Dividing by 0,what happens?

[Kirby71]

Ok,so i thought a number/0=∞,but randomly i learnt may not be like that,sooo,x/0 causes too many paradoxes like 7x0=0 but 7/0Ôëá0,but if you get into density formula,a D= m/v and a black hole singularity has 0 volume basically seeing how a singularity has theorically ∞ density,that means x mass/0=∞.

Things doesn't fit very well with division by 0 problems,also remember how 0x∞=0 (0+0+0+0+0+0... And on going)but ∞/0Ôëá0.

So basically,per example,if a character has 0 joules,if the Guy boosts and multiplies his 0 infinite times,he isnt going to even get 0.1 j of energy so even if you go 0+0+0 infinitely,you aren't increasing,this is relationed with higher-D characters,and character 1-D below you is >∞ times weaker than you,so he is 0 to you,even if he multiplies by infinite (high 3-A and you low 2-C per example) as he is 0 to you he isnt even going to be 1 to you,how is this relationed with dividing by 0 and special calcs by 0?

Why those things happen?

 

[Antoniofer]

0 per infinite isn't 0, is an idetermination, same way with 0/0 or infinite/infinite.

 

[Theglassman12]

0 per infinite isn't 0, is an idetermination, same way with 0/0 or infinite/infinite.

^^^

 

[Kirby71]

0 per infinite isn't 0, is an idetermination, same way with 0/0 or infinite/infinite.

Yet more confusing.

I mean if you are left forever adding 0s you won't ever reach 0.000001.,you will be left at 0

 

[ZacharyGrossman273]

Beyond infinite, technically. The number would have to be so high when you multiply it and zero, you get an actual number. Somehow.

 

[Antoniofer]

I mean if you are left forever adding 0s you won't ever reach 0.000001.,you will be left at 0

No you don't, the result is indefinite, that's means there's infinite solución in the real numbers range.

 

[Kirby71]

The thing is,that per example,the higher and below dimensional comparision high 3-A is infinite, low 2-C confronts not high 3-A,the tier 2 guy sees the opponent's powerlevel as 0 because he below dimensional so he is infinitely weaker,but then the 3-D boy multiplies his power by Infinity and becomes high 3-A so basically (the tier 2 sees the 3-D boy as 0) 0x∞,and after multiplying his power by infinite and being high 3-A,he is still infinitely weaker! So basically the 3-D being yet becoming infinite he still 0 to the tier 2,then laughs at the opponent seeing how doing 0x∞ he is still 0,so the high 3-A boy has a last option:become 4-D and with that,multiply himself beyond infinite (uncontable infinity) to do not be 0.

So thats why 0x∞ its still 0,0+0+0+0... And so forth,you get stuck at 0.

 

[Antoniofer]

Dimensional stuff works differently, mathematical principle do not apply between them.

 

[Kirby71]

I mean if you are left forever adding 0s you won't ever reach 0.000001.,you will be left at 0

No you don't, the result is indefinite, that's means there's infinite solución in the real numbers range.

So the indefinite results like x/0 and ∞/∞ can be any number?

And dimensional stuff works differently.

Also,if 0x∞ ocurres,how Will be indefinite? Is the infinite number breaking the rule of (number)x0=0?

 

[Antoniofer]

X/0 is not an indetermination, the result is ∞; the list of indeterminations are: 0*∞, 0/0, ∞/∞, and ∞-∞, any other equation has a determinated result.

 

[RRTheEndMan]

isn't ∞^0 indetermined too, since it is equal to ∞^1/∞^1?

 

[Antoniofer]

Yeah, ∞^0, 0^0 and 1^∞ are other indeterminations.

 

[Kirby71]

Ok thanks.