After 6 years I still don't understand this

How can X^0= 1

To me that still goes against laws of simple math.

10^2 = 10 x 10 = 100

10^0= 0 x 0 = 1

If you have 0 apples, where does that 1 apple come from?

I always wonder this too, how dose it work?

2^4 = 16
2^3 = 8
2^2 = 4
2^1 = 2

If you notice so far, the terms keep on dividing by 2 as the exponent goes down. So, 2^0 = 1. This also applies to negative exponents.

If 10^0 = 0 x 0 does that mean 10^2 = 2 x 2? The basic logic on which your question is based is flawed. Apart from that Corns post basically explains the concept I guess...

10^0=/= 0x0, because 2^3=/=2x3
Also, 0x0 doesn't exist, you have to use L'Hopital's law.

Think of it this way: The starting point for multiplication is 1, like the starting point for addition is 0. If you multiply by x 0 times, you are still at 1.

10^0 is not 0 x 0. That would be 0^2.

How can X^0= 1

To me that still goes against laws of simple math.

10^2 = 10 x 10 = 100

10^0= 0 x 0 = 1

If you have 0 apples, where does that 1 apple come from?

X^0=1 is only so because scientists across the world agreed that they will call it that way. It could be 0 too, but since that is not possible for humans to calculate at this moment (same way as dividing something by 0 is "impossible"), we all just agree on that is must be 1 to keep things simple.

It's the same as the time-travel paradox, it shouldn't be possible, that's why scientists agreed it isn't possible.

Math isn't about truth, it's about logic. Logic that isn't possible isn't logic, it's chaos. Although chaos is possible, we all agree that it isn't, just to keep things simple...

10^0=/= 0x0, because 2^3=/=2x3
Also, 0x0 doesn't exist, you have to use L'Hopital's law.

noooo..... 0x0 does exist.... it's zero.

L'hopital's law is only used when you're dealing with limits and you have functions where (in the limit) you get forms like 0*0, 0/0, inf/inf, or inf*inf. Technically you're only supposed to use it on the 2nd and 3rd forms i've written, but the first and fourth can possibly be rewritten to get you the 2nd/3rd form.

I used 0 x 0 just cause they one i compared to was 10 x 10. Maybe 10^0=0=1 would have said it better

X^0=1 is only so because scientists across the world agreed that they will call it that way. It could be 0 too, but since that is not possible for humans to calculate at this moment (same way as dividing something by 0 is "impossible"), we all just agree on that is must be 1 to keep things simple.

Uhhh, Devil. 10 ^ -1 x 10 ^ 1 = 10 ^ 0 = 1. Or rather, I should say 1/10 x 10 = 10/10 = 1. Obviously this doesn't work with 0, but I'm pretty sure that you're saying that 10/10 could be 0 and that's just flat out wrong.

Well.

3 times 0 translates into: I have ZERO number three, how much does that ADD up? 0 + non-existent 3 = 0
3 to the zero power translates into: I have ZERO number three, how much does that multiply into? 1 x non-existent 3 = 1

The non-existent 3 is just... not there, it doesn't modify the 0 and 1 (or any variable x) in any way.
x^0 = 1 is just that. x^0 is non-existent and unable to affect other components multiplied by it.
The representation of "something that doesn't change anything when multiplied by" is, conveniently, existent, and it is 1.

Dammit Corn.

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

Mind blown.

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

Never thought of it like that.

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

Then why is 0^0 = 1?

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

I totally just said that.

And, I'm pretty sure 0 ^ 0 is also undefined Locked, but:

http://www.math.hmc.edu/funfacts/ffiles/10005.3-5.shtml <- this.

x^y

y = 0

y = 4 - 4

x^(4 - 4) = x^4 / x^4 = 1 :O

Pictofied.

I totally just said that.

Mind recovered.

The 0^0 argument is like the why .999... = 1 argument (sort of). In that people tend to argue craploads for multiple sides and no one like to agree with the other side. 0^x = 0 for any nonzero, thus people say that 0 ^ 0 should be 0. But x ^ 0 = 1 for any non zero x, thus people say that 0^0 should be 1. There's a few more arguments about why it should be 1 though:


-lim x^x = 1. Thus if we want the x^x function continuous from the right side we'll need to define 0^0 as 1.
x->0+

(Had to look this one up since i forgot the details of it)
-Another way to view the expression m^n is as the number of ways to map an n-element set to an m-element set. For instance, there are 9 ways to map a 2-element set to a 3-element set. There are NO ways to map a 2-element set to the empty set (hence 0^2=0). However, there is exactly one way to map the empty set to itself: use the identity map! Hence 0^0=1.


-Look at the taylor series expansion of x^x:

f(x) = 1+ xlnx + (1/2) x^2 (lnx)^2 + (1/3!) x^3 (lnx)^3 +....

take the limit as x approaches zero from the left to get 1. Sadly you can't just plug in x =0.


Overall though, 0 ^ 0 is an indeterminate form. 0 ^ 0 has no actual answer. If you try it in your google search bar it'll give you 1, which isn't actually true... not sure why they did that.