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.
look at the earlier posts
0^0 does not equal 0/0
“0=0^1=0^(2-1)=0^2/0^1=0/0”
0/0 is indeterminate, has infinite values, not infinity
No matter how many times I explain 0^0 is different from 0/0,
there are people who say so.
lol
good argument for 0^0=1 maybe
e^x=0^x ; o^(x(ln(e)/ln(0) ))= 0^x ; x*(1/-inf))=x ; x*0=x ( x must be zero for this to be true, e^0=1, so 0^0 should be 1)
Why should e^x=0^x be true?
i rearranged e^x=0^x to x*0=x (go through my previous post proof thing, I think it checks out)
so what does x have to be to make the equation equal? ( assumed that ln(0) is -infinity and that 1/infinity is zero)
Why do you write the equation e^x=0^x?
or we could go the other way e^x=0^x e^x=e^x(-inf); x=x(-inf) ( i am not entirely convinced that if x was zero; x=x(-inf), so my question is whats zero times infinity? i think it is zero personally
e^(0*inf)=(e^0)^inf=1^inf= 1
The best explanation I have seen of why 0^0=1 goes as follows:
Consider two functions: f(x) = 0^x, and g(x)=x^0. In a large number of situations (such as polynomials), it turns out to be important that g(x) is continuous (and equal to 1). On the contrary, f(x) is relatively unimportant, and few people care about it.
Limit is not the only viewpoint of math.
Limit does not exist does not imply the fucntion is undefined at this point.
If 0^0 is defined as 1,x^0=1 is valid for all complex numbers.
0^x=0 is valid only for x>0.
Im not a matematician but I think like this:
(x^0)=x^(1-1)=(x^1)x^-1=x(1/x)=x/x=1
0^x=0(0)0(0)…=o+o+o+o+0…=0
0^-x=1/(0^x)=1/0
Now, I supose that 1/0=∞,because note that 1/(1/2)=2,1/(1/4)=4,1/(1/10)=10, in resumen 1/(1/x)=x,the smaller is the number with we divide 1 the bigger is the division.
So (1/0)^-1=0/1=0
And 1/0≈1/(1/∞),so 1/0 tends to ∞.
Now:
0^0=(0^1)(1/0)=0∞
Oh! The indeterminated formula,ok I think like this:
0∞=0(0)0(0)…=0+0+0+0+0…=0
0^0=0∞=0
So, what do you think?
0 to the 0 power is not 1
It is “undefined”
0 to any power other than zero(not a power) is 1
Get it? Got it? There is no controversy
BRUCE990,
Why does x^(1-1) equal x^1*x^(-1)?
It is invalid for x=0,
and can not be a reason not to define 0^0.
Power zero has nothing to do with division.
Cynthia,
0 is undefined before it is defined.
0 to only positive power is 1, not any power.
Defining 0^0 as 1 is reasonable.
+1 for Michael. Rarely do we care that 0^x is 0 for all values of x, but x^0 needs to be continuous for all values of x in order to avoid many messy and pointless special cases.
(1/0) is not infinity, its undefined
0 never changes from 0 while infinity implies it does ever so slightly
Continuity should not be a major consideration.
It is at most a minor consideration.
x^0=1 is very convenient.
Polynomial
a[0]+a[1]*x^1+…+a[n]*x^n=
a[0]*x^0+a[1]*x^1+…+a[n]*x^n=
n
Σ a[k]*x^k
k=0
It is a brief expression.
Binomial theorem
(x+y)^n=
n
Σ C(n,k)*x^(n-k)*y^k
k=0
C(n,k)=n!/k!/(n-k!)
(1+0)^2=
C(2,0)*1^2*0^0+C(2,1)*1^1*0^1+C(2,2)*1^0*0^2
0^0=1 is a must.
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But people said that the answer is infinity
explain why the 1 in 612 does not have the least value of the three digits.
I’m 14, and don’t understand any of this.
We say anything power 0 is equal to 1 i,e
x^0=1 but it is only possible when x is not equal to 0
therefore 0^0=infinity
asif rasool,
It depends on definition.
If 0^0 is defined as 1,
x^0=1 is valid for any number.
It is convenient in many fields.
mathamatics is fun also
asif rasool,
How did you obtain 0^0=infinity?
email new posts. thanks.
What you can do is go to the top of the page and look at the right side. We got us some twitter and rss and all kinds of stuff!
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This may sound dumb, but prove x^0=1
this how I see it, let x=5
example 5^2= 5X5
(exponent# -1= #of operation sign(multiplication if positive, division if negative))
so, 5^1 = 5 because (1-1= 0)
5^0= 5/5 because (0-1=-1)
5^-1= 5/5/5 because (-1-1=-2)
Now lets replace 5 with 0 and see what changes
0^1 = 0 because (1-1= 0)
0^0= 0/0 because (0-1=-1)
0^-1= 0/0/0 because (-1-1=-2)
I know zero to any negative number is undefined, but what about 0/0/0?
Patty,
There are no index laws like that.
If so,
then 0=0^1=0^(2-1)=0^2/0^1=0/0
How do you explain it?
0=0^1=0^(2-1)=0^2/0^1=0^1*(0^1/0^1)= 0*indeterminate value=0
a^(m-n)=a^m/a^n
is invalid for a=0,
because 0 can not be a divisor.
The answer is neither 0 nor 1 but undefined.
The answer depends on how people choose to define.
It is convenient to define 0^0 as 1.
Just because its convenient does not mean its right, its like saying slavery is ethnically right because it was more convenient for the economy. There is no clear prove for 0^0=1 other than its convince for other laws, so its indeterminate.
It is irrelevant to right and wrong.
Nothing needs to be proved.
patty, you’re really reaching here. Bringing slavery into the discussion as a comparison is ridiculous, to be frank. This is not a moral issue, you are not advancing your argument convincingly with that sort of rhetoric.
In any case, I agree with Yee in that it’s a matter of definition. The most common scenario, in my experience, is that 0^0 is defined and equal to 1. That’s not to say that some don’t choose to define 0^0 as 0 (or something else, or even leave it undefined). In fact, there are probably authors who will handle it differently depending on what they are working on.
As to your point about no proof existing that shows 0^0 = 1, why do you require one? I’m assuming you agree that 1+1 = 2, but have you ever asked for a proof of that? What if I’m a computer scientist so I define things such that 1+1 = 10 (and the symbol “2” has no meaning in my world)?
Math is, very broadly, about defining constructs on top of a set of axioms and applying logic to them to get useful results. As the original post points out, depending on your choice of definition of x^y for x,y > 0, the value of 0^0 may be a natural extension. Or you might just arbitrarily choose o^o = 1 to make things “nicer.”
0^0=0/0= (42*0)/(1*0) let 0=x (42*0)/(1*0)=42x/x=42. Satisfying?
So, 0^0 is what you get when you multiply six by nine!
0^0≠0/0
X^0 = 1
X = Any real number
4^0 = 1
3^0 = 1
2^0 = 1
1^0 = 1
0^0 = 1
Why? Because the math rule states that any real number to the zeroth power is equal to 1.
It is not 0. Even use a calculator. It equals 1.
let me re-do my last post, Just because its convenient for most people does not mean its right. Its good to stray from popular beliefs every once and a while, just look at Einstein and Galileo. the majority believe that 0^0=1 , yet it could be proved to be really any # we want
patty,
Please don’t use these words :”right”,”prove”.
Really? Why should the words, “right” and “prove” not be used in this discussion? What troubles you about these words, Yee?
Bounce has rightly pointed out that mathematics builds logical constructs out of axioms, but axioms are simply statements that are generally accepted without proof. Unfortunately, 0^0=1 is not as generally accepted as other axioms, else there would not be the controversy even amongst the professionals. So, while I myself see the usefulness of 0^0=1, I can at least appreciate Patty’s skepticism about this definition in lieu of any proof. In my opinion, her instinct favoring proof is rather natural for a mathemetician.
As for using the term “right” (and, by implication, the term “wrong”), I hope we can admit that some logical concepts force us to use terms such as these, without regard to paradigm or, to use Bounce’s example, numerical base. (1+1=10 in binary.)
For illustration, consider these quotes:
“If a = b and b = c, a = c.” This statement is right, even true(!), in all paradigms where equivalence relations hold.
“With identical qualifications, a=b AND a≠b.” This statement is clearly false, i.e. wrong, as it violates the law of non-contradiction.
Is 0^0=1 right? Can it be proven? I dunno, but I bet alot of the great mathemeticians gave these questions some attention instead of shying away from them.
It’s a matter of definition,
so it does not need a proof
and there is nothing right and wrong about it.
Defining 0^0=1 is convenient in many fields.
This is the reason to define it.
Many people consider it in terms of limit
and choose not to define it.
Thank you for explaining your comment, Yee. I do accept the fact that a definition, standing on its own, does not require proof and, on its own, is not right or wrong.
0^0 = 1 is indeed convenient in many situations, as you have said. I adopt this definition often. Even so, where many people struggle with this is that another definition exists for 0^0, and together these definitions represent contradiction in the very same way that a=b and a≠b cannot both be true definitions with the same qualifying circumstances. Calling them different definitions does not really remove the obviousness of their contradiction to one another.
I suppose that we might agree that it would be more convenient for people not to bring up troublesome questions, but that might just violate another imponderable — human nature. 😉
Thanks for the interesting discussion.
Concerning 0^0,
mathematicians dispute between two options:
defining it as 1 or not defining it.
anything divided by ‘0’ is infinity ………. so i think ‘o^0’=infinity …………. i think so ….. after seeing the site ……….
but all things are so cool …. i like it ….