Clever student:
I know!
= 
= 
= 
= 
.
Now we just plug in x=0, and we see that zero to the zero is one!
Cleverer student:
No, you’re wrong! You’re not allowed to divide by zero, which you did in the last step. This is how to do it:
= 
= 
= 
= 
which is true since anything times 0 is 0. That means that
= .
Cleverest student :
That doesn’t work either, because if 
then
is 
so your third step also involves dividing by zero which isn’t allowed! Instead, we can think about the function 
and see what happens as x>0 gets small. We have:
= 
= 
= 
= 
= 
= 
= 
= 
=
So, since = 1, that means that = 1.
High School Teacher:
Showing that approaches 1 as the positive value x gets arbitrarily close to zero does not prove that 
. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that is undefined. does not have a value.
Calculus Teacher:
For all 
, we have
.
Hence,
That is, as x gets arbitrarily close to (but remains positive), stays at .
On the other hand, for real numbers y such that 
, we have that
.
Hence,
That is, as y gets arbitrarily close to , 
stays at .
Therefore, we see that the function 
has a discontinuity at the point 
. In particular, when we approach (0,0) along the line with x=0 we get
but when we approach (0,0) along the line segment with y=0 and x>0 we get
.
Therefore, the value of 
is going to depend on the direction that we take the limit. This means that there is no way to define that will make the function 
continuous at the point .
Mathematician: Zero raised to the zero power is one. Why? Because mathematicians said so. No really, it’s true.
Let’s consider the problem of defining the function 
for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:
:= 
where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get
= 
.
However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving
=
which holds for any y. Hence, when y is zero, we have
.
Look, we’ve just proved that ! But this is only for one possible definition of . What if we used another definition? For example, suppose that we decide to define as
:= 
.
In words, that means that the value of is whatever 
approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.
[Clarification: a reader asked how it is possible that we can use in our definition of , which seems to be recursive. The reason it is okay is because we are working here only with 
, and everyone agrees about what equals in this case. Essentially, we are using the known cases to construct a function that has a value for the more difficult x=0 and y=0 case.]
Interestingly, using this definition, we would have
= 
= 
=
Hence, we would find that 
rather than . Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what means for all positive real numbers x and y, and it does preserve continuity of the function as we approach x=0 and y=0 along a certain line.
So which of these two definitions (if either of them) is right? What is really? Well, for x>0 and y>0 we know what we mean by . But when x=0 and y=0, the formula doesn’t have an obvious meaning. The value of is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what means for positive values is not enough to conclude what it means for zero values.
But if this is the case, then how can mathematicians claim that 
? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use 
or if we say that is undefined. For example, consider the binomial theorem, which says that:
= 
where 
means the binomial coefficients.
Now, setting a=0 on both sides and assuming 
we get
= 
= 
= 
= 
= 
where, I’ve used that 
for k>0, and that 
. Now, it so happens that the right hand side has the magical factor . Hence, if we do not use then the binomial theorem (as written) does not hold when a=0 because then does not equal .
If mathematicians were to use , or to say that is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to k=0. We gain elegance and simplicity by using .
There are some further reasons why using is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more “natural” to mathematicians. The choice is not “right”, it is merely nice.
Debankur Paul,
0^0 does not involve divison.
There is no reasonable reason to define 0^0=∞.
this is really fascinating, zero and infinity……..mathematicians cannot deal with zero anf infinity at the present moment, if the are any limits they must be on the mathematician not the math…..there is always a solution…………….its safe to say ONLY GOD KNOWS HOW TO DEAL WITH ZERO AND INFINITY
onke,
Well what do you think human can deal with?
Can human deal with 1?
Can human deal with 2?
Can human deal with 3?
…
Here a simple proof of why zero raised to the zeroth power equals 1
By definition: a + 0 = a
then it follows that (a + 0)^x = a^x if and only if ( a^x )( 0^x) = a^x
therefore: 0^x = 1 for any x.
(a + 0)^x = a^x if and only if ( a^x )( 0^x) = a^x
Why?
0 to the 0 is indeterminate because 0 squared over 0 squared is 0 over 0, and 0 over 0 is indeterminate. the exponet of the denominator(according to quotient rule) is subtracted from the exponet of the numerator, getting 0 to the 0.
*read this over a few times to avoid confusion*
annabeth,
0^0≠0/0
0/0 is indeterminate.
But 0^0=1 is reasonable.
okay this may be stupid but….
ok lets take an example
5 raised to 0 is proved as 1
now using basic common sense we can say that 5-(5-4)=1
=5-(4)=1
1=1
that way taking 0 raised to 0 we can say,
0-(0-1)
0-(-1)
=1
thus we can say 0 raised to 0=1
I may not be correct but please reply if this is even good 😀
Ananthu,
5-(5-4)=1
Why?
Sure, it is easy to argue:
4^0=1
3^0=1
2^0=1
1^0=1
therefore 0^0=1
However, it is easy to demonstrate a simple contradiction as well…
0^4=0
0^3=0
0^2=0
0^1=0
therefore 0^0=0
It is important to note that the CONTRADICTION is important in realizing why it is indeterminate. More importantly, limits do not prove equality.
4^0=1
3^0=1
2^0=1
1^0=1
(-1)^0=1
(-2)^0=1
(-3)^0=1
(-4)^0=1
It is reasonable to infer 0^0=1 .
0^4=0
0^3=0
0^2=0
0^1=0
0^(-1)≠0
0^(-2)≠0
0^(-3)≠0
0^(-4)≠0
It is unreasonable to infer 0^0=0 .
You are all wrong, except for some of you. Some of you are double wrong.
0^0=0/0
BECAUSE
0/0=(0^1)*(0^-1)=0^(1-1)=0^0
Now, given the fact that X/X is by definition always 1, no matter what, 0^0 is also 1. I had a particularly tough time with this one as I defined all alleged imaginary numbers in my head, but this was my conclusion.
I have prepared for a rebuttal, should you call me out on saying that I can divide zero by zero. Ha ha, just kidding, I’m going to give it to you anyway.
Obviously the main issue with 1/0 is that I can prove that 1=2 with the following.
0*1=0
0*2=0
0*1=0*2
0*1/0=0*2/0
1=2
This begins with an obvious truth, and ends in an obvious falsehood, all while sticking to the basic rules of mathematics. HOWEVER, our mistake in believing this is fact, is that we forget that not only does multiplication by zero cause all finite variables to lose value and become zero, division ALSO causes these finite variables to lose value, but instead they become one, since this is the identity for division. This means that for all finite non-zero X, X/0=1/0, and instead of the previous equations result, we end up with this.
0*1=0
0*2=0
0*1=0*2
0*1/0=0*2/0
0*1/0=0*1/0
0/0=0/0
Now, I’m going to pretend like we all don’t already think that 0^0 is 1, and also that 0/0 is an entirely spurious idea. I’m also going to pretend like the X/X=1 law doesn’t exist, for kicks and gags. This will go into logs, though, so pay attention.
We hold this as truth, except in the cases of X=0, so I will prove it, all while assuming X≠0:
X^0=1
If I take the log (Base zero), I get
log(X)/log(0)=0
Now, if I Divide by log(X) I have
1/log(0)=0/log(X)
1/log(0)=0
And take the reciprocal
log(0)=1/0
This means that in any non-zero base, log(0)=1/0. Now to prove the math is flawless in its foundation, I continue with the original reworked equation:
log(X)/log(0)=0
log(X)/(1/0)=0
log(X)0=0
log(X^0)=0
This is already an identity, as our non-zero finite X raised to 0 is one, and log(1)=0.
X^0=0^0
Since X^0 equals 1 in all non-zero finite cases, and we had already stated X fit these ideals, it goes to say that:
0^0=1
The Cool Dude,
x^(a-b)=x^a/x^b
is invalid for x=0
because 0 can not be a divisor.
0^0≠0/0
As for the discussion with log,
I can not find a reason for it to be correct.
How about instead of telling me what I can’t do, tell me why, as it seems to me that my math is flawless. First tell me why my logs are incorrect, then tell me why 0/0≠0^0. Because the logs prove that zero can be a theoretical divisor, in the same way as the imaginary unit is a theoretical multiplier.
In any case, your statement contains circular logic, as you claim that my proof for 0/0 can’t be true because there is no 0/0, and the result is 0/0. This argument can be used for basically anything, and therefor is invalid. I would be glad, however, for you to explain to me why I am wrong, even though I am not.
I will now explain exactly what I did with my logs, and why.
Given: X≠0
Fact: X^0=1
log(X)/log(0)=0
1/log(0)=0/log(X)
1/log(0)=0
log(0)=1/0
I prove that log(0)=1/0 to use in my next point, where I plug the result into X^0=1, after taking the log.
log(X)/log(0)=0
log(X)/(1/0)=0
log(X)0=0
The reciprocal of 1/0 is 0, and so dividing by 1/0 is the same as multiplying by zero.
log(X^0)=0
I move the zero into the log as an exponent, because this is how logs work.
X^0=0^0
Raise zero to each side, erasing the log, and causing 0 to be raised to 0.
0^0=1
X≠0, and X^0=1, and 0^0=X^0, so 0^0=1.
Tell me why any of this is wrong, without just saying that I can’t divide by zero, because that isn’t a reason, it’s an excuse.
The Cool Dude,
I have told you,
x^(a-b)=x^a/x^b
is invalid for x=0
so 0^0≠0/0 .
I cannot tell you why your logs are incorrect,
I can only tell you I cannot understand why they are correct.
log(X)/log(0)=0
I don’t understand where it comes from.
log(0) is undefined.
Zero cannot be a divisor in any case.
All you have to do is say why. Saying something is incorrect because the thing that it was designed to prove was proven isn’t a valid point. And even so, if you try to resolve the equation with the rules of the undefined, every single step ends up being a correct statement anyhow.
The rules being:
U=1/0
-U=U
0U=1
0/U=0
U/0=U
U+X=U
U*X=U
U/X=U
log(0)=U
log(U)=-U=U (Not base zero)
The whole point of U being considered an object is that it can still re-enter the real form of numbers, but the instant you use it, each side becomes either U, 0, or 1, and then it’s just an identity statement. (Depending on the usage of U, and the placement of zeroes on each side)
Basically the rules of U are all correct statements about the nature of U that lets you use U in an equation without breaking the equation. But it’s not like it really matters. The whole point is just so that you always end up with real equalities, even though both sides contain the exact same thing. If you have to change the rules based on the situation, your rules are wrong, wouldn’t you agree?
Your rules are strange.
I don’t agree.
Ha ha, yeah, but so is zero.
4^0=1
3^0=1
2^0=1
1^0=1
(-1)^0=1
(-2)^0=1
(-3)^0=1
(-4)^0=1
Does NOT demonstrate that 0^0=1 in precisely the same way that:
f(x)=x/x
f(3)=1
f(2)=1
f(1)=1
f(-1)=1
f(-2)=1
f(-3)=1
Does NOT demonstrate that f(0)=1. It clearly doesn’t. 0/0 is undefined. The LIMIT of a function, that is, the value of a function as it approaches something is NOT the same thing as the value of a function when it equals something.
The limit of f(x) as x approaches 0 does equal one, but the value of f(x) when x equals zero in undefined. The limit of x^0 as x approaches zero does equal one, but the value of x^0 is undefined.
Yee, you are absolutely right to say 0^0≠0/0.
the statement “undefined = undefined” would be complete mathematical gibberish.
Consider this function: f(x)=(3^x + 9^x)^(1/x) , x≠0
For increasingly larger values of x, we see that the function approaches 9.
f(1)=12
f(2) ≈ 9.48683298
f(5) ≈ 9.007395
f(10) ≈ 9.000015
f(20) ≈ 9.000000000129
etc…
We can evaluate the limit of the function as x approaches infinity as such:
lim(x→∞) f(x) = lim(x→∞) (3^x + 9^x)^(1/x)
=lim(x→∞) (9^x (3^x/9^x + 9^x/9^x))^(1/x)
=lim(x→∞) 9 (3^x/9^x + 9^x/9^x)^(1/x)
=lim(x→∞) 9 ((1/3)^x + 1)^(1/x)
= 9 (0+1)^0 = 9 (1)^0 = 9*1 = 9 Exactly as expected.
1^0=1 is uncontroversial, and we get the answer that we thought we would. However, we can do a solution that tests the idea of 0^0 as well by multiplying the expression with something larger than 9^x as follows:
lim(x→∞) f(x) = lim(x→∞) (3^x + 9^x)^(1/x)
=lim(x→∞) (12^x (3^x/12^x + 9^x/12^x))^(1/x)
=lim(x→∞) 12 (3^x/12^x + 9^x/12^x)^(1/x)
=lim(x→∞) 12 ((1/4)^x + (3/4)^x)^(1/x)
= 12 (0+0)^0 = 12 (0)^0 = ??
If we insist on 0^0=1, we end up with 12 as our solution which we KNOW is wrong. If we evaluate it with the right answer of 9, then 0^0 must equal 3/4 in this specific case.
If we do this evaluation again by multiplying the expression by 18^x outside the parentheses and dividing by 18^x inside the parenthesis… we end up with…
lim(x→∞) f(x) = 18 (0)^0 = ??
Of course, the answer is NOT 18. We have already proved that it was 9. In this case, 0^0 must equal 1/2 in order for the expression to evaluate correctly.
What this ultimately demonstrates, is that 0^0 is undefined. That doesn’t mean that it can’t equal one in some cases, it just doesn’t equal 1 in all cases. In these two examples, it equaled 3/4 and 1/2. In reality, depending on what we multiplied the function by, we could have attained an infinite number of possibilities on the value of 0^0 (between zero and one). That is what it means to be undefined.
Well yes, Ron, this is why nobody with a credible argument was taken seriously, because I could easily present the graphs 0^x and x^0, where they should CLEARLY intersect at x=0, but for every other value, differ by exactly one.
I, on the other hand, used only solid equalities, all that start from an absolute fact, and end in something that is already *generally* accepted as true. In fact, the only argument you could use against my math is “1/0 is unpredictable”, but my logic seems to remove all self conflicting statements, and therefor is MORE correct than no statement at all.
In any case, as I said before, saying something doesn’t work because it SHOULDN’T work is not a reason, it’s more along the lines of an excuse.
Oh wait, I retract my assumption. That was in fact YOU who said that, one HALF page ago.
It still, however, does not apply here, as again, really the only thing you can say is “That doesn’t conform to my current set of ideas about the reciprocal of zero”. Unless, of course, you would like to find a flaw somewhere in that mathematical storm I had posted earlier. I would be glad to hear about anything immediately wrong you can point out, so that I might fix it.
Ron,
There is no need for a precise demonstraton.
It is a matter of rationality.
Defining 0^0=1 is reasonable.
@Yee…..just so that we can say 0-(0-1)=1
People use things like that to work their way to RHS=LHS….similar situation….as I said I am not a genius and I am probably wrong 😐
Ron,
Limit is not a necessary consideration.
f(x,y)=x^y is not continous at (0,0).
However f(0,0) can still be defined.
It’s quite obvious that we can’t agree on an actual definition for 0^0, so why don’t we just agree on a largely accepted idea of 0^0. Oh wait, that’s 1. Cool, I solved it without even using math. (joke)
Anyway Ron, you can’t disprove an argument by saying that the LIMIT doesn’t prove that argument. The ONLY thing you can do is say that using limits as an argument is a BAD argument. It does not mean that the point is incorrect, AND, nobody used limits.
Ananthu,
0-(0-1)=1 is correct.
But 5-(5-4)=1 is incorrect.
And what do they have to do with 0^0 ?
Logarithms are only defined for input values x > 0. You cannot write log(0). If you have a hard time believing this, consider this:
Assuming a logarithm with any base a such that a > 0 and a is not equal to 1.
y = log(0)
means
a^y = 0
There is no exponent y that can be applied to a base and have the output = 0.
Trax, logs don’t have a limit, they just have a limit for REAL numbers. In fact, it was proven that ln(-1)= i(pi), because e^(ipi)=-1. So if ln(-1) has an imaginary, but existing value, then why doesn’t ln(0)? The answer is that it does. It is equal to 1/0, the undefined unit, U. It isn’t a real value, mind you, but it is a value. Other than the fact that I proved it with my equations, it’s just quite obvious that any graph with an asymptote intersects that asymptote at U, because that is the definition of asymptotes.
Similar in that x^2+1=0 at i, e^x=0 at U. It doesn’t actually matter if U is positive or negative, as -U=U, just as -0=0.
I like the idea of U identities, but I am bothered by the following:
-1 = -(oU) = (-0)U = 0U = 1
It seems that there should be some distinction between U and -U in the same way that lim(x\to 0) 1/x is positive or negative infinity depending on whether x is approaching 0 from the positive or negative direction.
And what about cardinality? Surely U is not equal to U*U, otherwise
1 = o*U = 0*U*U = 1*U = U.
Now 1/0 just seems like wretched notation; who wouldn’t confuse (1/0)^2 with 1^2/o^2 = 1/0?
If powers of U are distinct, we should expect log to work nicely with powers of U. Let log denote log base e. Then,
log(U) = U = 2 U = 2 log(U) = log(U^2)
so
U = e^log(U) = e^log(U^2) = U^2.
Hmm, and worse
U = e^log(U) = e^U = e^(-U) = 1/e^U = 1/e^log(U) = 1/U = 0.
Oh noes!
Your mistake was the lack of parenthesis. (oU)U=U, while 0(UU)=1, because UU=U, but oUU isn’t even a valid notation. It’s kind of like if you did this.
7=3+4
2*7=2*3+4
14=10
Technically you did everything correct, but when you multiplied by two, you didn’t mind the parenthesis.
Although it appears that U has many flaws and holes, really the only thing you need to do is simplify, and use parenthesis. I mean, isn’t that how math works already?
Good point though. It isn’t so much that U is confusing as much as it is perfectly ordinary, we’re just not used to using objects in math this way.
Cool dude,
You are right when you say “you can’t disprove an argument by saying that the LIMIT doesn’t prove that argument.” The ONLY thing you can do is say that using limits as an argument is a BAD argument.”
That is exactly what I have said. There are cases when the limit does equal the function at x, and there are cases when they don’t. Limits of a function at x cannot be used to prove the value of a function at x. I never said that the argument was disproved because they used limits, only that they didn’t have a valid proof that it was true.
Cool Dude, I’m sure you already understand the following, but for others reading, here are two examples:
For f(x) = 2x,
lim(x→2) f(x) = 4 and f(2) = 4…. hence, the limit of a function at x can be the same as the value of a function at x.
However, for f(x) = (x^2 – 9)/(x-3)
lim(x→3) f(x) = 6 and f(3) =”0/0″ or undefined and thus the limit of a function at x and the value of a function at x can be different.
However, you are wrong to say “… AND, nobody used limits.”
Every time someone uses the reasoning:
3^0=1
2^0=1
1^0=1
etc….
They are using the CONCEPT of limits, even if they have never studied calculus and consequently have no idea what a limit is, or how they are used in math. They are using the idea of a “trend” that seems logical to them. That “trend” is the same idea as a limit, more or less.
I suppose I should change my statement then. Nobody important used limits. (joke)
Yee,
When you say: “f(x,y)=x^y is not continous at (0,0), However f(0,0) can still be defined.” You are making one mistake…You are assuming that mathematical definitions can be arbitrarily defined by humans. That is incorrect(could we not just arbitrarily define it as 2?? Of course not!). Definitions are derived from axioms and must be proven true. Regardless of what you may have read in one text book(x^o=1), other text books get it right by being more precise.. (x^0=1, where x≠0). Is it convient to think of x^0 as equalling 1? Sure, but it is also just as convienient to think of it as zero x’s. After all, that is basically what x^0 means.
You have yet to provide one solid example of 0^0=1. Your binomial theorem argument fails because (x+0)^n simplifies to x^n before ever applying the theorem. However, in cases where y or x MIGHT equal zero in (x+y)^n, the terms that evaluate to the zero power simply disappear from the final equation, resulting in 0^0 never actually being evaluated.
As for your argument that 0^0=1 is “reasonable,” what does that mean anyway? Is it EVER reasonable to have a definition that gives wrong answers as I demonstrated half a page ago??? I don’t think so. You said that there is no need for a precise demonstration, but nothing could be further from the truth. Just ONE precise demonstration that 0^0 doesn’t equal 1 is all that is needed to prove that 0^0 cannot be defined as 1. Proof of ‘all cases’ is a demanding standard, dispoving ‘all cases’ only requires one contradictory example.
0^0 is INDETERMINATE.
If x^0=1 for all numbers,
then a polynomial a[0]+a[1]*x^1+…+a[n]*x^n
can be abbreviated as
n
Σ a[k]*x^k
k=0
If not,
it is not exactly correct.
Ron, technically speaking, nobody ACTUALLY invalidated my argument, so it still stands, if for no other reason than to simply harass anyone who can’t disprove it.
Actually Yee, if we assume the x^0 simply means there are no x terms within that portion of the equation rather than insisting that it must equal 1 in all cases, the summation equation for polynomials works just fine for all values of x.
Cool dude, I am assuming that you are referring to your log proof…
log(0) isn’t indeterminate, it is simply invalid as there are zero solutions to e^x that evaluate to 0. To that end, it can’t be shown to equal anything, let alone 1/0, even if that equality were possible. 1/0 can at least be thought of as infinity, yet log(0) could never be thought of as infinity. Your proof falls apart in the very first step. Also, saying that “you can’t divide by zero is an ‘excuse’ ” doesn’t change the fact that you can’t divide by zero.
That is like saying, “Hey guys, prove me wrong, but you are not allowed to use the rules of math.” Naturally, this is absurd.
Cool dude, I just read your earlier post where you said the log was base zero….
The short answer to why this argument fails is that log base zero is an invalid function. It fails the vertical line test. For a function f(x) to be valid, for each value of x, only one result may be returned.
1/0 can never be evaluated to equal 3, yet o^3 does equal 0 and is a valid result. Of course, for any value of x, where x>0, 0^x will evaluate to zero. There are an infinite number of solutions that are valid solutions to log-base-zero(0) yet are invalid solutions to 1/0. Given that they can not be equal, they are not equal.
Indeterminate does not equal undefined. log-base-zero(0) does not equal 1/0
The square root fails the vertical line lest as well, but it is still a valid function.
The square root is certainly referred to as a function in everyday language, and can also be used in proofs as well. However, it is important to note that in order to use the square root in a proof, you must demonstrate the consequences of both possible values it returns, or at least demonstrate that one set of results is irrelevant to the proof or domain of the analysis.
log-base-zero(0) can, and does, equal 3 along with an infinite number of other correct values. However, 1/0 does not ever= 3 or any other real value in the domain of (0,∞). If you are going to use log-base-zero(0) in your proof, you must deal with all possible values that the “function” returns.
Your “proof” starts off with: log(x)/log(0)=0…. base zero where x≠0, leaves us with a problem:
1. log-base-zero(any number ≠0) does not exit(the “empty set”). There are no solutions for x such that 0^x ever equals a non-zero number. The only “possible” exception would be to assume that 0^0=1, but that was what you are trying to prove in the first place…Generally speaking, proofs in the form of “assume that x=n, therefore x=n” aren’t all that convincing. Assuming that x≠1 and by definition in the proof x≠0 is fatal to the “proof.” The empty set can not be added, subtracted, multiplied, or divided, etc… Assuming that x=1 in this case “might be” fatal in this step and is definitely not convincing…
Your next problem comes in the form of 1/log(0)=0/log(X)
2. We have already determined that log-base-zero(x) can only be valid when x={0,1}. All other values of x give us the empty set. (0^x can not equal any number greater than zero) AND x equaling the value of one still requires us to pretend that 0^0=1 has already been established. However, even if we pretend 0^0=1, we have a problem in this step. First off, we already know that x≠0 based on the conditions of the proof…that means that if x≠1, then the right side of the equation is 0/”the empty set” which is invalid and fatal to the proof. If x=1 (still pretending that 0^0=1), that means the right side of the equation is “0/0” which is indeterminate and fatal to the proof. You are not going to be able to get away with multiplying by the reciprocal in this case, because the reciprocal of 0/0 is still 0/0…
Your next problem comes in the form of: 1/log-base-zero(0)=0:
3. The only way that 1/log-base-zero(0) could ever equal zero is if log-base-zero(0) equals infinity(yet 0^ infinity is also an indeterminate form and fatal to the proof) . I’ve already talked at great length why log-base-zero(0) can equal in infinite number of correct values between zero and infinity which means that you must deal with the fact that your equation can also = 1/x where x is a real number. In other words, 1/2, or 1/3, or 1/(any positive real number) are all valid results for the left side of the equation, yet none of them equal zero. 1/2=0 is fatal to your proof.
Ok, well I started by typing an enormous set of paragraphs that thoroughly re-disprove everything you just said, but I have a feeling that nobody is actually reading these, and also we are extremely off topic. I will just be very simple. The point is to prove that 0^0=1. If it JUST SO HAPPENS to also equal a lot of other things, so be it.
Where still X≠0, X^0=1.
X^0=1
Log(X^0)/Log(0)=Log(1)/Log(0)
Taken log base zero
Log(X^0)/Log(0)=0/Log(0)
Simplified Log(1)
Log(X^0)/Log(0)=0/U=0^2=0
ASSUMING Log(0)=U
X^0=0^0
1=0^0
So as long as Log(o) CAN equal U, 0^0 CAN equal one. So, let’s prove that Log(0) CAN equal U.
So?
1^0=1
Log(1^0)/Log(0)=Log(1)/Log(0)
0Log(1)/Log(0)=Log(1)/Log(0)
0/Log(0)=0U/Log(0)
0/Log(0)=1/Log(0)
Log(0)U=Log(0)
Now we have a clear issue. We can’t define Log(0). But we don’t need to. All I need is ONE result. If I plug in Log(0)=U…
UU=U
1/0^2=1/0
1/0=1/0
U=U
This means that Log(0) CAN be U. If Log(0) CAN be U, Log-base-zero(1) CAN be zero. If it CAN be zero, X^0 CAN be 0^0, and X^0 is one, because I stated that X doesn’t need to be zero. But it CAN.
Perhaps you could argue 0^0=(1,2,3,4,5,6,7….) But it doesn’t matter. It CAN be one, therefor, it is.
Sorry cool dude, but you haven’t proven that it CAN equal 1 because you had to break the rules of math to do so.
Any proof that has invalid steps is invalid. period.
2=1 does not make a proof valid in one crazy instance.
besides, even if you argued that it could equal one, You already know that it doesn’t always equal one.
That ALWAYS invalidates you using the equals sign. Unless, of course, you want to confess that it equals indeterminate….Naturally, I won’t disagree with you there…
You had these consecutive steps in your proof:
0/Log(0)=0U/Log(0)
0/Log(0)=1/Log(0)
Sorry, but U*0 does NOT simplify to 1
Well, it depends on how you define U. If you define U as equaling 1/0, then yes, then we might have an issue. But instead we define U as the solution to 0U=1. This issue also comes up in the imaginary unit. Instead of simply saying i=(-1)^.5, we say it is the solution to i^2=-1. This is because that the contents of multiplied square roots can only be combined if at least ONE of them is positive. If I combined two square roots of negative one, I would end up with positive one, and we know this as false, because the correct answer is negative one.