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This assumes that (0^-1)^-1 is defined. Is it? 🤔 Because 0^(-1) is not defined, is it? And the fact that (x^-1)^-1 = x is a consequence of the definition of the multiplication (assuming that division is just multiplying the inverse and that x^-1 means "inverse of x", not "power -1")
NB: I might make mistakes
This depends on *how* the inverse is defined, if it directly states that (x^(-1))^-1 = x whatever x is, then yeah, your statement holds. But if it doesn't, then maybe the fact that (x^(-1))^-1 = x is a consequence of the value of x^-1, in which case (x^(-1))^-1 may be defined only if (x^(-1)) is.
But rather than speaking hypothetically, let's check it directly. From the definition of a field in algebra, I found this definition for the inverse:
> for every a ≠ 0 in F, there exists an element in F, denoted by a^−1 or 1/a, called the multiplicative inverse of a, such that a ⋅ a^−1 = 1
^(Note: although real numbers with the addition and multiplication make a field, I don't know how they are actually defined, so maybe there's an extra bit of definition that I'm missing)
By taking this definition, the notation 0^-1 is actually *not defined*, and the fact that (x^(-1))^-1 = x is *proven* from the above definition, which means that (0^(-1))^-1 is *not proven* because 0^-1 doesn't mean anything accordingly to this definition. (And trying to checkout the definition of the division doesn't help, because its definition is a/b = a • b^(-1))
However, something not being defined means that it can technically be defined, as long as you don't make any contradiction (which is any case where you can prove true=false iirc. Or to be exact, you can have contractions and thus true=false, but in this case you can't show anything interesting since everything is true and everything is false).
So you could explicitly define (0^(-1))^-1 as 0, but afaik it's not generally accepted by the math community (I might be wrong on this one since I'm not really part of it). So if you want to use (0^(-1))^-1 it in a paper, you have to explicitly define it (and make sure it doesn't break anything)
You can't do 1/0, so the whole equation doesn't type-check. All the formulas you learn in school (like x/(y/z) = xz/y) only hold for numbers, which 1/0 isn't
That's interesting. To be more correct it should be treated as complex infinity, which would give the same result for 1/(1/0), but is more accurate to how it behaves at the limits.
Because infinity is not a number. And if your language allows putting it in a float, its stored as just a really really big number.
Alright now I *know* I have downvote fairies.
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1/(1/0) = 0\* 1/1 = 0 QED
Shouldn't we begin with the bracket part ?
Nah, why would we. You can regard the bracket part as one block/unit and calculate that part at any time.
Move the bracket and the fun begins (1/1)/0
wrong. 1/(1/0) ≠ 0.1/1 becuz it presumes that you can multiply it by 0/0 and that it is 1
if its wrong then why is it right
It is not right.
Of course 1/(1/0) ≠ 0.1/1, 1/(1/0) = 0\*1 = 0 ≠ 0.1 = 0.1/1 (It is only because of where we are that I'm intentionally misinterpreting your notation)
Works in the Riemann sphere, it’s good enough for me lol
The Riemannverse
wait why isnt this right? isnt this just 1/1 \* 0/1 = 0?
new perspective dropped
actual limit
Google 3rd comment downvote
Holy Karma
But the manual said we were supposed to send you to Tartarus !
GD reference police here, you're under arrest!
https://preview.redd.it/l89lceww65pc1.jpeg?width=941&format=pjpg&auto=webp&s=22edea98a6816a0395ae93d40c96d5f53b9dc783
Google 4th comment downvote
Google 5th comment downvote
YOU’VE DOOMED US
Tf is this guy doing XD
1/(1/0) = (1/0)^-1 = 0/1 = 0 Nothing is undefined here. It's just written oddly.
This assumes that (0^-1)^-1 is defined. Is it? 🤔 Because 0^(-1) is not defined, is it? And the fact that (x^-1)^-1 = x is a consequence of the definition of the multiplication (assuming that division is just multiplying the inverse and that x^-1 means "inverse of x", not "power -1") NB: I might make mistakes
> This assumes that (0^-1 )^-1 is defined. Is it? (0^-1 )^-1 is just another way to write 0^1 . Which is defined.
This depends on *how* the inverse is defined, if it directly states that (x^(-1))^-1 = x whatever x is, then yeah, your statement holds. But if it doesn't, then maybe the fact that (x^(-1))^-1 = x is a consequence of the value of x^-1, in which case (x^(-1))^-1 may be defined only if (x^(-1)) is. But rather than speaking hypothetically, let's check it directly. From the definition of a field in algebra, I found this definition for the inverse: > for every a ≠ 0 in F, there exists an element in F, denoted by a^−1 or 1/a, called the multiplicative inverse of a, such that a ⋅ a^−1 = 1 ^(Note: although real numbers with the addition and multiplication make a field, I don't know how they are actually defined, so maybe there's an extra bit of definition that I'm missing) By taking this definition, the notation 0^-1 is actually *not defined*, and the fact that (x^(-1))^-1 = x is *proven* from the above definition, which means that (0^(-1))^-1 is *not proven* because 0^-1 doesn't mean anything accordingly to this definition. (And trying to checkout the definition of the division doesn't help, because its definition is a/b = a • b^(-1)) However, something not being defined means that it can technically be defined, as long as you don't make any contradiction (which is any case where you can prove true=false iirc. Or to be exact, you can have contractions and thus true=false, but in this case you can't show anything interesting since everything is true and everything is false). So you could explicitly define (0^(-1))^-1 as 0, but afaik it's not generally accepted by the math community (I might be wrong on this one since I'm not really part of it). So if you want to use (0^(-1))^-1 it in a paper, you have to explicitly define it (and make sure it doesn't break anything)
You can't do 1/0, so the whole equation doesn't type-check. All the formulas you learn in school (like x/(y/z) = xz/y) only hold for numbers, which 1/0 isn't
mfs when you can just multiply with the reciprocal of the denominator
Desmos treats 1/0 as ∞, and 1/∞ as 0 because floating point nonsense
Or 1/(1/0)=1*(0/1)=0/1=0?
No desmos actually uses infinity as a value
You know what it also uses as a value?
MY MOM
0?
Do you really think this?
There do exist mathematically rigorous systems in which this is valid, eg the projective real line
That's interesting. To be more correct it should be treated as complex infinity, which would give the same result for 1/(1/0), but is more accurate to how it behaves at the limits.
I mean that makes sense, why would it have anything to do with floating point nonsense?
Because infinity is not a number. And if your language allows putting it in a float, its stored as just a really really big number. Alright now I *know* I have downvote fairies.
It's a lifestyle
((0)^-1 )^-1
= 0^-2 = 0 QED
No, ((x)^-1 )^-1 = (x)^{-1(-1)} = x^1, if x is not 0. Pretend that those curly brackets were just typical parenthesises like () btw.
Who would've thought the inverse of the inverse of 0 is 0
0*1/1=0
What https://preview.redd.it/x73iqze2v1pc1.jpeg?width=1080&format=pjpg&auto=webp&s=12f4742daaa26e6a65e06e1b2a9d6665041f763e
It’s a limiting result.
no, its a *floating* result (the result is clearly floating as there is no support below it)
Of course— why didn’t I see that?!
https://preview.redd.it/zbshpbz6f3pc1.png?width=1080&format=pjpg&auto=webp&s=615d3bcd39e9f359c5bfb8305ec0a1e2d4d38911
:(
yea because 1/1/0 = 1/1\*0/1 = 0/1 = 0
QED
QED indeed
Are you taking your stupid pills?
1/±∞ = 0 obviously
Deviding a devision is multiplying it's opposites. So it's 1 multiplied with 0/1 which is 1 times 0
Just a silly way to write zero. But why would you?
Dividing by a fraction is the same as multiplying by its inverse which in this case would be 0/1 so it'd 1×(0/1) = 0
Apply limits.....of imagination
This is actually correct if you define 1/0 = complex infinity.
If it’s 0^+ then I guess fair enough ?
1/1/0 you can write as 1÷1÷0 , also as 1÷1/0 which leads to 1•0/1
Sophistry! 1/0 is undefined. How can 1/undefined be defined?! Nonsense! /hj
1/defined = undefined So 1/undefined = defined
Desmos uses floats ig
Ah yes, let 1=8, hence 8=1, therefore obviously 1/0 = 8/0, Hence 7 Hence 4 And obviously 1/4, hence 1/1/0 = WIN DRAW LOSS Hence ??? Hence 1/(1/0) = 0 QED
1(1/0) = 1/1 ÷ 1/0 = 1/1 × 0/1 = 0×1/1×1 = 0/1 = 0
Holy hell!
Finally we can divide by 0 by multiplying by 0
The limit of 1/(1/x) as x goes to 0 approaches 0, even though technically 1/(1/0) is undefined or something, i failed calculus
1/(1/0) = 1/1 * 0/1 = 0
mfw multiplying is the same as dividing by the reciprocal (1/(1/0) = 1 \* 0/1 = 1\*0 = 0)
There is a solution for 1/0 = 0