/sci/ - Science & Math

[Anonymous No.9862100]

If any number raised to the zero power is 1 but zero raised to any power is 0, what does 0^0 equal?

[Anonymous No.9862112]

>>9862100
Who is stupid enough to think that zero raised to any power is 0? Even on /sci you cannot find any takers, and that says a lot.

[Anonymous No.9862119]

>>9862100
Both of your premises are false

[Anonymous No.9862120]

>>9862100
If your mum is an 18-wheeler, and your mum is a schoolbus, then how many wheels has she got?

[Anonymous No.9862125]

>>9862100
There's a contradiction in your question, retard

[Anonymous No.9862126]

>>9862119
Maybe if you could present such a number for which its zero'th power is not 1. You would have made a sensational new discovery.

[Anonymous No.9862128]

>>9862100
Define exponentiation as 1*x repeated n times and the contradiction goes away.

1*0*0...= 0, but 1 = 1

[Anonymous No.9862131]

>>9862126
0

[Anonymous No.9862138]

>>9862128
What about 2*2=4 but 2=2?

[Anonymous No.9862139]

>>9862131
>>9862128

[Anonymous No.9862144]

>>9862138
2^2 = 1*2*2 = 4
2^1 = 1*2 = 2
2^0 = 1 = 1
Go back to /pol/

[Anonymous No.9862145]

>>9862131
Prove it

[Anonymous No.9862147]

x^0 = 1, x =/= 0
0^x = 0, x =/= 0

[Anonymous No.9862150]

>>9862138
>>9862144
Also,
2^(-1) = 1 / 2
2^(-2) = 1 / 2*2
etc

If you didn't figure out that this is how exponents work back in middle school you are mentally retarded

[Anonymous No.9862151]

>>9862144
Nigger

[Anonymous No.9862154]

>>9862151
>being dumber than a nigger
What does that say about you, wh*Toid dog?

[Anonymous No.9862158]

>>9862154
Nigger

[Anonymous No.9862172]

>>9862120
Oh, Oh! 20?

[Anonymous No.9862877]

0^0 = 0. 1/x is a continuous function when x > 0. So, limit as n to infinity of x^(1/n) = x^0. now, 0^(1/2) = 0. 0^(1/3) = 0 ... 0^(1/n) = 0. thus limit as n to infinity of 0^(1/n) = 0.

[Anonymous No.9862895]

>>9862877
1/x is continuous, but n^(1/x) is not

[Anonymous No.9862912]

>>9862877
prove 1/x is continuous at x = 0 or your proof doesn't mean shit you retard

[Anonymous No.9862913]

0^0=1 because [math]\lim_{x \to 0^+}x^x= 1[/math]

[Anonymous No.9862919]

>>9862912
It's not continuous at 0, but he didn't say or use contuinity at 0

[Anonymous No.9862933]

>>9862919
thanks.
>>9862912
apply yourself

[Anonymous No.9862945]

>>9862933
0/10 either underage faggot who doesn't know what a limit is or a shit bait

[Anonymous No.9862959]

>>9862945
where did I misuse limits?

[Anonymous No.9862964]

That's a good question op and since everyone in here is retarded I will try to not be. Zero to the zero is defined contextually as either one or zero when it makes things nice to work with, but is often left undefined due to the contradiction you've pointed out. The limit of x to the x certainly goes to one as x goes to zero from above so one seems like the best definition if it is to be defined at all. This result shows in some sense that one of the rules "wins out" in a reasonable race between them. Anyone making "exponentiation is repeated multiplication" arguments doesn't actually know what exponentiation is.

[Anonymous No.9862965]

>>9862945
He's taking to limit as x -> infinity, not x -> 0

[Anonymous No.9862966]

>>9862877

[Anonymous No.9862967]

>>9862964
What is exponentiation?

[Anonymous No.9862971]

>>9862966
unless you or anyone proves me wrong, this is the end of the thread.

[Anonymous No.9862985]

>>9862100

*frantically nervous about this thread*

[Anonymous No.9862990]

>>9862913
\thread

[Anonymous No.9863004]

>>9862967
Your opinion about what it truly "is" is sort of subjective because there are a handful of equivalent ways to define it. The way that I think is most "true" in the sense that it most accurately depicts what is happening is a geometric definition. The complex exponential function is a transformation from rectangular to polar coordinates. It takes vertical lines to circles and horizontal lines to lines radiating from the origin. A more common analytic perspective is to define the function by a power series (mobile so not typing it). We can restrict our attention to the reals for this series to see the e^x most people are familiar with and all other exponentials can be described in terms of this function and it's inverse, i.e., a^b = e^(blna).

Of course, you could argue what "is" anything in math besides definition, and certainly you can define integer exponents by repeated multiplication, but when talking about limits and such you really need that continuum of definition.

[Anonymous No.9863116]

>>9862971
You're defining x as 1/n so you have to have both x variables represented in that way. You then run into the same problem. You then have (1/n)^(1/n) and as that alproches infinty I can tell you it does not equal 0

[Anonymous No.9863121]

>>9862971
>>9862964

[Anonymous No.9863154]

>>9862990
yep, that's the pic we need. we can't say that 0^0=1 exactly, but a small number to the small number approaches 1, so that's what it is in a non-rigorous sense. definitely less wrong than 0^0=0.

[Anonymous No.9863163]

>>9863154
What makes that graph more right than this graph?

[Anonymous No.9863206]

>>9863116
>You're defining x as 1/n
no I'm not. x = 0. 0 = 0^(1/n) = x^(1/n) for all n. Thus as n to infinity, 0= lim 0^(1/n) = 0^0

[Anonymous No.9863207]

>>9863163
It doesn't give one of the rules a chance in the race, just as x^0 is always one

[Anonymous No.9863227]

>>9862877
>>9862959
First, you gave no argument whatsoever for why [math] \lim_{x \to \infty } 0^{1/x} = 0 [/math], you just pulled it out of your ass. Second, and this is the big one [math] \lim_{x \to a} f(x) \neq f(a) [/math].

[Anonymous No.9863238]

>>9863206
Okay,
x=1. 1= (1/n)^0.
As n approaches infinity, (1/n) approaches zero, therefore 0^0 = 1

[The Lord No.9863279]

what is the value of a function at the point where its slope is infinity?

[Anonymous No.9863330]

>>9862877
why 0^(1/n) = 0 as n approaches infinity? why do you assume 0^x to be continuous at 0?

[Anonymous No.9863353]

>>9863238
you have to prove x^0 is continuous at 0 to use that argument.
>>9863227
>you gave no argument whatsoever for why [math] \lim_{x \to \infty } 0^{1/x} = 0 [/math],
let x > 0. then 0^(1/x) =0, and this holds for all x. Thus, as x -->infinity, 0 = lim 0^(1/x).
>[math] \lim_{x \to a} f(x) \neq f(a) [/math].
this is true for continuous functions, which 0^(1/x) is for x positive

[Anonymous No.9863356]

>>9863330
>why do you assume 0^x to be continuous at 0?
because 0^x is 0 for all x positive. thus, x is an arbitrary positive number, so can let x --> infty

[Anonymous No.9863378]

>>9862100
context dependent based on a limiting process. could be 0. could be 1. could diverge. could not approach any value. if you don't give a context, you identify it as an indeterminate form. wiki's actually a pretty good article

[Anonymous No.9863392]

>>9863353
>this is true for continuous functions, which 0^(1/x) is for x positive
Right, which implies at the limit as x->a of 0^(1/x) = 0^(1/a) for positive x. It does NOT mean that limit as x->infinity of 0^(1/x) = 0^(limit as x->infinity of 1/x). For that to be the case, you first need to prove that 0^x is continuous at x=0. Which it isn't.

[Anonymous No.9863396]

>>9863378
OP is not talking about any limiting process, just the algebraic term. What behavior particular limiting processes have is completely unrelated to that.

[Anonymous No.9864215]

>>9863392
>It does NOT mean that limit as x->infinity of 0^(1/x) = 0^(limit as x->infinity of 1/x).
yes it does since x is an arbitrary positive real number

[Anonymous No.9864292]

>>9863396
Taking powers is an algebraic process which is only well-defined for non-zero rationals, until you extend to the reals by taking some limit so that x^y is continuous at your specific value (x,y). This function does not converge at (x,y)=(0,0) so you have to do something more arbitrary.

[Anonymous No.9864461]

>>9863353
>let x > 0. then 0^(1/x) =0, and this holds for all x. Thus, as x -->infinity, 0 = lim 0^(1/x).
This is right.
>this is true for continuous functions, which 0^(1/x) is for x positive
Yes and infinity is not a positive real number.

[Anonymous No.9864685]

[eqn]lim_{y \to 0^{+}} lim_{x \to 0^+} y^x = 1 [/eqn]
[eqn]lim_{x \to 0^{+}} lim_{y \to 0^+} y^x = 0 [/eqn]
Therefore it is not continous at (x,y)=(0,0)

[Anonymous No.9864771]

>>9862100
If you define x^n to mean "multiply x by 1 for n times" then 0^0 means "multiply 0 by 1 for 0 times" therefore it's 1.

If you write the power series for e^x out, then plug in x=0, you can also consider this as a legitimate case here 0^0=1 is a handy convention but of course it is an indeterminate form.

[Anonymous No.9865015]

>>9864685
yes. i think this also shows 0^0 can't be well-defined.

[Anonymous No.9865021]

>>9865015
>i think this also shows 0^0 can't be well-defined.
It shows no such thing.

[Anonymous No.9865031]

>0^0
>0/0
>equal

what did he mean by this

[Anonymous No.9865042]

>>9864215
>>9863356
NO, positive x excludes 0. 0^x is 0 for all x positive, therefore f(0+) = Lim x->0 f(x) = 0,
it doesn't give you anything about the actual value of f(x) at 0, what you got is the right side limit.

[Anonymous No.9866148]

>>9865042
He's letting x tend to infinity. Nowhere does he use f(0+). He's still wrong, but you are too.

[Anonymous No.9866550]

>>9866148
Lim x->inf (0^(1/x)) = 0^(0+)

[Anonymous No.9866712]

>>9866550
Yes, 1/x goes to 0 from the right. x is still going to infinity, not 0.

[Anonymous No.9866730]

0 IS NOT A NUMBER.

[Anonymous No.9866749]

>>9862145
0 multiplied by 0 0 times is still 0.

[Anonymous No.9866779]

>>9866730
good answer

[Anonymous No.9866808]

>>9863163
0^x is not mathematically useful

[Anonymous No.9866817]

>>9862100
I believe this problem would require to divide by 0. The reason why a number raised to the 'zeroth' power is almost always one is because that number is dividing by itself. So 100^0 is the same as 100/100, or 1. In this case, 0^0 would be equivalent to 0/0, which is undefined. Therefore, there would be no real (or imaginary) answer to this problem.

You may or may not disagree with this and I will be open to any critical feedback.

[Anonymous No.9866878]

>>9864771
this is the worst post I've ever seen

[Anonymous No.9867068]

>>9862131
Fuuuuuuug
Go home mathematicians he's done it
No more grant money for anyone else

[Anonymous No.9867111]

>>9862144
So 0^2 = 1*0*0 = 0
0^1 = 1*0 = 0
0^0 = 1 = 1
>Forgetting everything is *1 anyway and is uterly useless to include

[Anonymous No.9867114]

>>9864771
I award you no points, and may God have mercy on your soul

[Anonymous No.9867157]

>>9866878
>>9867114
if you don't multiply 1 by 0, then it's 1, what's so hard to grasp?

[Anonymous No.9867453]

Better question guys.
Why is x^x not defined for x<0?

[Anonymous No.9867460]

>>9862100
0.9999....

[Anonymous No.9867469]

>>9867453
It's perfectly defined.

For x =/= 0
x^y = exp(y Log(x))
For negative values of x
Log(x) = i pi + log(|x|)

So for x<0
x^x = exp(x i pi + x log(-x))

[Anonymous No.9867474]

>>9867469
OK, I was just wondering why wolfram doesn't show x^x for x <0.

[Anonymous No.9867479]

>>9867474
if you base your math knowledge on wolfram exclusively I have nothing but contempt for you.

[Anonymous No.9868594]

>>9862100
x^1 = 1*x
x^0 = 1

[Anonymous No.9868799]

>>9862145
a^n/a = a^(n-1)

0^1/0 = 0^(1-1)

a^1 = a

0^1 = 0

0^1/0 = 0^0

0/0 = 0

0^0 = 0

[Anonymous No.9868815]

It's generally accepted that 0^0 is 1, not because it is the actual right way but because some formulas would be less pretty if this wasn't true (Newton's binomial formula is one of them IIRC). It's undetermined though, so you could say 0^0="your mom's dick" and you would not be wrong.

[Anonymous No.9868851]

>>9862100
0^0 = 1 because this is what algebra requires

[Anonymous No.9869071]

>any number multiplied by itself no times equals one.

I don't get this.

[Anonymous No.9869073]

>>9869071
It's the multiplicative identity

[Anonymous No.9869079]

5 > 2
x)6$8
< 1-1=2
((((exp)
▲
▲ ▲

[Anonymous No.9869108]

>>9862100
https://www.wolframalpha.com/input/?i=y+%3D+x%5Ex

Looks like 1 to me.

[Anonymous No.9869130]

>>9862913
Prove that

[Anonymous No.9869340]

>>9869073
the multiplicative identity is 1

[Anonymous No.9869344]

>>9869340
I know

[Anonymous No.9869393]

>any number multiplied by itself no times equals 1


This has nothing to do with the multiplicative identity

[Anonymous No.9869407]

>>9869393
It makes sense that a number multiplied by itself no times should equal the multiplicative identity

[Anonymous No.9869551]

>>9862100
x^0 represents x/x.
You can't divide by zero.
>zero raised to any power is 0
Zero can't be raised to a power of 0 or below because it requires division by zero.

[Anonymous No.9870532]

>>9869407
No it doesn't

[Anonymous No.9870579]

>>9862120
24

[Anonymous No.9870923]

Exponents would be defined as a recurrance where the a(k+1) depends on a(k) and a(0)=1 so I'd assume that anything to the zero is.one