Math geeks are too stupid to fix this

azurepudding · 2018-09-29 02:18:12

LukeM wrote:

ILikeTofuuJoe wrote:
Mathmeticans say 0/0 is invalid only because they don't want to violate that /0 is invalid. Proof 0/0 = 0:
0^2/0^1 = 0^1 = 0.
They aren't just doing it for the sake of it, 0/0 physically has to be undefined or the whole of maths doesn't work
I'm not so sure about 0^0, I haven't been able to find any concrete evidence for why it can't be 1 and a lot of people do seem to say they think it should be, but 0/0 definitely has to be undefined

0^0 = 0/0 tho

x^3/x^2 = x
x^3/x^3 = x^0 = 1
x^3/x^4 = x^-1 = 1/x

Right? But what if x = 0?

0^3/0^2 = 0/0 = undefined
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined

Anything to the power of zero equals itself over itself (x/x), so you can't have 0 to the power of zero cuz it equals 0/0. You also cannot divide by any power of 0, unless you have some form of addition or subtraction to keep the denominator from totaling 0.

ILikeTofuuJoe · 2018-09-29 02:21:08

azurepudding wrote:

LukeM wrote:
ILikeTofuuJoe wrote:
Mathmeticans say 0/0 is invalid only because they don't want to violate that /0 is invalid. Proof 0/0 = 0:
0^2/0^1 = 0^1 = 0.
They aren't just doing it for the sake of it, 0/0 physically has to be undefined or the whole of maths doesn't work
I'm not so sure about 0^0, I haven't been able to find any concrete evidence for why it can't be 1 and a lot of people do seem to say they think it should be, but 0/0 definitely has to be undefined
0^0 = 0/0 tho
x^3/x^2 = x
x^3/x^3 = x^0 = 1
x^3/x^4 = x^-1 = 1/x
Right? But what if x = 0?
0^3/0^2 = 0^1 = 0
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined
Anything to the power of zero equals itself over itself (x/x), so you can't have 0 to the power of zero cuz it equals 0/0.

0^0 is 1 okay. But they really need to get rid of the undefined values. Not like theres not enough information.

azurepudding · 2018-09-29 02:23:17

ILikeTofuuJoe wrote:

azurepudding wrote:
LukeM wrote:
ILikeTofuuJoe wrote:
Mathmeticans say 0/0 is invalid only because they don't want to violate that /0 is invalid. Proof 0/0 = 0:
0^2/0^1 = 0^1 = 0.
They aren't just doing it for the sake of it, 0/0 physically has to be undefined or the whole of maths doesn't work
I'm not so sure about 0^0, I haven't been able to find any concrete evidence for why it can't be 1 and a lot of people do seem to say they think it should be, but 0/0 definitely has to be undefined
0^0 = 0/0 tho
x^3/x^2 = x
x^3/x^3 = x^0 = 1
x^3/x^4 = x^-1 = 1/x
Right? But what if x = 0?
0^3/0^2 = 0^1 = 0
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined
Anything to the power of zero equals itself over itself (x/x), so you can't have 0 to the power of zero cuz it equals 0/0.
0^0 is 1 okay. But they really need to get rid of the undefined values. Not like theres not enough information.

I made an edit as I made an error in one example and added some clarification.

And.. why? If something cannot happen, it cannot happen- it's not because we just don't feel like it, it's because it cannot happen.

LukeM

ZeldaXD · 2018-09-29 02:31:50

This is probably the most retarded post I've ever seen.

0 to the power of 0 is undefined, because 0 to the power of anything is 0, but anything to the power of 0 is 1, so 0^0 presents a conflict. Depending on the function, 0^0 could be either 0 or 1. Take this as an example of a function returning 1.

Take f(x) = x^x
lim (x -> 0) of f(x) = lim (x -> 0) of x^x = 0^0
Thus take M = lim (x -> 0) of x^x
ln M = ln [lim (x-> 0) of x^x]
ln M = lim (x -> 0) of (ln x^x)
ln M = lim (x -> 0) of (xln x)
ln M = lim (x -> 0) of [(ln x) / (1 / x)]

Take lim (x -> 0) of [(ln x) / (1 / x)] and apply L'Hôpithal's rule
lim (x -> 0) of [(1 / x) / (-x^(-2))]
lim (x -> 0) of -x = 0

thus
ln M = 0
for ln M = 0, we need e^0 = M
M = 1

In this case, x^x = 1, but could be anything else for another problem, thus being undefined.

PS: Square root of -1 does not exist, sqrt(-1) = i, i standing for "imaginary", it's not a real solution, it's an imaginary number.

Zumza, Minimania

HeyNK · 2018-09-29 02:42:13

We really need to make more math threads so that interneet teenagers can show off how much they know about math before being humiliated by a virtual anime girl.

ILikeTofuuJoe · 2018-09-29 02:52:38

ZeldaXD wrote:

This is probably the most retarded post I've ever seen.
0 to the power of 0 is undefined, because 0 to the power of anything is 0, but anything to the power of 0 is 1, so 0^0 presents a conflict. Depending on the function, 0^0 could be either 0 or 1. Take this as an example of a function returning 1.
Take f(x) = x^x
lim (x -> 0) of f(x) = lim (x -> 0) of x^x = 0^0
Thus take M = lim (x -> 0) of x^x
ln M = ln [lim (x-> 0) of x^x]
ln M = lim (x -> 0) of (ln x^x)
ln M = lim (x -> 0) of (xln x)
ln M = lim (x -> 0) of [(ln x) / (1 / x)]
Take lim (x -> 0) of [(ln x) / (1 / x)] and apply L'Hôpithal's rule
lim (x -> 0) of [(1 / x) / (-x^(-2))]
lim (x -> 0) of -x = 0
thus
ln M = 0
for ln M = 0, we need e^0 = M
M = 1
In this case, x^x = 1, but could be anything else for another problem, thus being undefined.
PS: Square root of -1 does not exist, sqrt(-1) = i, i standing for "imaginary", it's not a real solution, it's an imaginary number.

You don't understand so get out.

Pingohits

ZeldaXD · 2018-09-29 02:53:15

Regarding 0/0... take these two examples.

f(x) = (1-(x^2))/(2x-2)
lim (x -> 1) of f(x)
lim (x -> 1) of [(1-(x^2))/(2x-2)] = (1-1^2)/(2-2) = (1-1)/(2-2) = 0/0 which is undefined.
Now, apply L'Hôphital's rule and derivate:

lim (x -> 1) of [(0-2x)/(2-0)]
lim (x -> 1) of (2x/2)
lim (x -> 1) of (x/1)
lim (x -> 1) of x = 1

0/0 in this case is 1.

Let's look at another example where 0/0 is 0.

g(x) = (1 - cos(x))/(sin (x))
lim (x -> 0) of g(x)
lim (x -> 0) of [(1 - cos(x))/(sin (x))] = (1 - cos(0) / sin(0)) = (1 - 1) / 0 = 0/0
Now, apply L'Hôphital's rule and derivate:

lim (x -> 0) of [(sin(x))/(cos x)] = sin(0) / cos(0) = 0 / 1 = 0

0/0 in this case is 0.

ILikeTofuuJoe

ILikeTofuuJoe · 2018-09-29 03:01:20

ZeldaXD wrote:

Regarding 0/0... take these two examples.
f(x) = (1-(x^2))/(2x-2)
lim (x -> 1) of f(x)
lim (x -> 1) of [(1-(x^2))/(2x-2)] = (1-1^2)/(2-2) = (1-1)/(2-2) = 0/0 which is undefined.
Now, apply L'Hôphital's rule and derivate:
lim (x -> 1) of [(0-2x)/(2-0)]
lim (x -> 1) of (2x/2)
lim (x -> 1) of (x/1)
lim (x -> 1) of x = 1
0/0 in this case is 1.
Let's look at another example where 0/0 is 0.
g(x) = (1 - cos(x))/(sin (x))
lim (x -> 0) of g(x)
lim (x -> 0) of [(1 - cos(x))/(sin (x))] = (1 - cos(0) / sin(0)) = (1 - 1) / 0 = 0/0
Now, apply L'Hôphital's rule and derivate:
lim (x -> 0) of [(sin(x))/(cos x)] = sin(0) / cos(0) = 0 / 1 = 0
0/0 in this case is 0.

Who knew a little 0 could cause so much trouble.

Zumza · 2018-09-29 11:09:13

ILikeTofuuJoe, each mathematical operation has a domain for the values involved.

For the dividing operation a / b, a ∈ ℝ, b ∈ ℝ*

For the exponential operation a ^ 0 = 1, a ∈ ℝ*

Where ℝ* is (-∞, 0) ∪ (0, ∞)

There are rules for mathematical operations. This rules aren't arbitrary but deduced. If you violate a rule you violate the principles and the logic behind the operation thus the result won't be valid.

LukeM, drunkbnu

LukeM · 2018-09-29 11:10:30

azurepudding wrote:

0^0 = 0/0 tho
x^3/x^2 = x
x^3/x^3 = x^0 = 1
x^3/x^4 = x^-1 = 1/x
Right? But what if x = 0?
0^3/0^2 = 0/0 = undefined
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined
Anything to the power of zero equals itself over itself (x/x), so you can't have 0 to the power of zero cuz it equals 0/0. You also cannot divide by any power of 0, unless you have some form of addition or subtraction to keep the denominator from totaling 0.

Thats just a pattern though, not the definition, you could also define x^n as 1 * x * x * x (n times), in which case 0^0 would be 1. You could also take it to the extreme and use that to say that 0^1 is undefined (x^3/x^2 = 0/0 = 0^1 = undefined), which obviously isn't true.

ZeldaXD wrote:

This is probably the most retarded post I've ever seen.
0 to the power of 0 is undefined, because 0 to the power of anything is 0, but anything to the power of 0 is 1, so 0^0 presents a conflict. Depending on the function, 0^0 could be either 0 or 1. Take this as an example of a function returning 1.
Take f(x) = x^x
lim (x -> 0) of f(x) = lim (x -> 0) of x^x = 0^0
Thus take M = lim (x -> 0) of x^x
ln M = ln [lim (x-> 0) of x^x]
ln M = lim (x -> 0) of (ln x^x)
ln M = lim (x -> 0) of (xln x)
ln M = lim (x -> 0) of [(ln x) / (1 / x)]
Take lim (x -> 0) of [(ln x) / (1 / x)] and apply L'Hôpithal's rule
lim (x -> 0) of [(1 / x) / (-x^(-2))]
lim (x -> 0) of -x = 0
thus
ln M = 0
for ln M = 0, we need e^0 = M
M = 1
In this case, x^x = 1, but could be anything else for another problem, thus being undefined.
PS: Square root of -1 does not exist, sqrt(-1) = i, i standing for "imaginary", it's not a real solution, it's an imaginary number.

What about 0^-1? (its not 0)

Minimania · 2018-09-29 11:32:53

HeyNK wrote:

We really need to make more math threads so that interneet teenagers can show off how much they know about math before being humiliated by a virtual anime girl.

I don't get what the point of being both off-topic and toxic on a thread about math is going to accomplish. Either try to help come up with a solution or just leave. Where's "what you know doesn't matter" or "let's let all the kiddoes try before the real professionals get here" getting you in this thread? Nothing. Goodbye.

ILikeTofuuJoe wrote:

You don't understand so get out.

This isn't helping anyone either.

ILikeTofuuJoe · 2018-09-29 15:14:14

The whole point of this topic is I'm trying to say: Since x^m / x^n = x^(m-n), say x=0, m1=1, m2=2, and n=1. We get two equations: 0^1 / 0^1 = 0^(1-1), 0^2 / 0^1 = 0^(2-1). Simply both equations, we get: 0 / 0 = 0^0, 0 / 0 = 0^1. Since 0^0 = 1, So 1 = 0.

Trytu · 2018-09-29 15:33:03

azurepudding wrote:

Trytu wrote:
▼I wroted
never divide by 0
if you divide x=1/(5+3-8)
x = 1/5 + 1/3 - 1/8 = 3/15 + 5/15 - 1/8 = 24/120 + 40/120 - 15/120=64/120 - 15/120 = 49/120 = x
you'll have different results
when you divide same thing by x=1/(2+3-5)
x = 1/2 + 1/3 - 1/5 = 5/6 - 1/5 = 25/30 - 6/30 = 19/30 = x
19/30 = 76/120 =/= 49/120
you'll get another result, even if 2+3-5 = 5+3-8
because 2+3-5=0, and 5+3-8=0
that's why you can't divide by 0
0^1/0 = 1, because 0^0 = 1.
0^2/0^1 = 0^1, becuse of a rule of exponents.
so in both examples you divided by 0
LukeM wrote:
The great thing about Maths is that you can take take a statement and prove that it is true given some original statements, the reason this works is that the rules are constructed in a way that means that there will never be any contradictions, so there is no other option than to say that 0/0 and 0^0 are undefined, otherwise the whole of Maths won't work It isn't us saying 'we don't know the answer', its just the way things are, there *cannot* be a single answer.
my opinion about 0^0 is 1
because 0^3 = (infinity amount of 1s)... 1*1*1*0*0*0*1*1*1...(infinity amount of 1s) = 1
so 0^0 = (infinity amount of 1s)...1*1*1*1*1...(infinity amount of 1s)
because 1 times everything = it self
that's how x^0 work according to my opinion
and that makes it not undefined
noone said that you can't do powers of 0, you only can't divide
even google say that 0^0 = 1, and 0/0 is undefined, so you shoudn't say that 0^0 is undefined
Here is how I understand it:
x^0 is a simplified way of writing x^2/x^2 (or any other exponent over the same). You subtract the exponents and remove the denominator and end up with x^0. Tho an even simpler (and more obvious) way is simplifying this to just 1. BUT what if x equals 0? 0^0 alone seems like it checks out, but that'd mean you could also write it as 0^2/0^2, which is also equal to 0/0 which is division of zero. So I'd think 0^0 would be undefined. Anything to the 0th power equals itself over itself, which results in 1 in seemingly every case except for 0.

azurepudding wrote:

0^3/0^2 = 0/0 = undefined
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined

ILikeTofuuJoe wrote:

The whole point of this topic is I'm trying to say: Since x^m / x^n = x^(m-n), say x=0, m1=1, m2=2, and n=1. We get two equations: 0^1 / 0^1 = 0^(1-1), 0^2 / 0^1 = 0^(2-1). Simply both equations, we get: 0 / 0 = 0^0, 0 / 0 = 0^1. Since 0^0 = 1, So 1 = 0.

one simple mistake
0^3 and 0^2 and 0^x (when x<0) equals to 0, same as in my brackets (2+3-5) and (5+3-8)
that's why these transformations doesn't work
in that cases, if you want get right result, you have to use "order of operations"
that mean firstly do brackets, and powers/roots, ect...
else it won't work,
0^0 isn't undefined, it's equals to 1
btw you compared 0/0 to 0^0, it's quite like my 76/120 = 49/120
but i naturaly agree with 0/0 is undefined

ILikeTofuuJoe · 2018-09-29 15:36:18

Trytu wrote:

azurepudding wrote:
Trytu wrote:
▼I wroted
never divide by 0
if you divide x=1/(5+3-8)
x = 1/5 + 1/3 - 1/8 = 3/15 + 5/15 - 1/8 = 24/120 + 40/120 - 15/120=64/120 - 15/120 = 49/120 = x
you'll have different results
when you divide same thing by x=1/(2+3-5)
x = 1/2 + 1/3 - 1/5 = 5/6 - 1/5 = 25/30 - 6/30 = 19/30 = x
19/30 = 76/120 =/= 49/120
you'll get another result, even if 2+3-5 = 5+3-8
because 2+3-5=0, and 5+3-8=0
that's why you can't divide by 0
0^1/0 = 1, because 0^0 = 1.
0^2/0^1 = 0^1, becuse of a rule of exponents.
so in both examples you divided by 0
LukeM wrote:
The great thing about Maths is that you can take take a statement and prove that it is true given some original statements, the reason this works is that the rules are constructed in a way that means that there will never be any contradictions, so there is no other option than to say that 0/0 and 0^0 are undefined, otherwise the whole of Maths won't work It isn't us saying 'we don't know the answer', its just the way things are, there *cannot* be a single answer.
my opinion about 0^0 is 1
because 0^3 = (infinity amount of 1s)... 1*1*1*0*0*0*1*1*1...(infinity amount of 1s) = 1
so 0^0 = (infinity amount of 1s)...1*1*1*1*1...(infinity amount of 1s)
because 1 times everything = it self
that's how x^0 work according to my opinion
and that makes it not undefined
noone said that you can't do powers of 0, you only can't divide
even google say that 0^0 = 1, and 0/0 is undefined, so you shoudn't say that 0^0 is undefined
Here is how I understand it:
x^0 is a simplified way of writing x^2/x^2 (or any other exponent over the same). You subtract the exponents and remove the denominator and end up with x^0. Tho an even simpler (and more obvious) way is simplifying this to just 1. BUT what if x equals 0? 0^0 alone seems like it checks out, but that'd mean you could also write it as 0^2/0^2, which is also equal to 0/0 which is division of zero. So I'd think 0^0 would be undefined. Anything to the 0th power equals itself over itself, which results in 1 in seemingly every case except for 0.
azurepudding wrote:
0^3/0^2 = 0/0 = undefined
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined
one simple mistake
0^3 and 0^2 equals to 0, same as in my brackets (2+3-5) and (5+3-8)
that's why these transformations doesn't work
in that cases, if you want get right result, you have to use "order of operations"
that mean firstly do brackets, and powers/roots, ect...
else it won't work,
0^0 isn't undefined, it's equals to 1
btw you compared 0/0 to 0^0, it's quite like my 76/120 = 49/120
but i naturaly agree with 0/0 is undefined

My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution

LukeM · 2018-09-29 15:39:09

ILikeTofuuJoe wrote:

My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution

It really isn't, you can prove that there CAN BE NO SINGLE SOLUTION (at least for 0/0), so it physically has to be undefined, its not just us saying we don't know what it is, we know for certain that there is no answer.

drunkbnu

ILikeTofuuJoe · 2018-09-29 15:41:30

LukeM wrote:

ILikeTofuuJoe wrote:
My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution
It really isn't, you can prove that there CAN BE NO SINGLE SOLUTION (at least for 0/0), so it physically has to be undefined, its not just us saying we don't know what it is, we know for certain that there is no answer.

You are saying things that have multiple solutions is undefined, you are wrong.

Trytu · 2018-09-29 15:44:03

ILikeTofuuJoe wrote:

My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution

just stop dividing by 0 forms

ILikeTofuuJoe wrote:

You are saying things that have multiple solutions is undefined, you are wrong.

i agree with it
sum of 1-1+1-1+1-1+1-1+1-1+1-1+1-1...
has 3 results
first is 0, second is 1, and third is 1/2
because for example in 1-1+1-1+1-1+1-1+1-1+1-1... we can put brackets like (1-1)+(1-1)+(1-1)+(1-1)... and 1-1 = 0, so 0+0+0+0...=0
we can also put brackets one number later like 1+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1) so 1+0+0+0+0+0...=1
or for example S=1-1+1-1+1-1+1-1+1-1+1-1+1-1...
so 1 - S = 1 - (1-1+1-1+1-1+1-1+1-1+1-1+1-1...) that one minus reverse all operators (-1+1-1+1-1+1-1+1-1+1-1+1...), and changes into +, so 1-1+1-1+1-1+1-1+1-1+1-1+1-1+1-1... = S
we got same thing that we started so 1 - S = S, so 1 = 2S, so S =1/2

ILikeTofuuJoe · 2018-09-29 15:47:25

Trytu wrote:

ILikeTofuuJoe wrote:
My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution
just stop dividing by 0 forms

No I will not.

Minimania · 2018-09-29 17:11:31

ILikeTofuuJoe wrote:

Trytu wrote:
ILikeTofuuJoe wrote:
My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution
just stop dividing by 0 forms
No I will not.

And this is your mistake

Trytu, drunkbnu

HeyNK · 2018-09-30 00:23:00

guys I can prove that 1 = 2

watch this:

0 / 0 = 2
but 0 / 0 = 1

therefore 1 = 2

drunkbnu

Norwee · 2018-09-30 00:52:44

HeyNK wrote:

We really need to make more math threads so that interneet teenagers can show off how much they know about math before being humiliated by a virtual anime girl.

What the heck? Are you referring to me? Are you saying you want me to humiliate them?

Pingohits · 2018-09-30 01:32:09

ILikeTofuuJoe wrote:

ZeldaXD wrote:
This is probably the most retarded post I've ever seen.
0 to the power of 0 is undefined, because 0 to the power of anything is 0, but anything to the power of 0 is 1, so 0^0 presents a conflict. Depending on the function, 0^0 could be either 0 or 1. Take this as an example of a function returning 1.
Take f(x) = x^x
lim (x -> 0) of f(x) = lim (x -> 0) of x^x = 0^0
Thus take M = lim (x -> 0) of x^x
ln M = ln [lim (x-> 0) of x^x]
ln M = lim (x -> 0) of (ln x^x)
ln M = lim (x -> 0) of (xln x)
ln M = lim (x -> 0) of [(ln x) / (1 / x)]
Take lim (x -> 0) of [(ln x) / (1 / x)] and apply L'Hôpithal's rule
lim (x -> 0) of [(1 / x) / (-x^(-2))]
lim (x -> 0) of -x = 0
thus
ln M = 0
for ln M = 0, we need e^0 = M
M = 1
In this case, x^x = 1, but could be anything else for another problem, thus being undefined.
PS: Square root of -1 does not exist, sqrt(-1) = i, i standing for "imaginary", it's not a real solution, it's an imaginary number.
You don't understand so get out.

me irl

HeyNK · 2018-09-30 02:18:47

NorwegianboyEE wrote:

HeyNK wrote:
We really need to make more math threads so that interneet teenagers can show off how much they know about math before being humiliated by a virtual anime girl.
What the heck? Are you referring to me? Are you saying you want me to humiliate them?
https://i.imgur.com/93eYNPo.png

No, I'm talking about zeldaxD. But i'd love to watch you, my favourite everybody edits virtual anime girl, humiliate internet teenagers for money.

azurepudding · 2018-09-30 07:34:49

Trytu wrote:

azurepudding wrote:
Trytu wrote:
▼I wroted
never divide by 0
if you divide x=1/(5+3-8)
x = 1/5 + 1/3 - 1/8 = 3/15 + 5/15 - 1/8 = 24/120 + 40/120 - 15/120=64/120 - 15/120 = 49/120 = x
you'll have different results
when you divide same thing by x=1/(2+3-5)
x = 1/2 + 1/3 - 1/5 = 5/6 - 1/5 = 25/30 - 6/30 = 19/30 = x
19/30 = 76/120 =/= 49/120
you'll get another result, even if 2+3-5 = 5+3-8
because 2+3-5=0, and 5+3-8=0
that's why you can't divide by 0
0^1/0 = 1, because 0^0 = 1.
0^2/0^1 = 0^1, becuse of a rule of exponents.
so in both examples you divided by 0
LukeM wrote:
The great thing about Maths is that you can take take a statement and prove that it is true given some original statements, the reason this works is that the rules are constructed in a way that means that there will never be any contradictions, so there is no other option than to say that 0/0 and 0^0 are undefined, otherwise the whole of Maths won't work It isn't us saying 'we don't know the answer', its just the way things are, there *cannot* be a single answer.
my opinion about 0^0 is 1
because 0^3 = (infinity amount of 1s)... 1*1*1*0*0*0*1*1*1...(infinity amount of 1s) = 1
so 0^0 = (infinity amount of 1s)...1*1*1*1*1...(infinity amount of 1s)
because 1 times everything = it self
that's how x^0 work according to my opinion
and that makes it not undefined
noone said that you can't do powers of 0, you only can't divide
even google say that 0^0 = 1, and 0/0 is undefined, so you shoudn't say that 0^0 is undefined
Here is how I understand it:
x^0 is a simplified way of writing x^2/x^2 (or any other exponent over the same). You subtract the exponents and remove the denominator and end up with x^0. Tho an even simpler (and more obvious) way is simplifying this to just 1. BUT what if x equals 0? 0^0 alone seems like it checks out, but that'd mean you could also write it as 0^2/0^2, which is also equal to 0/0 which is division of zero. So I'd think 0^0 would be undefined. Anything to the 0th power equals itself over itself, which results in 1 in seemingly every case except for 0.
azurepudding wrote:
0^3/0^2 = 0/0 = undefined
0^3/0^3 = 0/0 = 0^0 = undefined
0^3/0^4 = 0^-1 = 1/0 = undefined
ILikeTofuuJoe wrote:
The whole point of this topic is I'm trying to say: Since x^m / x^n = x^(m-n), say x=0, m1=1, m2=2, and n=1. We get two equations: 0^1 / 0^1 = 0^(1-1), 0^2 / 0^1 = 0^(2-1). Simply both equations, we get: 0 / 0 = 0^0, 0 / 0 = 0^1. Since 0^0 = 1, So 1 = 0.
one simple mistake
0^3 and 0^2 and 0^x (when x<0) equals to 0, same as in my brackets (2+3-5) and (5+3-8)
that's why these transformations doesn't work
in that cases, if you want get right result, you have to use "order of operations"
that mean firstly do brackets, and powers/roots, ect...
else it won't work,
0^0 isn't undefined, it's equals to 1
btw you compared 0/0 to 0^0, it's quite like my 76/120 = 49/120
but i naturaly agree with 0/0 is undefined

But anything to the power of 0 is the same as that same number over itself. That's why in almost every case, x^0 is 1, because x^0=x/x. Plug in 1, 2, 3, 5, 10, a million, you will always end up with 1.. except when you plug in 0, as it results in a division of 0. So 0^0=0/0, making both undefined.

I also don't see the error on my part.. but one you made was "0^3 and 0^2 and 0^x (when x<0) equals to 0".. if x is less than 0 there, say -1, you've have 0^-1 which is the same as 1/0, and later on you said division of 0 gets you undefined.

ILikeTofuuJoe wrote:

My point is undefined is absolutely them being lazy and not think of a solution an ACTUAL solution

Again, it's not "because we don't feel like it," it literally has no answer.

Take this example:

You take any number and divide it by 3. 6/3=2, and 2*3=6 to get 6 again.
Now replace 3 with the number 0.
You take 6 again and divide it by a 0. 6/0=x, and x*0=6
If x is not undefined, what is x? What number times 0 will get you 6?

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#26 2018-09-29 02:18:12, last edited by azurepudding (2018-09-29 02:22:02)

Re: Math geeks are too stupid to fix this

#27 2018-09-29 02:21:08

Re: Math geeks are too stupid to fix this

#28 2018-09-29 02:23:17

Re: Math geeks are too stupid to fix this

#29 2018-09-29 02:31:50

Re: Math geeks are too stupid to fix this

#30 2018-09-29 02:42:13

Re: Math geeks are too stupid to fix this

#31 2018-09-29 02:52:38

Re: Math geeks are too stupid to fix this

#32 2018-09-29 02:53:15

Re: Math geeks are too stupid to fix this

#33 2018-09-29 03:01:20

Re: Math geeks are too stupid to fix this

#34 2018-09-29 11:09:13

Re: Math geeks are too stupid to fix this

#35 2018-09-29 11:10:30, last edited by LukeM (2018-09-29 11:20:11)

Re: Math geeks are too stupid to fix this

#36 2018-09-29 11:32:53

Re: Math geeks are too stupid to fix this

#37 2018-09-29 15:14:14

Re: Math geeks are too stupid to fix this

#38 2018-09-29 15:33:03, last edited by Trytu (2018-09-29 15:37:26)

Re: Math geeks are too stupid to fix this

#39 2018-09-29 15:36:18

Re: Math geeks are too stupid to fix this

#40 2018-09-29 15:39:09

Re: Math geeks are too stupid to fix this

#41 2018-09-29 15:41:30

Re: Math geeks are too stupid to fix this

#42 2018-09-29 15:44:03, last edited by Trytu (2018-09-29 15:55:16)

Re: Math geeks are too stupid to fix this

#43 2018-09-29 15:47:25

Re: Math geeks are too stupid to fix this

#44 2018-09-29 17:11:31

Re: Math geeks are too stupid to fix this

#45 2018-09-30 00:23:00

Re: Math geeks are too stupid to fix this

#46 2018-09-30 00:52:44

Re: Math geeks are too stupid to fix this

#47 2018-09-30 01:32:09

Re: Math geeks are too stupid to fix this

#48 2018-09-30 02:18:47

Re: Math geeks are too stupid to fix this

#49 2018-09-30 07:34:49

Re: Math geeks are too stupid to fix this

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