So, after two glasses of sweet red wine, I was thinking about

>>9023477
Are you the 1/0 guy?

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>>9023477
Whoops, I meant to say, the Identity property, NOT the associate property...

>>9023478
lol no... I'm assuming he's a troller on here?

>>9023477
But you can also argue that

ln (x^0) = ln (1)
=0*lnx = 0

So really, defining x^0 =1 is just an identity for exponents.

>>9023486
Or just remember the property: x^(a)*x^(b)=x^(a+b)
So that:

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

>>9023477
I'm not gonna read all your bullshit because it's wrong. Exponents are repeated multiplication or division, and they are defined by how many times you multiply or divide 1 by the base. x^2 = 1 * x * x, x^-2 = 1/(x * x) and so forth. An exponent of zero means we are multiplying or dividing 1 by x zero times, so x^0 just equals 1.

>>9023495
Who the fuck defines powers that way? I've never seen it defined as 1 * x ...

Either way there are different ways to prove it depending on which branch of math you like the most. My proof is simpler: it's explained on Wikipedia and in relevant math books. QED.

>>9023513
>no citations
>QED

kill yourself

>>9023477
x^0 = 1 BY DEFINITION. why is that ? because then x^a * x^b = x^(a+b) holds. in other words, because it's really fucking convenient, that's all, nothing esoteric going on. can you do math without it ? certainly. but try doing any group theory without this notation a you're gonna cry.

Take [math]c \in \mathbb{R}[/math]
Consider [math]c^{\frac{1}{n}} [/math] for some [math]n \in \mathbb{N}[/math] . What happens when n is large? What is the limit as [math]n \to \infty[/math] ?

>>9023574
correction make it [math]c \in \mathbb{N}[/math]

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>>9023477
Hopefully you'll agree that [math](x^a)(x^b)=x^{a+b}[/math]
Let a=0
[math](x^0)(x^b)=x^{0+b}=x^b[/math]
Let [math]x^0 = y[/math]
[math]y(x^b)=x^b[/math]
[math]y=1[/math]
[math]x^0=1, \ \forall \ x \neq 0[/math]
0^0 is an indeterminate form, but I'm sure there's some retard lurker waiting in the wings to debate that one too.

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>>9023574
maple agrees

Btw the binomial theorem also proves that x^0 = 1.

\sum_{k=0}^n {{n}\choose{k}} = 2^n

We see that 2^0 = 1 because of the empty set o.

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>>9023537
>>9023486

OP Here. Wow, didn't expect any intelligent response to this! Thanks guys! And this explanation (and everyone elses! :P) makes sense...So basically this convention that x^0 is more 'artificial', just to let us make use of other identities/properties etc?

What other conventions are like this? I mean to ask, what other "rules" do we have in math that are not 'natural' and sort of created to let us continue on doing math the way we have?

>>9023702
I was following you up until the
x^0=1

How did you make the leap that because y(x^b) = x^b that this means x^0 equals 1?

Could you use real numbers and not varibles?

>>9024474
???
he stated earlier that x^0 = y, ∴ because y=1, x^0 must also = 1

if that isn't the bit you're stuck with, then look at:
y(x^b) = (x^b) || here if you take both sides of the equation and divide them by (x^b), you can cancel the (x^b) from either side, right?
As in, you are taking some value and dividing it by itself; e.g. 5/5, 3/3, (2^2)/(2^2), things to this extent simply equal '1', and thus the (x^b) must also equal equate to '1' on either side, you can omit this on the LHS since it does nothing (try any number multiplied by 1) and the RHS is left at just 1.

hopefully this isn't bait, lmao

>>9024474
The easier way as already staTed is that x^0 = x^(1-1)
= x/x = 1 for all nonzero x

>>9024438
Numbers aren't real. It's all purely based on convention and practicality.

>>9023537
Yup, this.
>>9024438
A lot of things in math are defined to produce certain results that we want. If you want another example, look at arithmetic on the complex numbers. One of the things that we wanted was some way to satisfy x^2=-a, which we couldn't do with the real numbers. But we also wanted to preserve a lot of the nifty properties that the reals have, and so we defined a unit i and defined operations on numbers of the form a + bi.

>>9023477
>Without thinking that there is a real 1 in exponential multiplication like 3*3*3*3, then that would mean 3^0 is really zero
Why?

Without writing that implicit 1* there, completing the pattern of removing the 3s leaves you with nothing on the right hand side. That does NOT mean the number zero; it means an incomplete equation.

An (integral-power) exponentiation is the multiplicative product of a collection of factors. If you place zero factors there, what remains is the product of zero terms. The result is the multiplicative identity, which is 1.

If you ADD a number of terms, and there happen to be zero such terms, the result is the additive identity (the term you can add to a sum without changing the result), which is 0. If you MULTIPLY a number of factors, and there happen to be zero such factors, the result is the multiplicative identity, which is 1.

All the "this way is more convenient" arguments are technically true, but not really relevant here. The sensible pattern has a 1 here, not a 0. Sometimes there are multiple sensible ways to define something, and the answer really is "this was convenient"; but here, 6^0 = 1 is just what makes sense.

>>9024507
Except the properties we imbued them with actually show up everywhere in nature and solve lots of unpredicted problems. That makes them real enough to me.

>>9023702
Out of sincere curiosity, why is 0^0 indeterminate

>>9024540
I'm not really making any argument over what mathematical concepts are "real" or not. Just giving an example of how mathematicians wanted to produce certain results, and so they introduced concepts that would do so.

>>9023477
We say that because the limit of anything to the power of x as x approaches 0 is one. There's no reason to make it different than the limit just because some autist thinks it's illogical.

>>9024632
cuz 0 times itself any amount of times is equal to 0 thus 0^x should be zero however x^0 is always 1 now 0^0 is both which means it would have to be both 0 and 1 to satisfy it. You can also say that x^0 is defined as x/x but that too is undefined when x = 0

>>9024632

Because the basic idea of extending exponentiation, raising a number to a power, depends on being able to /divide/ by that first number.

We start out with exponentiation, where the exponent is natural (that is, strictly positive). This is easy enough to think of as repeated multiplication, in the simplest examples. One also notices very quickly that when you multiply together a thing which is raised to two different powers, you get the same thing by simply adding the exponents.

Now suppose we want to try something wild. Just say that one of the exponents /is a negative integer/, say -1. Well again, starting from the above, you conclude very, very quickly that a^-1 must mean precisely 1/a, a^-2 must mean precisely 1/(a^2), and so on. You can see this in the sketch below.

This leads very quickly to an understanding that in general, a^0 = 1. /But notice that this whole argument rests on the ability to divide by a/. As any good boy knows, dividing by zero is, in standard analysis, RIGHT OUT. That is why 0^0 is sometimes correctly characterized as an indeterminate form-because the argument making such a form possible /breaks/.

[eqn] a^m a^n = a^{m+n} \rightarrow a^{m-1} = a^m a^{-1} = \frac{a^m}{a} \rightarrow a^{-1} = \frac{1}{a} [/eqn]

[eqn] a^0 = a^{1-1} = a^1 a^{-1} = \frac{a}{a} = 1 [/eqn]

It is also true that in certain situations, 0^0 is simply defined to be equal to one, to patch the hole depending on what's being done. But the above shows why some people have a stick in their ass about 0^0, and why they are quite right to have it lodged there.

>>9024699
>That is why 0^0 is sometimes correctly characterized as an indeterminate form-because the argument making such a form possible /breaks/.
Your explanation is correct for the most part, but this one line is wrong. All of what you say, while true, has nothing at all to do with 0^0 being an indeterminate form, which is an unrelated concept entirely. "indeterminate form" does not mean "shit people argue about because naive intuitions break down".

X^1=x/1
X^-1=1/x
X^0=x/x

The question you should ask is why 0^0=1

>>9023477
If 6^n is 6*6*...*6 n times then 6^0 is 6*...*6 zero times, and empty product is 1

This guy explains it very very well... Check it out, this is a visual as one can make it..

https://www.youtube.com/watch?v=rmAaFohGkvc&t=181s