the smallest)

Do you consider 1-i, -1+i, and 0 to be equal size?

They have the same sum so I would consider their absolute values. If two numbers have the same absolute value then I consider the fact that -1+I has a negative real part.

I rank them -1+i, 0 , 1-i.

Do I understand correctly that your size function is required to be single valued, so you use competing definitions of size as tie-breakers? First sum of components, then magnitude, then real component cardinality? Why use this complicated system rather than choosing one and allowing same size numbers?

The complex are not ordered. Any answer other than N/A would be incorrect.

No one said they had to be ordered to have a size

Does this resolve to a poll?

This resolves to whatever I think is the biggest, but yall are able to try and change my mind before it happens.

If biggest = cardinality, then 4i+19

If biggest = magnitude, then -98i

If you count infinity as a real number rather than an undefined limit, then 1/0

Giving 1/0 any value, including "infinity", means that multiplication no longer has a zero. That's a little bit of breakage.

what do you mean by no longer having a zero?

@TheAllMemeingEye I think what they mean is that if 1/0 = thing, then thing * 0 = 1, so multiplication with zero would cease to make sense. Additionally, it would lead to the conclusion that infinity = - infinity

In any case, even if you count infinity as some number you can do math with, I don't think 1/0 is the answer because the limit of 1/0 would depend on how you approach 0

I haven't studied the academic arguments for and against infinity, but my layman intuition is that

1/0 = ± infinity

+infinity*0 = 1

1*0 = 0

+infinity != -infinity

cardinality(+infinity) >> cardinality(4i+19)

magnitude(+infinity) >> magnitude(-98i)

Zero has a special property in multiplication: X × 0 = 0 for all X. ("Anything times zero is zero.") This is related to how multiplication relates to addition: if you add together zero copies of a number, we expect the result to be zero.

Division and multiplication are inverses: if A = B ÷ C, then C × B = A. ("If you divide A by B, then multiply the result by B, you get the original A back.") This is how division is normally defined.

If 1 ÷ 0 = ∞, then the inverse rule means ∞ × 0 = 1. But this contradicts the zero rule.

So if you allow 1 ÷ 0 = ∞ then you either have to give up the zero rule or the inverse rule. Either division is no longer the inverse of multiplication (weird choice) or multiplication no longer relates to addition in the obvious way (also a weird choice).

Right, thanks for explaining

My intuitive response is to discard the addition relation rule, since dividing by decimal or negative numbers already violates its obvious relation to physical division

@KarlK Even if you reject ∀x, x * 0 = 0, accepting the inverse rule still leads to absurdity, right?

Assuming that we're treating infinity as a number and can apply standard arithmetic operations to it, all the dividing by zero 1 = 2 proofs suddenly work, for instance. Additionally, if 1/0 = both + and - infinity, then 2 infinity = 0 assuming you allow addition and subtraction for that number

What would be an example of a 1=2 proof that follows from this without repeatedly jumping between rule sets?

@TheAllMemeingEye Could you specify what following rules you would like? Based on what I've interpreted, if we treat infinity as a number

1) if we apply the reflexive property, or a = a, to infinity and say that infinity = infinity

2) if we apply a = b, b = c => a = c

3) if a/b = c, cb = a must be true

You could conjure up something really simple like

1) 1 / 0 = infinity, so 0 × infinity = 1
2) 1 / (0.5 × 0) = 1 / 0 = infinity
3) 1 / 0 = 1 / (0.5 × 0)
4) 1 / (0.5 × 0) = 1/0.5 × 1/0
5) 1/0.5 × 1/0 = 1/0
6) 2 = 1

I suppose you could deny that 0.5 times 0 is 0 and insist that it's actually 0.5/inf, but I don't think that works because you'd have to deny that 1 times 0 is 0, and that it's actually 1/inf, and that 1/inf isn't zero, which would lead to a contradiction

Or actually actually last response this time, I suppose you could simply define 0 as 1/infinity and only 1/infinity, not 2/infinity, not -4/infinity, ...

But you'd have to acknowledge the property that two non-zero numbers which have the same sign can add to 0, and two nonzero things can multiply to zero

Fair enough, that makes sense, my intuition that you can have different multiples of infinity only works if you treat zero as an infinitesimal rather than true zero I guess