/sci/ - Science & Math

[Anonymous No.5171044]

Im just a stupid college sophomore still bottled by the fact why any number to the 0 is 1. I havent been provided with any proof of this.

[Anonymous No.5171050]

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

x^0 = 1

[Anonymous No.5171056]

It's just a convention, it makes formulas like the Newton binomial easier to write.

[Anonymous No.5171060]

>>5171056
it's not.

x^0=exp(0*ln(x))=exp(0)=1.

[Anonymous No.5171061]

>>5171056 here
Didn't notice you were asking about ANY number.
Yeah, like the guy above me said, it extends nicely to the real domain as well.

[Anonymous No.5171062]

x^a * x^b = x^(a+b)

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

[seraphnb No.5171077]

Well, you should know that (x^y) / (x^z) = x^(y-z), correct? Now, let's plug in numbers. (2^4)/(2^4) should equal 2^(4-4), or 2^0. Now, let's get a real answer. 2^4=16. With the substitution property of equality, 16/16=2^0. 16/16=1. 2^0=1. It's true for any real number, as far as I know.

[Anonymous No.5171111]

>>5171044
What else would it be? Look at the graph of any exponential function.

[Anonymous No.5171124]

>>5171060
So x^0 = 1 because e^0 = 1?
That's kinda circular reasoning.

[Anonymous No.5171128]

You understand that it is us who define what happens in mathematics? If we feel x^0 should be 1, then so it will be. We also like to make our definitions such that they make sense. We like to write x*x as x^2. At the same time, we like 1/x to be denoted x^-1. Notice how when we multiply powers of x, the exponents are added together? What a convenient way to multiply powers! But then in order for this to work properly, we must have 1 = x*1/x = x^(1 + -1) = x^0. So we define, x^0 = 1. We did so, because it makes sense.

[seraphnb No.5171136]

>>5171128
Although there are a few exceptions. The difference between equality and congruence, for instance. There needs to be an extra element in a definition for two things to be different, and there is no extra element in the definition of congruent that equal does not have. I've been fighting that battle since 9th grade.

[Anonymous No.5171147]

>>5171128
This is not entirely true. Apart from the fundamental axioms, everything is proved using exact logic (except model definitions of course). So when you say <span class="math">x^{0} = 1[/spoiler] because we just defined it to be that way, it doesn't really make sense. We made up the notation, but from the definition of the notation and the aforementioned axioms, it's clear that x^0 in fact is equal to 1.

Saying that it's 1 because we defined it to be 1 is crude and not very mathemathical.

[Anonymous No.5171161]

>>5171044
It's basically just defined that way. It makes a lot of sense, really. It makes the exponential function continuous, and lets many properties of exponents hold in all cases rather than all cases except for zero (which would just be dumb). It's very similar to the reasoning behind 0! = 1. It's the fact that 1 is the multiplicative identity.
What do you get when you multiply 5 by nothing (not 5x0, but 5 times not anything)? 5, right? If you don't multiply it but anything, it's unchanged. So multiplying by nothing is identical to multiplying by one. From there, what is 5 x 2^4 x 3^0? Well a stack of one 5 times a stack of four 2's times a stack of no 3's--5 x 2 x 2 x 2 x 2 = 80. So the 3^0 must be equal to 1.

[Anonymous No.5171168]

>>5171147
lookit the freshman math major

[Anonymous No.5171174]

>>5171044
x^1=x
x^-1=1/x
(x^1)(x^-1)=x^(1-1)=x^0
x^0=x*(1/x)=1.

There you have it, x^0=1 for all x except x=0(because that would require dividing by 0)

[seraphnb No.5171175]

>>5171147
Gotta love logic. Is it mathematics? Is it philosophy?

[Anonymous No.5171188]

Okay I think I get it now

and I know that 0 to the power of any integer is a 0. But what happens if I do 0^0?

What do I get? Is the answer 1 or 0?

[seraphnb No.5171199]

>>5171188
Heh. Good question. Then again, what's 7/0?

[Anonymous No.5171200]

>>5171188
Neither, it indeterminate.

[Anonymous No.5171201]

>>5171168
I didn't even go to college.
If you disagree with something, say what. Going "lookit the freshman math major" just makes you a dick.

Also foundations of math. There's basically only two things we humans made up. Axioms, and definitions that define new models (also entails notation). There's also sort of "meta-math" that includes arbitrary conventions, like order of operations or that x is horizontal and y is vertical. That's it. Axioms define entirely new field of mathematics and everything is further only proven using axioms or theorems (theorems are statements proven using axioms or other theorems).

You cannot just go 1+2=3, 1+3=4, but 1+4=5757 because I feel like it.

[Anonymous No.5171203]

>>5171147
What fundamental axioms do you propose for proving x^0 = 1? At some point after defining the fundamental axioms, one must define what x^n means. In many different algebraic structures we simply define x^n to be x*x*....*x (n times) for natural n, where x^-n is the product of n copies of inverse x. Could we prove that x^0 equals 1 from this? I do not think so, not unless we go in circles and attempt stuff like x^n*x^-n = 1, "and since exponents are added on multiplication, we must have x^0 = 1". We do not even know that exponents should be added on multiplication at that point.

[Anonymous No.5171216]

>>5171203
You can use everything that has been already proven, you don't have to just stick with axioms. Also you pretty much agreed to what I said, we start with the definition of x^n and we're to discover that x^-n is multiplicative inverse of x^n (keep in mind that we can use division to expand x^n to negatives). If we know what x^-n is, it's not circular logic

[Anonymous No.5171219]

>>5171203
Actually would know that exponents would be added on multiplication. Knowing that x^n=x*x*x*..n times,and x^-n is the same but with the inverse of x, is enough information to solve the problem in it's entirety.

[Anonymous No.5171228]

x^n=x*x^(n-1)

By definition.

So x^1=x
x^0=x^1/x
x^0=x

This also explains why x^-1=1/x and so on.

[Anonymous No.5171239]

>>5171219
>>5171216
In my opinion, no. We have simply defined what x^n and x^-n means for natural n. We would like the exponents to be additive on multiplication. Hence, we MUST have x^0 = 1. Indeed, for exponents to be additive on multiplication, x^0 must equal 1. But it is not, per say, a "proof" that it is so. It simply must be so for the exponent adding to work out. So, we let x^0 = 1.

I am aware we are not going anywhere with this discussion, so I am signing out. In any case, many important results do later become definitions. That is the case with topological axioms which we derived from our work in analysis. If something is nice, and works, we can define it to always be true. Trying to trace back through all of mathematics to understand why x^0 = 1 is a worthless endevaour, so I decide to leave it at that.

[Anonymous No.5171240]

>>5171228

x^0=1, goddammit.

[Anonymous No.5171246]

>>5171203
Actually in a space where every element x has multiplicative inverse such that <span class="math">x*x^{-1}=x^{-1}*x=1[/spoiler], and we define exponentiation by <span class="math">x^n = x*x*x...[/spoiler] n times and <span class="math">x^{-n}=x^{-1}*x^{-1}*x^{-1}...[/spoiler] n times, we have enough to prove that x^0 must equal one.

[Anonymous No.5171264]

>>5171062

[Anonymous No.5171267]

>>5171239
I'm sure that knowing what x^n and x^-n is enough to understand why x^0 = 1. If I find something assuming only these two things, I'll post it here.

[Anonymous No.5171424]

>>5171060
>>5171124

Your circularity makes me dizzy

[Anonymous No.5172755]

>>5171188

We had a whole thread dedicated to this a few days days ago. It was a good time.

[Anonymous No.5172766]

>>5171246
You also have to either define exponentiation to be additive or define x^0 separately. So no, you are wrong.

[Anonymous No.5172799]

[Anonymous No.5172951]

>>5171050
Explain how you made the jump from

>x^0 * x = y *x
to
>x = x * y
given
>x^0 = y

[Anonymous No.5172967]

>>5172951

Doesn't matter, multiplying both sides of an equation by x is not proper mathematics.

For example,

if x = 5, then x*x is not 5x because that would imply that x is 5 or zero which it isn't.

[Anonymous No.5173086]

>>5172967
>if x = 5, then x*x is not 5x because that would imply that x is 5 or zero which it isn't.
what

[Anonymous No.5173112]

>>5173086
never mind i didn't think about it enough/his statement wasn't worded appropriately for me

I would have preferred

>if x = 5, then x*x is not 5x because that would imply that x 'can be' 5 or zero which it isn't.