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.
My physics teacher says it’s infinity….
Physicists say a lot of things. Can’t be trusted.
well, if 0^0 = 0/0 , then 0^0 = any real number, because if you use cross multiplication, and multiply 0 on both sides, you will get 0=0(x) , x being any real number.
i believe it is zero…………look at it this way
to solve exponents in fractions you subtract them-top exponent minus bottom exponent-, if u put 0^3 over 0^3 when you subtract them it is 0^0, if you you just solve 0^3 over 0^3 then it comes out as zero
Would like to see explanation using hyperreal numbers.
Any takers?
0^0 is what my face is doing right now…
0^0 doesn’t really mean anything. It is undefined.
However you can take the limit of x^x as x approaches 0 using this method…
(assume x->0 means “The limit as x approaches 0 from the right side” since
approaching 0 from the left side in this context makes no ‘real’ sense.)
given y = x^x
x->o y = x->0 x^x
x->0 ln (y) = x->0 ln(x^x)
x->0 ln (y) = x->0 x * ln(x) Use of logarithm rule
x->0 ln (y) = x->0 ln(x) / (1/x)
x->0 ln (y) = x->0 (ln(x))’ / (1/x)’
Taking the limit at this point would result in -Infinity/Infinity
Since -Infinity/Infinity is an indeterminate form, use of L’Hôpital’s rule is permitted
x->0 ln (y) = x->0 (1/x) / -(1/x^2)
x->0 ln (y) = x->0 -x
x->0 ln (y) = 0
x->0 e^(ln(y)) = e^0
x->0 y = 1
Since x->0 y = x->0 x^x
x->0 x^x = 1
0^0 is essentially multiplying zero by itself never, or not multiplying zero at all. Hence, would 0^0 not simply equal zero?
It’s simply o….why…anytime you have nothing…and you add…subtract..multiply or divide by 0…it’s in fact still 0…. the only time you will have anything other than nothing…is when the value of what you are adding or subtracting is something other than 0… which in this case…. it is not. It’s simple logic and mathmatical fact.
My Calculator is wrong then? I tried 0^0 and it yielded “0”.
Any calculator that doesn’t say “in deference, I respectfully decline to answer this query” is wrong.
But, if it has to say something, “0” is among the top two answers.
Any number x to the nth power (x^n, say 5^2) can be written as 5*5*1*1*1*…*1 where the number of 5’s represented is determined by the exponent. therefore 5^0 would be _*1*1*…*1, or 1. therefore, there are no 0’s represented in 0^0, so it is just _*1*…*1 and thus, 1.
L’Hopital’s rule, end of story
Nore is 0^0 = 1 or 0 but if we look very close to the graph of x^x
http://www.wolframalpha.com/input/?i=x^x+x+from+-0.1+to+0.1
we see that it coult be 1 but it isn’t just of the fact that you would realy have to divide by zero to get an result. 0^(1-1) or 0^(2-2) … turns out to be 0*0…/0*0…
the difference is that you dont have to divide by zero when you want to calculate 0^x x>0 you will always make this out of it.. 0*0… the result then is zero
My calculator said “A Suffusion of Yellow.”
Ha!
Good times.
It’s indeterminate! You’re using different definitions and getting different answers, 0/0 is NOT undefined, it’s INDETERMINATE. for example, if you look at lim x->3 (x^2-3)/(x-3), if you just plug in 3, you’ll get 0/0, which is an indeterminate expression, meaning it still likely has a limit. Now, if you do take the limit, lim x->3 (x^2-3)/(x-3)=> lim x->3 (x-3)(x+3)/(x-3)=> lim x->3 (x+3)=6
If an expression is indeterminate at a point, it can(not necessarily does(correct me if I’m wrong)) have a limit at the point, but an undefined form, ie. 1/0, can’t have a limit, unless it’s a one sided limit.
Really, the difference between indeterminate though and undefined is fairly simple: undefined: having no meaning, ie. 1/0
indeterminate: having multiple solutions to an arithmetic expression, ie. 0/0, 0^0, 1^(infinity), infinity^(infinity), etc.
Or, as wikipedia defines an indeterminate form: “More formally, the fact that the functions f and g both approach 0 as x approaches some limit point c is not enough information to evaluate the limit
lim x->c f(x)/g(x)
That limit could converge to any number, or diverge to infinity, or might not exist, depending on what the functions f and g are.”
So basically, in layman’s terms, 0/0 has infinite values, ie. 0/0=3, 0/0=4, 0/0=infinity, etc.
It is a fairly complicated concept, and hard to explain, but to clear up any confusion, 0^0 is indeterminate and can be shown to have any value from 0 to 1.
If anyone can explain this better, please do so, I’m only a calculus student at this point, and I realize some of the way I explained it may have been bad, but I tried to make it as straightforward as possible.
I plotted the modulus of x^x from -1 to +1. From there it is evident that |x^x| (modulus because negative number to the power of a negative rational number may be imaginary) tends to 1 as x tends to 0.
However, this is the same as finding the limit and doesn’t really equal to 0^0, unless of course we assume that |x^x| is continuous.
Anyway, another reason I could think of is, lets say I have
x = a^a
=a^(a-0)
[=a^a/a^0]
ie a^a = a^0 * a^a
let a become zero, and substitute x = a^a or x = a^0
x = x^2
this equation has two solutions, 0 and 1. If it is zero, then I must have divided be zero in the step b/w [].
Another reason is binomial expansion of (1 + 0)^0.
Usually, an indeterminate form is viewed as a concept relating to limits. The question was not “what is 0^0 when it occurs after I have substituted expressions with their limits.” In that case, it would be an indeterminate form, by definition.
Wikipedia explains this nicely:
“In calculus and other branches of mathematical analysis, an indeterminate form is an algebraic expression obtained in the context of limits. Limits involving algebraic operations are often performed by replacing subexpressions by their limits; if the expression obtained after this substitution does not give enough information to determine the original limit, it is known as an indeterminate form.”
nothing times nothing never is still nothing – you can’t take knickers off a bare arse
Unfortunately to everyone who claims that “anything times nothing is nothing” everyone already agrees that 2^0 = 1, and that 1^0= 1, and that 0.0001^0=1, and that every *other* number raised to 0 is actually 1.
Same way that we know that x^1=x, no matter the x, we know that x^0 = 1, no matter the x. Zero isn’t be an exception.
Where people intuition fails is that they think they start by zero and have to multiply it. But “starting with zero” would mean having one zero, namely 0^1. 0^0 means that we *don’t* start with zero — and that means we start with the default 1.
If you repeat SumYungGai’s analysis but start with allowing x to be a complex number in polar form r*exp(i \theta), then take the limit as r -> 0, you find that lim r->0 x^x = 1 approaching from any direction \theta. This may explain why it is best from a mathematical perspective to define 0^0 = 1.
The way I usually tackle this problem is with the natural log (ln).
Step One: 0^0 = x
In this step I am saying that I do not know the answer to the problem and I want to find the solution… So how do we do that? Keep reading….
Step two: ln(0^0) = ln(x)
In this step I am taking the ln of both sides (any algebra student would do this).
Step three: 0*ln(0) = ln(x)
This step is just taking out the exponent from the log. Once again, any algebra student would do this.
Step four: 0 = ln(x) 0r ln(x) = 0
This step is my favorite as it gets us close to the answer. I know that ln(0) is irrational, what have you… but multiply it by 0 and it becomes 0. So then I just rearrange the equation to be easier on the eyes.
Step five: x = 1
This step comes out because I needed to find a value of x, whose ln is 0. I know this number to be 1. So, the answer to what is x, is 1.
This was a specific solution, meaning that is not generalized. But notice that if you replace the equation with x^0 = y, then you find that for all x, x^0 = 1. Just follow the steps and you’ll see in steps two and three that the power (0) comes out of the ln and reduces the left side to 0. Amazing how math works out like that 😀
Any questions, please leave a reply. Any revisions, please leave a reply. Any complaints… do whatever.
Hope this clears up the situation:
0^0 = 1 and x^0 = 1 for all x (for all x was not proven in this post).
ln(0) is negative infinity. So you can’t just do 0*ln(0) = 0.
I don’t understand. Sorry but I’m not a clever student, so please state it simple. Thanks.
Here’s a bit more on why 0^0 = 1 is “natural.” It’s more or less the same as the Binomial Theorem reason, but with more words.
Take a sheet of paper and draw 3 red dots on the left side and 4 blue dots on the right side. For each red dot, draw a single line connecting it to one of the blue dots on the right, allowing the possibility of picking the same blue dot multiple times or not at all. Be sure that every red dot has one and only one line, though. How many diagrams can we draw? Each red dot has 4 choices, so it’s 4*4*4 = 4^3. If we have r red dots and b blue dots, the total number of possibilities is b^r.
Now, what if we had 0 red dots? Then we can’t draw any lines, but we have a perfectly fine diagram already — every red dot has exactly one line, so it’s valid and satisfies the rules (can you find a red dot that fails?). Then b^0 = 1. Meanwhile, if we have some red dots but 0 blue dots, we can’t draw anything to satisfy the rules, so 0^r = 0. Finally, what if we have 0 red dots and 0 blue dots? Again, the “empty” diagram is acceptable because each and every (nonexistent) red dot has it’s own line. Thus it makes sense that 0^0 should be 1.
Is it not true that the multiplicative identity is one?
Therefore the idea of nothing happening with multiplication would leave us with one by definition. It is just like 0! (factorial).
0^{0} is undefined. 0^{0} does not have a value.
0^0 (using a superscript) is a 3 dimensional infinity sign.. the farther circle is just farther away, yeah? and maybe the infinity sign is just a couple of sine waves anyway, one over the other.. but infinity is just two 0s anyway, huh? split; multiplied.. still equals zero, yeah… but if you get 360 of those 0^0s (with superscripts again) you get 360 degrees.. which is one circle.. or a zero (which is a circle) so now you have 1^0 which still equals zero, but now you can pretend the number one even exists.. but not because it doesn’t, but it also equals 1 and 0 and both, and only each one separately all at the same time.. or maybe it doesn’t matter except when it does..
(I think I got the order of some of that wrong.. oh well..)
Maybe the entire resin we have not been able to discover certain things is because we always choose the more simple or elegant way. Who knows what we could be doing if any scientist, not just a mathematician, stepped outside the air of comfortability and chose to do something the harder, less elegant way, like making 0^0=0 instead of 1.
Mathematicians/scientists are a weird bunch, there are plenty of them who just have a knack for doing things the hardest way possible…
(any number pow zero) is defined as 1
a E R pow 0 = 1
so 0 pow 0 = 1
this is part of the definition of (any number)(to the power of)(any number)
so we have no reason to wonder what (0 pow 0) is, as it is defined.
But o * o = o !!!
Yes, so (0 pow 2) = 0
So what?
We are talking of the power of 0
Hi All,
I’m not sure if there is a term in math like below 🙂
but in the function x^y if we consider both x and y getting closer to zero we got result as “1”.
Lim(x^y) = 1
X->0 & y->0
Issue here is with how exponents work fundamentally. Look at this simple pattern for how exponents work.
X ^ 3 = X times X times X times 1
X ^ 2 = X times X times 1
X ^ 1 = X times 1
X ^ 0 = X times 1 times 1/X
X ^ -1 = 1 times 1/X
X ^ -2 = 1 times 1/X times 1/X
X ^ -3 = 1 times 1/X times 1/X times 1/X
There is two ways you can view Zero to the Zero Power now.
1. Accept that Anything divided by zero is either undefined, impossible, falsity, or non existent.
2. Then X/1*1/X = X/X = 1
Just depends on where you are on the fence with something that is divided by zero.
then again, x/x = 1
but 0/0 = 0 so that rule doesnt apply
You didn’t read what I said.
I said choice 1 or 2.
You would be choice 1. Which says anything divided by zero doesn’t exist, is impossible, a falsity etc.
For people who think it does exist and we just can’t solve it. There is choice 2.
You could also look at the fact that everything getting closer to 0 such as
+/- .000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001.^0 is practically 1 and only get closer to one as it gets near zero.
Not to mention in my opinion if 0/0 even if they say it is impossible. Would in fact be 1 because infinite nothing goes into infinite nothing one time. So 0/0=1.
You define y^x as:
y^x := 1 * y * y … * y
and say if you have zero ys th result would be one. But how can you multiply 1 with nothing? And why would 1 * = 1 ?
You can think about it this way:
1 * y * y is applying the operation “multiply by y” to 1 two times.
1 * y is applying the operation “multiply by y” to 1 one time.
1 is applying the operation “multiply by y” to 1 zero times. When you apply an operation zero times it does nothing.
An elegant and quite comprehensive explanation why 0^0 = 1. Thank You 🙂
I believe that 0^0 does not exist. Maybe we should do some voting?
No amount of opinion can settle a mathematical question! Only argumentation.
As always, Calvin & Hobbes answered this question a long time ago: http://blog.onbeing.org/post/250746172/calvin-and-hobbes-math-is-a-religion
Any number, including 0, multiplied by 0 equals 0.
I can see one fault with much of the thinking here…
yes if x>0 or x (0,0)) x^y = 1 for all y in the real numbers
but unfortunately in order for a limit to exist, it needs to approach the same thing from ALL angles… which means
(lim(0,y) -> (0,0)) 0^y = {-infinity, infinity, or 0} = 1 = (lim(x,y) -> (0,0)) x^y
but no… 1 does not match any of these solutions… thus 0^0 is undefined under the general definition.
0 raised to the power 0 is 1
According to the rule of converting ’10’ base 2 (binary number) into base 10 (decimal number)
0 raised to power ‘0’+1 raised to power ‘1’
=1+1
=2.which if converted back into base 2 (binary number) will be 10.
if 0 raised to the power 0=0, then 10 base 2(binary number)=1 which is incorrect.
So, 0 raised to the power 0=1.
latex path not specified
What about 0↑↑0? How about higher rank hyper-operations?
According to Wikipedia, a↑↑b is only defined for positive a, but I am thinking (hoping, actually) that perhaps this is erroneous (which is quite possible as there are claims in multiple places that 0^0 is undefined as well).
It’s one because the song says so. (“Any number under the sun, raised to the zero power is one”) This proof is most convincing with sung with fervor.
I’m getting lots of “latex path not specified” errors.