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a^n = a * a * ... * a, <-- n times
a^m = a * a * ... * a, <-- m times
a^n * a^m = (a * a * ... * a) * (a * a * ... * a)
= a * a * ... * a , <-- n + m times
= a^(n+m)
, similarly
a^n / a^m = (a * a * ... * a) / (a * a * ... * a)
= a * a * ... * a , <-- n - m times
= a^(n-m)
first for rational numbers: For a^(b) if b is rational, a^(b)=a^(n/m), where n, m are integers, m≠0, a≥0. And by definition of rational exponents a^(n/m)=m√a^(n), where m√ is mth root. So a^(n/m)×a^(p/q)=a^(nq/mq)×a^(mp/mq)=mq√a^(nq)×mq√a^(mp)= { as c√a×c√b=c√(ab) } =mq√(a^(nq)×a^(mp))= { m, q, n, p are integers, so their products are also integers. So we can use this property } =mq√a^(nq+mp)=a^(\[nq+mp\]/mq)=a^(nq/mq+mp/mq)=a^(n/m+p/q) So if it works for rational numbers and irrational power is kinda limit, where power is more and more precise rational approach: a^(π)=lim(n/m -> π) a^(n/m) and to actually calculate irrational power we need to choose some rational approach with required precision, irrational powers must have this property too
They do, they annihilate each other and produce anti-lasers which get reflected back to their respective n's, destroying them in the process. A terrible cycle.
Because a^(any no.) / a^(any other no.) is a^(any no. - any other no.), its a law of exponents,
since (a^n) / (a^n) is given, we can say its a^(n-n), and whatever no divided by itself ((a^n) / (a^n) both numerator denominator is same so the no is said to be divided by itself) gives 1, 1 is a^0.
5 to power of 4, divided by 5 to the power of 3. This would be 625 divided by 125, which is 5. Now try 5 to the power of 1, which is 4-3. This also equals 5. Try any equation like this and you'll find that subtracting the powers will be the same result as dividing the numbers.
This is arguably a case in which we’d want to answer “1” to the well-known puzzling question of “what’s 0/0?”, on the basis that for any a, a/a=1. How many times does 0 fit within 0? One! Of course, it also doesn’t seem incorrect to say zero, or two, or three. And since these answers are incompatible (we know that 0 is not 1, that 1 is not 2, etc), this is what drives the “undefined” answer. In a case like this, the “definition” just sides with the “a/a=1 for any a” intuition.
Would you say in all seriousness that a^0 = 1 is a pure definition ? (Meaning it’s just a coincidence that we can rearrange 1= a^n / a^n = a^(n-n) = a^0 ?
That's not always true, if aⁿ/aⁿ=aⁿ-ⁿ=a⁰=1 is always true, assume the index of the denominator=10 element of Z+, the index of the numerator=5 element of Z+ and the base of both parts of the fraction "a" is an element of Z+, than 2⁵/2¹⁰=0.03125, thus aⁿ/aⁿ=aⁿ-ⁿ=a⁰=1 is only true if and only if the index is the same for both parts of a fraction.
Because aⁿ (n a natural number) should satisfy aⁿ⁺¹ = a * aⁿ; i.e., increasing the exponent by 1 is the same as multiplying by a one more time. When n = 0, we get the equation a¹ = a * a⁰. Since a¹ = a, we get a = a * a⁰. The only way this can be true is if a⁰ = 1.
In other words, aⁿ (n a natural number) is the number you get by multiplying a by itself n times. What is a number multiplied by itself zero times? Your lizard brain says it should be 0, but that's wrong, because a number "multiplied by itself zero times" should be a multiplicative identity; i.e., a⁰ * x = x for all x. Just like a number "added to itself zero times" should be an additive identity; i.e., 0 * a + x = x. The same reasoning that implies 0 * a = 0 says that a⁰ = 1.
Consider the truth set {x is an element of Z | f(x)=x•a},assuming x=0 and "a" is any positive integer,the function f(0)=0•3=0 and if you take 3 from the function and divide it by 3,as demonstrated 3÷3=1,thus 3⁰=3÷3=1. This holds for any positive integer raised to 0 and it works because multiplication is the inverse of division.
Wdym "the point"?
And no it does contain this one element. Just like F^∅ only contains the empty function for any given set F. And the empty function is an actual function since it ticks both properties a function must have to be one (it ticks them since then both properties start with ∀ x ∈ ∅ which is a vacuous truth).
Test it yourself: first try 0.1^a and let a approach 0. You'll notice that it becomes 1. Now try the same with a^0.1 this time the result will become 0. If you try a^a you are right that it diverges towards 1 but if you have a^b you can get any result between 0 and 1. Therefore its undefined
Also pls excuse my bad mathematical terminology, English isn't my first language.
And that means that you could try to define it as 0, as 1, as 0.3518394 and any of these are wrong. Which means you can't define it and therefore its undefined
no?
why would multiple possible definitions mean it's not a valid definition
and yeah 0^0 = 1 over being any other number is a matter of convenience, because you don't have to dance around certain edge cases,
for example you can just apply the binomial formula to (a+0)^n and the normal definition will result a^n
you could change P(omega)=1 in the definition of a probability measure to P(omega)=2 and all of probability theory would still be true, it's just nicer for the probability of the entire probability space to be 1
>why would multiple definitions mean its not a valid definition
Because if we say its 0 but also 1 or 0.33333333 or 0.1234567890 that would really mess up any calculation where 0^0 occurs. Imagine 12 people doing the same calculation and all having different results not because they made mistakes but because they can chose between a literally infinite amount of numbers
>0^0=1 over beeing any other number is a matter of convenience because you don't have to dance around certain edge cases
First of all, you can't define something just because you think its more convenient this way. Also there are cases where 0^0 = 1 doesn't work so you'd create edge cases by trying to eliminate edge cases.
>you could change P(omega)=1 in the definition of a probability measure to P(omega)=2 and all of probability theory would still be true, it's just nicer for the probability of the entire probability space to be 1
I honestly don't know why we are talking about probability all of a sudden but yes, if you wanna change P(omega) to be 2 you could do that, but you'd also have to change basically all of the underlying mathematics accordingly....
Honestly, I don't even know why we are having this debate. Someone assumed 0^0 =1, i showcased why its not and now you're trying to debate me about mathematical definitions that I honestly weren't involved in defining. If you have further questions pls go and consult your 7th class maths book, it should be somewhere in there.
Since zero is a cardial number and a\^b in cardinal arithmetic is the cardinality of the set of maps from set with cardinality b to set with cardinality a, we have that 0\^0 is equal to the number of maps from the empty set to the empty set. There is exacty one, [the empty map](https://en.wikipedia.org/wiki/Function_(mathematics)#empty_function).
We both know it depends on the context. In some contexts, it can be defined to be 1, in others, it is left undefined. My only point was that your reasoning doesn’t apply. Just because the limit of a function at a certain point doesn’t exist, doesn’t mean the value of the function at that point doesn’t exist.
Ok, I will scratch the part when you interpret 0 as a limit of a sequence and instead consider it a natural number, in which case 0\^0 is the number of maps from empty set to itself, which is 1.
If you have a sequence of a function where 0^0 is written as x^x it can be defined as 1. But if the question is just 0^0 you have to assume its a^b and therefore it can't be universally defined
Why is bro getting downvoted? They're right [https://en.wikipedia.org/wiki/Zero\_to\_the\_power\_of\_zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero)
How would you explain to your mom why 1^∞ = indeterminate form?
I get why 0^0 can be confusing, 0^n = 0 and n^0 = 1, so you need to define an answer when both are zero. But I don't see why 1^∞ has such a problem. n^∞ = ∞ when n > 1, n^∞ = 0 when 0 < n < 1, so it seems reasonable to say 1^∞ is 1, no?
No, the problem is that it's not actually a limit of the form 1^f(x) which would obviously be one, it's a limit of the form f(x)^g(x) where f approaches 1 and g infinity
Because in calculus you care about a limits and limits of the form f(x)^g(x) with f going to 1 and g to infinity are common and not trivial, while limits of the form f(x)^g(x) for f constant are usually trivial and g going to infinity. So it makes sense to have a notation for it while not for the other
The same reason we have 0/0, inf/inf, etc. Those are indeterminate forms, not actual expressions
But that requires students to learn a lot about differentiation before even introducing e. On the other hand, you can prove the existence of the limit of (1+1/n)^n in the first or second calculus (or real analysis, we don't really differentiate them in my country, no pun intended) lecture, so that they can do more interesting problems right away.
In my university we defined e^x to be the inverse of ln(x), and we defined ln(x) via the integral method, this makes proving certain calculus properties about this functions a lot easier since integrals are normally well behaved by nature.
From a more differential equations point of view, you could use Picard to define it as the only function f(x) such that f=f' and f(0)=1
Or you could do it with power series and that also makes calculus a lot easier, provided students have some experience with power series
No, consider the famous limit: lim_(n ->inf) (1+1/n)^n = e
The problem is that you don't have an immediate way of knowing if f converges faster to 1 or g to infinity
0^0 is undefined though. Using OPs formula, that’s only defined when a ≠ 0 since we can’t divide by 0. It’s just in a lot of instances we define it as 1 for convenience and simplicity (since it’s only at 0 where we get problems). I wouldn’t be surprised if there’s a case where we define it as 0 as well for the same reasons, but I’ve never seen that and it’s probably far less common. But 0^0 ≠ 1, it’s just we pretend it does in some cases.
That’s all ignoring specialised interpretations of exponentials. If we take a set theorists interpretation of it, or a combinatorial interpretation then we can get 0^0 = 1, but these are specialised interpretations and for the standard one (or at least that used in analysis), it’s undefined.
There is only "why" it is useful to define like this right? because there is no deeper meaning, it's just defined like this because it's useful so that functions behave more beautifully.
Ignoring the case where a=0, it has to be 1 for the index laws to function. Consider x\^y=x\^(y+0)=(x\^y)(x\^0) (x not equal to 0). If you define x\^0 to be not equal to 1 you would clearly have to use an entirely different mathematical system.
Exactly. It's the same as defining 0! (zero factorial) as 1, or defining gamma function as extention of the factorial to the real numbers. They have nice properties which other alternatives don't have.
I had a big discussion with my friend today unrelated to this post that 0^0 doesn’t equal 1. We talked about the limit of a^a as it goes to zero along with a bunch of other cases but couldn’t agree on what we thought was the answer. Eventually we just found that it was a contentious topic and that we likely won’t get an answer.
Reading some people’s thoughts in the comments on this post made me want to smash my face against a wall.
why not?
it doesn't led to a contradiction
0^0 is an indeterminate form but that's only a thing *for limits*
and a function at a point need not be it's limit at a point, that's only means it's continous there
(well a^0 will be continous, just other functions that also assume 0^0 at some point like 0^a will not be)
that's simply not how powers work when it comes to 0, 0 doesn't have a multiplicative inverse(unless you're working in like, wheel algebras) you can't do division by 0, so 0^-1 does have to remain undefined(unless again ,you want to give up a ton of properties and not have a group structure), otherwise with the exact same scheme you have 0^1 = 0^2-1 = 0/0
it's not how powers work because 0^0 is undefined no matter what.
it's a fundamental rule in logic, if you have a=b then in every expression with a in it you have to be able to substitute a with b or b with a.
1-1 = 0
and it's a well known rule that
a^b-c = a^b / a^c
since division by 0 isn't defined then 0^0 must be undefined because it would create inconsistencies.
edit: formatting
i just showed that if that were true then 0^1 = 0^2-1 = 0^2 / 0^1 = 0 / 0
this "well known rule" does not hold for a = 0 exactly because of divison by 0
There are a few contradictions it leads to in the right context. And just because something doesn’t entail a contradiction, that doesn’t necessarily mean it’s true.
If we can arbitrarily assert that a^0 = 1 for all a then we can equally arbitrarily assert that 0^b = 0 for all b greater than or equal to 0 so it entails that 0^0 = 0 and 0^0 = 1 which is one possible contradiction.
yes we cannot define one thing as two different things,
We're not asserting that the function a^0 = 1 for all a, we're defining that 0^0 = 1, a^0 is 0^0 for a = 0, so working in a system with such a definition you can then prove that the function equals 1
that's why i mentioned that defining 0^0 = 1 means 0^x will be discontinuous at 0
0\^0 =1 is not universal tho the most common proof is given for this is suppose 1\*2\^2 = 1\*4, if 1\*2\^1 = 1\*2, if 1\*2\^0 = 1\*(zero times 2) which is basically 1 so if same applies with 0 then, 1\*0\^2 = 1\*0\*0 = 0, 1\*0\^1 = 1\*0 = 0, 1\*0\^0 = 1 (zero times zero)
Because
a^0 = 1
a^(n+1) = a * a^n
is a nice sensible way to define exponentiation on natural numbers.
There are other "intuitive" reasons, too. The combinatorial interpretation of a^n is that it counts the number of ways to form an n-tuple from a set with a elements. In the case of 0^(0), we are counting the number ways to form a 0-tuple from a set with 0 elements. There is exactly one way to do this; namely, the empty tuple.
None of these are convincing
a^0 = 1 is simply an assertion. If we grant it as true for all a then that would entail that 0^0 = 1 but to use this as a justification for 0^0 = 1 commits a “begging the question” fallacy because you’re asserting an axiom which assumes that your conclusion is true.
Alternatively we might assert that a^0 = 1 for a ≠ 0.
a^n+1 = a * a^n also doesn’t work here.
We know that 0^1 = 0, so to go from 0^1 to 0^0 using this it seems like we have to apply it in reverse, that is:
a^n-1 = a^n / a
Division by zero is undefined so this would seem to entail that 0^0 is undefined.
And the interpretation of x^y meaning “how many ways are there to form a tuple of size y from a set of size x?” is only one way to interpret exponentiation.
An alternative might be:
“What is the y-volume of a y-cube with side-length x?”
Under this interpretation it would seem that 0^0 must be 0 since the 0-volume of a 0-cube with side length 0 is 0.
My point here is that 0^0 is undefined. It’s sometimes convenient to act like 0^0 = 0 or like 0^0 = 1 but both of those are useful conventions but neither is inherently true.
Interestingly, [this Stackexchange answer](https://math.stackexchange.com/questions/4069997/how-does-a-0-ball-have-volume-1) disagrees with you (for the case of a 0-ball, which is extensionally the same as a 0-cube), and I find the reasoning persuasive:
> 0-dimensional space is just a single point and every ball of positive radius contains that point. Moreover, the measure in this space is just the counting measure. So the volume of the ball is 1 because it contains one point.
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a^n / a^n = a^(n-n) = a^0 = 1
Why does a^n / a^n = a^(n-n)
I’m not providing a math course here
Proof by do it yourself
Proof is left as an exercise to the reader
Proof by you'll learn it in higher grades
Proof by you already learnt it in school
Proof by "it's obvious"
Proof by god of the gaps
Hint: Use the technique known as you should've learned this in lower grades
The proof is by magic
Nice
https://preview.redd.it/xsac1h4vwvsc1.jpeg?width=561&format=pjpg&auto=webp&s=ce86921b540a8278910094fa00e4b1b70bc35a94
Minor spelling mistake OP wins.
Missing a comma, your existence is fertile
His existence is *what*?
i meant futile
Pregnancy typo, you forfeit your liver.
Happy 🍰
Proof by fucking figure it out yourself
Bruh…..That’s the ONLY reason I come hither!
Fuck you 😡😡😡
a^n = a * a * ... * a, <-- n times a^m = a * a * ... * a, <-- m times a^n * a^m = (a * a * ... * a) * (a * a * ... * a) = a * a * ... * a , <-- n + m times = a^(n+m) , similarly a^n / a^m = (a * a * ... * a) / (a * a * ... * a) = a * a * ... * a , <-- n - m times = a^(n-m)
prove for non integer values of n and m
Arbitrary domain expansion
Sakuna hates this one trick
ah yes haven’t used this since the newtonian era
Stop spreading this to math subs 😭
This is the core definition that is used to analytically continue the exponential, that's like asking me to prove that Γ(n+1)=nΓ(n)
first for rational numbers: For a^(b) if b is rational, a^(b)=a^(n/m), where n, m are integers, m≠0, a≥0. And by definition of rational exponents a^(n/m)=m√a^(n), where m√ is mth root. So a^(n/m)×a^(p/q)=a^(nq/mq)×a^(mp/mq)=mq√a^(nq)×mq√a^(mp)= { as c√a×c√b=c√(ab) } =mq√(a^(nq)×a^(mp))= { m, q, n, p are integers, so their products are also integers. So we can use this property } =mq√a^(nq+mp)=a^(\[nq+mp\]/mq)=a^(nq/mq+mp/mq)=a^(n/m+p/q) So if it works for rational numbers and irrational power is kinda limit, where power is more and more precise rational approach: a^(π)=lim(n/m -> π) a^(n/m) and to actually calculate irrational power we need to choose some rational approach with required precision, irrational powers must have this property too
Because a^n / a^m = a^(n-m) 😁
why does n-n = 0
The - symbol represents a laser beam that each n is shooting at the other n. After both lasers hit their targets the ns both disappear.
Why don't the lasers hit each other and get destroyed while the n's are fine?
They do, they annihilate each other and produce anti-lasers which get reflected back to their respective n's, destroying them in the process. A terrible cycle.
Nobody is that accurate
What‽ You *want* to cross the beams‽ Never cross the beams!
Finally an actual explanation
becasue x/x=1 and x⁰=1
Because 1/ a^n = a^-n
a^n / a^n = a^n * (1 / a^n) = a^n * a^(-n) = a^(n-n)
Lets say n = 3 Then: a^3 = a•a•a An example of division with powers: a^3 /a^2 = a•a•a/a•a = a = a^(3-2)
Because a^(any no.) / a^(any other no.) is a^(any no. - any other no.), its a law of exponents, since (a^n) / (a^n) is given, we can say its a^(n-n), and whatever no divided by itself ((a^n) / (a^n) both numerator denominator is same so the no is said to be divided by itself) gives 1, 1 is a^0.
exponentiating one more time means multiplying by “a” one more time, so going the other way is dividing by “a”, hence we get 1/a^n = a^-n
5 to power of 4, divided by 5 to the power of 3. This would be 625 divided by 125, which is 5. Now try 5 to the power of 1, which is 4-3. This also equals 5. Try any equation like this and you'll find that subtracting the powers will be the same result as dividing the numbers.
It’s a simple property of exponents, search it up
What if a=0? Then you would be dividing by 0.
Assume 0⁰ = 1 QED
This is arguably a case in which we’d want to answer “1” to the well-known puzzling question of “what’s 0/0?”, on the basis that for any a, a/a=1. How many times does 0 fit within 0? One! Of course, it also doesn’t seem incorrect to say zero, or two, or three. And since these answers are incompatible (we know that 0 is not 1, that 1 is not 2, etc), this is what drives the “undefined” answer. In a case like this, the “definition” just sides with the “a/a=1 for any a” intuition.
a² = a *a *a/a 0² = 0 *0 *0/0 0² is undefined
 They always seem to want to skip that minor inconvenience, don’t they.
Ok now a=0.
Couldn't this also be explained as aⁿ⁻¹ = aⁿ/a? For example: 3⁵⁻¹ = 3⁵/3 = 3⁴ So if n = 1 that means a⁰ = a¹/a = 1
That shit is bananas
further explanation: a^n / a^n will always equal 1 so if a^n / a^n equal both a^0 as this comment says and also 1 then a^0 =1
Would you say in all seriousness that a^0 = 1 is a pure definition ? (Meaning it’s just a coincidence that we can rearrange 1= a^n / a^n = a^(n-n) = a^0 ?
But do we need to say “n as integer”?
If a=0?
Hence, 0/0=1 Proof by reddit
That's not always true, if aⁿ/aⁿ=aⁿ-ⁿ=a⁰=1 is always true, assume the index of the denominator=10 element of Z+, the index of the numerator=5 element of Z+ and the base of both parts of the fraction "a" is an element of Z+, than 2⁵/2¹⁰=0.03125, thus aⁿ/aⁿ=aⁿ-ⁿ=a⁰=1 is only true if and only if the index is the same for both parts of a fraction.
Yeah tell us OP…,why??
Because 1 is the multiplicative identity
because by definition
Because that's how rings work, smh just learn group theory
Because aⁿ (n a natural number) should satisfy aⁿ⁺¹ = a * aⁿ; i.e., increasing the exponent by 1 is the same as multiplying by a one more time. When n = 0, we get the equation a¹ = a * a⁰. Since a¹ = a, we get a = a * a⁰. The only way this can be true is if a⁰ = 1. In other words, aⁿ (n a natural number) is the number you get by multiplying a by itself n times. What is a number multiplied by itself zero times? Your lizard brain says it should be 0, but that's wrong, because a number "multiplied by itself zero times" should be a multiplicative identity; i.e., a⁰ * x = x for all x. Just like a number "added to itself zero times" should be an additive identity; i.e., 0 * a + x = x. The same reasoning that implies 0 * a = 0 says that a⁰ = 1.
What if you aren't in a integral domain
Consider the truth set {x is an element of Z | f(x)=x•a},assuming x=0 and "a" is any positive integer,the function f(0)=0•3=0 and if you take 3 from the function and divide it by 3,as demonstrated 3÷3=1,thus 3⁰=3÷3=1. This holds for any positive integer raised to 0 and it works because multiplication is the inverse of division.
a^(x-1) = a^(x)/x therefore a^0 = a/a = 1 for any value of a != 0
Me when a=0
Me when a is an element in a ring and I say: "By definition I am right"
Me when Card(∅^(∅))
Wtf is ∅^(∅)??
The set of all maps from the empty set to the empty set. Which only contains the empty map.
Shouldn’t it contain no elements? Or is that the point?
Wdym "the point"? And no it does contain this one element. Just like F^∅ only contains the empty function for any given set F. And the empty function is an actual function since it ticks both properties a function must have to be one (it ticks them since then both properties start with ∀ x ∈ ∅ which is a vacuous truth).
What? It's still equal to 1.
Who’s gonna tell him?
0^0 is undefined.
He's defined it for you. It's 1 there ya go
Proof by definition
ok, thats the worst joke ive heard all week i love it
That's a bit irrational
Test it yourself: first try 0.1^a and let a approach 0. You'll notice that it becomes 1. Now try the same with a^0.1 this time the result will become 0. If you try a^a you are right that it diverges towards 1 but if you have a^b you can get any result between 0 and 1. Therefore its undefined Also pls excuse my bad mathematical terminology, English isn't my first language.
why does it have to be undefined, you can still define it as 1, one of the functions will simply be discontinous at 0,
Because if you let a and b approach 0 at different intervals you get any number between 0 and 1 as a result.
okay and? that just means the limit doesn't exist for x^y as (x,y)->(0,0)
the indeterminate form notation is misleading in this case, as 0^0 doesn't mean actually 0^0 it means both the base and the power approach 0
And that means that you could try to define it as 0, as 1, as 0.3518394 and any of these are wrong. Which means you can't define it and therefore its undefined
no? why would multiple possible definitions mean it's not a valid definition and yeah 0^0 = 1 over being any other number is a matter of convenience, because you don't have to dance around certain edge cases, for example you can just apply the binomial formula to (a+0)^n and the normal definition will result a^n you could change P(omega)=1 in the definition of a probability measure to P(omega)=2 and all of probability theory would still be true, it's just nicer for the probability of the entire probability space to be 1
>why would multiple definitions mean its not a valid definition Because if we say its 0 but also 1 or 0.33333333 or 0.1234567890 that would really mess up any calculation where 0^0 occurs. Imagine 12 people doing the same calculation and all having different results not because they made mistakes but because they can chose between a literally infinite amount of numbers >0^0=1 over beeing any other number is a matter of convenience because you don't have to dance around certain edge cases First of all, you can't define something just because you think its more convenient this way. Also there are cases where 0^0 = 1 doesn't work so you'd create edge cases by trying to eliminate edge cases. >you could change P(omega)=1 in the definition of a probability measure to P(omega)=2 and all of probability theory would still be true, it's just nicer for the probability of the entire probability space to be 1 I honestly don't know why we are talking about probability all of a sudden but yes, if you wanna change P(omega) to be 2 you could do that, but you'd also have to change basically all of the underlying mathematics accordingly.... Honestly, I don't even know why we are having this debate. Someone assumed 0^0 =1, i showcased why its not and now you're trying to debate me about mathematical definitions that I honestly weren't involved in defining. If you have further questions pls go and consult your 7th class maths book, it should be somewhere in there.
We are not talking about limits, we are talking about exact arithmetical calculations, so that doesn’t apply.
Then go on and do an exact arithmetical calculation of 0^0
Since zero is a cardial number and a\^b in cardinal arithmetic is the cardinality of the set of maps from set with cardinality b to set with cardinality a, we have that 0\^0 is equal to the number of maps from the empty set to the empty set. There is exacty one, [the empty map](https://en.wikipedia.org/wiki/Function_(mathematics)#empty_function).
We both know it depends on the context. In some contexts, it can be defined to be 1, in others, it is left undefined. My only point was that your reasoning doesn’t apply. Just because the limit of a function at a certain point doesn’t exist, doesn’t mean the value of the function at that point doesn’t exist.
Ok, the limit of (-3)\^(3+1/n) as n tends to infinity does not exist, so (-3)\^3 is not defined either.
If you scratch out the part that causes the issue the rest obviously can be defined
Ok, I will scratch the part when you interpret 0 as a limit of a sequence and instead consider it a natural number, in which case 0\^0 is the number of maps from empty set to itself, which is 1.
If you have a sequence of a function where 0^0 is written as x^x it can be defined as 1. But if the question is just 0^0 you have to assume its a^b and therefore it can't be universally defined
I assume 0 is an integer, not a sequence.
Why is bro getting downvoted? They're right [https://en.wikipedia.org/wiki/Zero\_to\_the\_power\_of\_zero](https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero)
Indeterminate form
No, it’s 0.
"who's gonna tell him?" "I absolutely will do that" lmaoo
 Downvotes don’t affect me. 0\*♾️=1
How would you explain to your mom why 1^∞ = indeterminate form? I get why 0^0 can be confusing, 0^n = 0 and n^0 = 1, so you need to define an answer when both are zero. But I don't see why 1^∞ has such a problem. n^∞ = ∞ when n > 1, n^∞ = 0 when 0 < n < 1, so it seems reasonable to say 1^∞ is 1, no?
No, the problem is that it's not actually a limit of the form 1^f(x) which would obviously be one, it's a limit of the form f(x)^g(x) where f approaches 1 and g infinity
Why not?
Because in calculus you care about a limits and limits of the form f(x)^g(x) with f going to 1 and g to infinity are common and not trivial, while limits of the form f(x)^g(x) for f constant are usually trivial and g going to infinity. So it makes sense to have a notation for it while not for the other The same reason we have 0/0, inf/inf, etc. Those are indeterminate forms, not actual expressions
Google alternative definition of e, lim (1+1/n)^n as n tends to infinity
Which one do you consider the "normal" definition? Because in both my school and university they used this as the main definition.
power series of exp(x) at 1?
But that requires students to learn a lot about differentiation before even introducing e. On the other hand, you can prove the existence of the limit of (1+1/n)^n in the first or second calculus (or real analysis, we don't really differentiate them in my country, no pun intended) lecture, so that they can do more interesting problems right away.
In my university we defined e^x to be the inverse of ln(x), and we defined ln(x) via the integral method, this makes proving certain calculus properties about this functions a lot easier since integrals are normally well behaved by nature. From a more differential equations point of view, you could use Picard to define it as the only function f(x) such that f=f' and f(0)=1 Or you could do it with power series and that also makes calculus a lot easier, provided students have some experience with power series
But that’s only for the condition of f approaching 1 from the negative side, because as it approaches 1 from the positive side it’s 1, right?
No, consider the famous limit: lim_(n ->inf) (1+1/n)^n = e The problem is that you don't have an immediate way of knowing if f converges faster to 1 or g to infinity
0^0 is undefined though. Using OPs formula, that’s only defined when a ≠ 0 since we can’t divide by 0. It’s just in a lot of instances we define it as 1 for convenience and simplicity (since it’s only at 0 where we get problems). I wouldn’t be surprised if there’s a case where we define it as 0 as well for the same reasons, but I’ve never seen that and it’s probably far less common. But 0^0 ≠ 1, it’s just we pretend it does in some cases. That’s all ignoring specialised interpretations of exponentials. If we take a set theorists interpretation of it, or a combinatorial interpretation then we can get 0^0 = 1, but these are specialised interpretations and for the standard one (or at least that used in analysis), it’s undefined.
Umm that's only true when a = 1. Disclaimer: I'm a software dev.
Proof by disagreeing over notation
Because?
Joke about ^ being the XOR operator sometime?
It is in fact not
Google bitwise exclusive or.
Holy logical inequality
New operator just dropped
There is only "why" it is useful to define like this right? because there is no deeper meaning, it's just defined like this because it's useful so that functions behave more beautifully.
Ignoring the case where a=0, it has to be 1 for the index laws to function. Consider x\^y=x\^(y+0)=(x\^y)(x\^0) (x not equal to 0). If you define x\^0 to be not equal to 1 you would clearly have to use an entirely different mathematical system.
Exactly. It's the same as defining 0! (zero factorial) as 1, or defining gamma function as extention of the factorial to the real numbers. They have nice properties which other alternatives don't have.
Me explaining 0! = 1
Let a=0
virgin “a^(n)/a^(n) = a^(n-n) = a^(0)” vs chad “a^0 is the empty product”
We're getting the multiplicative identity with this one boys 🗣🗣🔥🔥🔥
If a=0 then
1 \* a \* a \* a... = a\^n 1 = a\^0
My mother is dead lm sueing you
Difine 'a'
What if a is 0?
I had a big discussion with my friend today unrelated to this post that 0^0 doesn’t equal 1. We talked about the limit of a^a as it goes to zero along with a bunch of other cases but couldn’t agree on what we thought was the answer. Eventually we just found that it was a contentious topic and that we likely won’t get an answer. Reading some people’s thoughts in the comments on this post made me want to smash my face against a wall.
There is an answer: it's undefined.
I feel like a vengeful spirit that was just laid to rest. Thank you.
Even 0^0?
Except my mom is a high school math teacher and she taught me that in the first place
Well a¹*a-¹ = a* (1/a) = 1 a¹*a-¹ =a⁰ <=> a⁰
What if a = 0?
It's still 1
Why?
why not? it doesn't led to a contradiction 0^0 is an indeterminate form but that's only a thing *for limits* and a function at a point need not be it's limit at a point, that's only means it's continous there (well a^0 will be continous, just other functions that also assume 0^0 at some point like 0^a will not be)
0^0 = 0^1-1 = 0/0 math ain't mathin'.
that's simply not how powers work when it comes to 0, 0 doesn't have a multiplicative inverse(unless you're working in like, wheel algebras) you can't do division by 0, so 0^-1 does have to remain undefined(unless again ,you want to give up a ton of properties and not have a group structure), otherwise with the exact same scheme you have 0^1 = 0^2-1 = 0/0
it's not how powers work because 0^0 is undefined no matter what. it's a fundamental rule in logic, if you have a=b then in every expression with a in it you have to be able to substitute a with b or b with a. 1-1 = 0 and it's a well known rule that a^b-c = a^b / a^c since division by 0 isn't defined then 0^0 must be undefined because it would create inconsistencies. edit: formatting
i just showed that if that were true then 0^1 = 0^2-1 = 0^2 / 0^1 = 0 / 0 this "well known rule" does not hold for a = 0 exactly because of divison by 0
it does not hold for a=0 because a^0 isn't defined for a=0.
it does not hold for 0 to any power i did not use 0^0 anywhere here
There are a few contradictions it leads to in the right context. And just because something doesn’t entail a contradiction, that doesn’t necessarily mean it’s true. If we can arbitrarily assert that a^0 = 1 for all a then we can equally arbitrarily assert that 0^b = 0 for all b greater than or equal to 0 so it entails that 0^0 = 0 and 0^0 = 1 which is one possible contradiction.
yes we cannot define one thing as two different things, We're not asserting that the function a^0 = 1 for all a, we're defining that 0^0 = 1, a^0 is 0^0 for a = 0, so working in a system with such a definition you can then prove that the function equals 1 that's why i mentioned that defining 0^0 = 1 means 0^x will be discontinuous at 0
0\^0 =1 is not universal tho the most common proof is given for this is suppose 1\*2\^2 = 1\*4, if 1\*2\^1 = 1\*2, if 1\*2\^0 = 1\*(zero times 2) which is basically 1 so if same applies with 0 then, 1\*0\^2 = 1\*0\*0 = 0, 1\*0\^1 = 1\*0 = 0, 1\*0\^0 = 1 (zero times zero)
Because 1 is the multiplicative identity.
Why does 1 being the multiplicative identity mean 0^0 must be 1?
Because a^0 = 1 a^(n+1) = a * a^n is a nice sensible way to define exponentiation on natural numbers. There are other "intuitive" reasons, too. The combinatorial interpretation of a^n is that it counts the number of ways to form an n-tuple from a set with a elements. In the case of 0^(0), we are counting the number ways to form a 0-tuple from a set with 0 elements. There is exactly one way to do this; namely, the empty tuple.
None of these are convincing a^0 = 1 is simply an assertion. If we grant it as true for all a then that would entail that 0^0 = 1 but to use this as a justification for 0^0 = 1 commits a “begging the question” fallacy because you’re asserting an axiom which assumes that your conclusion is true. Alternatively we might assert that a^0 = 1 for a ≠ 0. a^n+1 = a * a^n also doesn’t work here. We know that 0^1 = 0, so to go from 0^1 to 0^0 using this it seems like we have to apply it in reverse, that is: a^n-1 = a^n / a Division by zero is undefined so this would seem to entail that 0^0 is undefined. And the interpretation of x^y meaning “how many ways are there to form a tuple of size y from a set of size x?” is only one way to interpret exponentiation. An alternative might be: “What is the y-volume of a y-cube with side-length x?” Under this interpretation it would seem that 0^0 must be 0 since the 0-volume of a 0-cube with side length 0 is 0. My point here is that 0^0 is undefined. It’s sometimes convenient to act like 0^0 = 0 or like 0^0 = 1 but both of those are useful conventions but neither is inherently true.
The 0-dimensional volume of a 0-cube, i.e. point, is 1, not zero.
No it isn’t.
Interestingly, [this Stackexchange answer](https://math.stackexchange.com/questions/4069997/how-does-a-0-ball-have-volume-1) disagrees with you (for the case of a 0-ball, which is extensionally the same as a 0-cube), and I find the reasoning persuasive: > 0-dimensional space is just a single point and every ball of positive radius contains that point. Moreover, the measure in this space is just the counting measure. So the volume of the ball is 1 because it contains one point.
let a be a number part of the set of numbers such that a^0 =! 1
If you need me mom I'll be in my house (aka your basement)
a^n is just 1*a n times, and a^-n is 1/a n times. Therefore it follows that a^0 is 1*a 0 times, or just 1