Which one is larger?

this aint math this is meth

Exponential graphs increase much faster than quadratic graphs

Before watching the video, im guessing that (if they are well defined) 2^infinity is WAYYYYY bigger than infinity^2 😂

Let me explain: Infinity is not a number, as it is endless, both equations shown in the video are endless, so you cannot conclude with one of them being larger than the other. As infinity is not a number, both of the products are equal to infinity. For the people who ask if they are equal, no they are not, as they don’t represent a value. This is like saying infinity multiplied by infinity is larger than infinity. Edit: I would also will like to confirm that there was an another saying where a half circle was equal to its diameter whereas the proof, they made smaller semicircles that go infinitely to prove pi was equal to four, but this is just completely false, as dividing something by infinity, which does not represent a value, will not give a clear result of what you are looking for. Example: Calling 1/infinity = 0 is the same thing with calling 0 x infinity = 1

I think the answer is based on a wrong assumption - you deal with infinity as a definitive number as 10. I don't think that is mathematically correct, accordingly the proof is incorrect.

2^infinity would be uncountable infinity

From the limit, the left one is larger(oof with needing to correct this).

It’s more dependent on the context in which you see this problem. For example, in set theory, 2^infinity is technically qualitatively superior in terms of cardinality as it is a power set. Outside of bigger infinites, they are equal as in the end shown by desmos, they both ultimately reach infinity either way.

Math isnt mathinnng

Infinity is not a concept that we can define so easily but from my perspective, by definition infinity is a number greater than all positive numbers, infinity multiplied by itself is just infinity and 2 to the power of infinity will always be a natural positive number since we are just multiplying by 2 every time so infinity is just larger.

When I studied limit and continuity mathematics, I was taught that if I want to make a comparison of this type, the best idea is to try very large numbers with tendencies to infinity as long as they are calculable and allow it to be applied to all cases. I checked by making a comparison between 2^1000 and 1000^2 and it turns out that 2^1000 is greater. If we constantly raise the numerical value of the exponent of "2^1000" and compare it with the numerical value resulting from constantly raising the base of "1000^2" the first case will always be greater. For those who are confused, the value "1000" in both cases represents infinities since when graphing and playing with infinities, you usually take very large values ​​close to where a function tends to infinities. So, in my opinion, without having seen the video yet, I think that 2 to the power of infinity is greater than infinity squared.

So in general, when we get a problem, which is larger 10^100 or 100^10 power, we can conclude that 10^100 power will always be larger? And sure enough 10^100 = 1^100 while 100^10 = 1,000,000,000,000,000,000,

This isn't math, this is DESMOS.

what if you set infinity in the right term as 2^∞? this is also infity and in this case the right term would be larger, right?

they are both undefined as infinite has no answer.

I think you're mistaken. There is a 1-1 correspondence between the real numbers in the interval [0, 1) and the infinite cross-product Z1 x Z1 x Z1 ... (Z = {0, 1}). The former set has cardinality aleph-sub-C. There is a 1-1 correspondence between Z x Z and Z itself, which has cardinality aleph-sub-0. The rules for counting tell us that the former set should have 2^infinity elements and the latter infinity^2 elements. So, 2^infinity = alephC > aleph0 = infinity^2.

Man is so underrated, great video!🔥

so youre telling me that the count of infinitely many people in infinitely many lines is smaller than just infinitely many people in one line

Technically, indeterminate. However, in CS, O(n) is considerably smaller than O(a^n)

Thanks for this explanation.

I agree with your assessment to a point, but each of these will be infinite, furthermore I can create a 1-1 correspondence between them. So in fact they are the same infinity.

IF you think about it, nothing is larger. Its basically the same as comparing infinity to infinity. Infinity cannot be measured because its magnitude can expand... Infinitely. Infinity is not a constant unlike the ones substituted in the video, so regular concepts cannot be applied.

The biggest possible mistake is that it is assumed two infinity are equal. If not, the answer would be uncertain.

Use monotonicity of x^1/x ... It would be a simple question

This better be a parody or something.

lim[x->∞] x^2/2^x = lim[x->∞] ln(x^2)/ln(2^x) = lim[x->∞] 2ln(x)/xln(2) = 2/ln(2) lim[x->∞] ln(x)/x = 0. Therefore what's said in the video is true Also, if you think about it with computability theorems, ∞^2 could refer to any infinite set multiplied (cartesian product) by itself. For example N×N. Then 2^∞ could refer to any infinite sequence of pairs of elements (it has to be the same infinite as before) which is basically [a,b]^N. You could make an bijective function from N×N to {a,b}^N because that last set is a numerable union of numerable sets, therefore N×N~{a,b}^N, so if you think about it this way, what's said in the video is not true I use N because I want to mantain coherence when going from one set to another, and 2^∞ only makes sense in the context of infinite sequences of pairs, therefore that ∞ must be numerable. Also, I assume both infinities have to be the same; if you use R×R, suddenly 2^∞ is smaller than ∞^2 One less Orthodox approach I'd like to use is thinking about 2^∞ and ∞^2 as the cardinals of infinite sets. Given an infinite set A such that |A|=∞, it's provable that ∞^2=∞ (or A×A~A), and that 2^∞ is the cardinal of the parts set of A, i.e. |P(a)|=2^∞, because for finite sets of size n, |P(a)|=2ⁿ. And it's also provable that there's no injective function from P(A) to A for any set A, therefore 2^∞ is bigger than ∞^2 in this context As a final note, it's 6 am, I have to give a class in 1 hour and take a test in about 4 hours. I would greatly hate myself in the future for using so much brain power for this comment

fuck yeah, i love this! abstract math at its finest

Debate of the Century.

It depends on what you mean by infinity, but if you meant lim x->inf 2^x / x^2, the result goes to inf

You can also do this with limits. If you have a fraction with x squared as the numerator and 2 raised to the power x as the denominator, if the function approaches 0 as x approaches infinity then 2 raised to the power x is greater. However, if you have a fraction with 2 raised to the power x in the numerator and x squared in the denominator, if the function approaches infinity as x approaches infinity, 2 raised to the power x is larger still. Using this method, We can see that the limit of 2 raised to the power x as x approaches infinity is larger than the limit of x squared as x approaches infinity. Now the question is by how much?

Obviously equal. Infinity means it never stops, so no matter what configuration you choose, the result of both equations is infinity. By your explanations one should be larger than the other, but if you put actual numbers in there, they will be equally inifinite

From how I understand it, with your method both are countably infinite if you first restrict them to natural numbers: if you specify a certain member of an infinite set where each member is represented as 2^n or n^2 with any natural n, you can count through all the natural numbers until you reach n, and thus when n→∞, you can still count up to there, even if it takes an infinite amount of time. A bigger infinity you wouldn't be able to count up to even with unlimited time (kinda like how you can't count up from every real in-between 0.23 and 0.24 since you can simply add more digits). Because both sets are one-to-one correspondent with the set of natural numbers, they are equally infinite. Countable infinities are considered the smallest between infinities of different sizes because they have the cardinality of natural numbers (aleph-null). Now, since you counted them up with the set of real numbers, we can use similar logic to conclude both have the cardinality of real numbers and, thus, should be the same size. Do correct me if I'm wrong.

I appreciate the explanation. Compared to most recent math videos on here, it is understandable and MAY be agreed with. Notice I said "may". It doesn't mean, that I agree or disagree with it.. The truth is where I was raised, we were taught not to treat the "lazy 8" symbol an individual number. Most people do, and hence the explanation in this video. Where I was brought up, the infinity symbol basically meant a value that increases without bounds. Since the exponent in increasing without bound in the first expression, and the base is in the second, there is no way to tell which value is in really higher. But since the video indicates hints that an expression with a lower base and a higher exponent. than reversing it using the same numbers (in other words the same higher exponent now the base, and the same lower base now the exponent), it appears that the first expression would be higher. The catch is as shown, the same numbers have to be experimented with each as base and exponent. If one is changed for the second expression, this could make a difference.

Either one can be bigger than the other. The only way to have a definitive answer, is within a limit. It entirely depends on how fast the infinity on the left increases relative to the infinity on the right. If we set both uses of infinities to be x as x approaches infinity, then you do indeed get the result in the video. However, If we set the infinity on the left (I’ll notate with L) to be log2(R) (where R is the infinity on the right), L and R still both approach infinity. And yet the expression on the right hand side will be greater instead.

Thanks for posting this. It illustrates on of great difficulties when thinking about infinities. For example, which is larger infinity or infinity + 1? Your reasoning must conclude infinity + 1. Just draw the graphs for y = x and y = x + 1

Simply do it using mathematical induction /using derivative to find rate of increase

it depends on what that ∞ represents. if that ∞ represents limit, 2^∞ > ∞^2. lim(x→∞) (2^x)/x² = ∞. if that ∞ represents transfinite cardinial, like Aleph0, 2^Aleph0 > Aleph0 = (Aleph0)^2 2^Aleph0 is cardinality of all real numbers while (Aleph0)^2 is still Aleph0. if that ∞ represents transfinite ordinal, like omega, 2^omega = omega < omega². the limit ordinal of {2, 4, 8, 16, ...} is a number that comes right after all natural numbers, which is equivalent to omega. omega² is defined as limit ordinal of {omega, omega2, omega3, ...}.

Infinity squared = infinity * infinity 2 ^ infinity = 2 ^ (2 ^ (infinity/2)) = 2^2^infinity since halfing an infinite number is still infinite, it gives you 4^infinity If you repeat this cycles, you get X^infinity where X goes towards infinity, which gives you Infinity ^ Infinity which is bigger then infinity ^ 2

Let n approach infinity, if you multiply 2 n times, and that n approaches infinity, 2^n would also approach infinity. n squared, as n approaches infinity is also well approaching infinity. And that’s also shown on the graph, they aren’t limited upper-bound, so they will continue to increase, thus neither of them is bigger than the other one.

Logic is flawed. Infinity is not finite

Take limit of x to infinity, and calculate the derivative of (2^x) / (x^2) using L'Hospital's rule. Using this method 2^x is larger, so 2^ infinity is larger

Infinity to the power of two is larger because you cant get to infinity starting with a finite number like two. Two to the power of infinity will just keep multiplying, and never get to infinity while infinity to the power of two is already infinity, meaning that infinity to the power of two is bigger.

Use more different colors in graphs, with emphasized intersection points

Alternative ways to think 1. Go through limit approach and apply L-H rule and find value of limit 2. Remember exponential always dominant with respect to polynomial

Except they both go to infinity, so they are the same.

You could transform this into a limits question, lim x--> ∞ 2^x/x^2. If answer is greater than 1 then 2^x is greater

Problem: Infinity is not a number. But if you want to use it as a number then both sides evaluate to infinity. Therefore they're equal. The only way you could show otherwise would be to show that one infinity has a higher cardinality than the other. This wan't addressed in this video. Georg Cantor showed that the number of positive integers (1, 2, 3, ...) is EQUAL to the number of infinitely dense rational numbers, even though there are an infinite number of rational numbers between any two poitive integers. Both infinite sets have the same cardinality and therefore considered equal.

It depends. Are we using ordinal exponentiation or cardinal? In the former they are equal, but in the latter they are different. (Or is infinity² larger in the former case? I don't remember...)

Personally, I'd prove the limit as x approaches infinity of 2^x/x^2

Me when continuum theorem:

Place an infinite amount of points on a circumference of a circle. Then pick any point of your choice on the circumference. Add one to that point or subtract one from that point. How far have you moved on the circumference in radians?

They are both infinity, but if we talking about a countable infinity, 2^inf is larger because it starts off slower and gets really big really fast. 2^15 is 32768. And 15^2 is only 225

on if the value of x=3 in X² and 2^X The value of X² is greater by just 1 rather than that in all the values 2^X is greater

Infinity is not a number, it's a concept

It is a Similar question to the question in jee advanced as it was to Compare π^e and e^π

very interesting video! i wish tho that you played a little bit with your proof there because going a little further you could prove that any number larger than 1 to the power of infinity is larger than the infinity to the power of any real number. Which means 1,0000000001^infinity is larger than infinity^999999999999 and that’s really fucking impressive lol.

When you compare Megabit (MB) and Mebibit (MiB)

I won't split hairs about the use of infinity. I assumed that the video is about the limit. The video has a nice, reasonable start with "let's try a few exmples to get a better idea", but there is no real proof at the end. For a real proof, you can start with the hypothesis 2^x > x^2 for x>1. Then take the natural logarithm on both sides to get x*ln(2) > 2*ln(x) or approximately x > ln(x)*2.89. It is already known that ln(x) eventually grows slower than x (that's why logarithmic amplifiers are used for signal compression). Therefore, eventually x will be larger than ln(x)*2.89, which proves the hypothesis for large enough x, in particular when approaching infinity.

Don't treat infinity like a number though, since there's a lot of rules in many equations that wouldn't work if infinity was a number. Infinity breaks the system, so it's treated instead as a concept.

Bro took"infinity war" literally

These are both equal to infinity, what is presented in the video is more a question of limits. For example if you had (2ˣ)-(x²), after the 3rd intersection point, it would always be a positive number, meaning the first is larger. This means that as x goes to infinity, the left side grows faster than the right side, and thus would overpower it, which would make it "larger"(for any finite number, x-y is positive if and only if x>y, so this is a reasonable pattern to carry over to infinite numbers.) In calculus this can be applied more generally as things that aren't the highest degree not mattering(eg. x³ will overpower x², because 3>2.) Cool video, but the presentation of the problem is a bit misleading.

in the end they're both equal... there is no number larger than infinity

How tf bro only have 160 subscribers

Well yeah my thought process tho is that, 2^infnity never ever ends you just get bigger and bigger numbers while infinity² is just infinity i think and infinity is always bigger than any number you get no matter how large, if you keep mutiplying by 2 even to the end of time

This is easy, just looking at the growth 2 to the power of x will always excess the growth of x squared. So 2 to the infinity power is bigger than infinity squared. Video over

Good question... As infinity itself undefinable... Comparing two such may lead to illusion though mathematically logically true

2 multiplied by itself equals infinity. Infinity squared is an unbound set of unbound sets. Largeness is not relevant when something is unbound.

It’s probably me ngl ( Im just built like that )

Actually, in cardinal, 2^infinity is bigger than Infinity^2. However, in ordinal, 2^infinity is smaller.

2^x > x^2 if x>4 so infinity is clearly higher than 4

Rule 1 = No 2 infinities are equal But if we are comparing with same infinity Then limn→∞2ⁿ/n²=∞ .... Therefore 2^∞ is absolutely larger in comparison

Well, they're the same Infinity isn't a number, its a concept, it just means it goes on forever

but in in fisrt one the maximum number u can obtain is infinite, as in the second one is infinite times infinite. Not the same

Infinity can’t be treated as a number it’s a concept, there is no difference between infinity^2 and 2^infinity, this video is fundamentally wrong to even compare the two. Of course if said infinity was a number 2^n would be larger than n^2

*"But bro ∞ is an unstoppable thing. So 2^∞ is ∞ and ∞² is also ∞ so both are equal. For 2^x and x², if x is finite then only we can say 2^x>>>x² (only if x is finite). But if x is ∞ they are non comparable as ∞,∞²,∞³,∞^∞,2^∞... All are equal as all tends to ∞. If u considered ∞ as a very large number then only 2^∞>>∞². But the definition of ∞ says it's an unstoppable value so all ∞,∞²,∞³,∞^∞,2^∞... Are equal"👍*

Technically they are equal Infinity is very strange number, so it don't have to act like others. No matter, ho many times me multiply infinity, it will be infinity. So, W^2 is infinity. 2×2×2×2.... Infinity times is obviously infinity. So, 2^W is olso infinity. W^2=2^W

In my logic if you add or substract anything or multiply or divide anything by infinity it equals infinity

It is impossible to guess which is bigger

2^infinity is infinity infinity^2 is infinity of infinities easy

They equal to other, greater than other and less than other at the same time

How about ∞^2 ?

Our first question should be "what is infinity"

Its technically 2^infinity If you were to replace infinity with a countable number You’ll see that 2^(insert number) is larger than (insert number)^2

this doesn't make any sense, how can infinity be smaller or larger?

Solving is Easier by log

There is not such thing as larger or not when they are both infinitely big...

Simple -1/12 = infinity, so -1/12 squared is indeed less than 2 powered to -1/12

If we put both of these numbers into a grid perspective infinity to the power of two is a grid in which there is and infinite line and that infinite line is repeated infinitely to the side of the infinite line but two to the power of infinity is a grid with two dots double infinitely and is two line of infinity so infinite to the power of two is bigger

This is exactly how I saw it with 2^10 > 10^2

Missed the chance to make it 8^∞ vs ∞^8 😂

2^infinity = 2^aleph-0 = aleph-1 infinity^2 = aleph-0^2 = aleph-0 WHICH ONE DO YOU THINK IT IS BIGGER? THE CARDINALITY OF THE REAL NUMBERS OR OF THE INTERGERS??

This shouldn’t matter I think… if infinity is involved the number is ALWAYS growing or infinity… so they both equal the same thing inherently. I haven’t done hardcore math in a long time, but since every number is being squared while 2 is being raised to the power of every number they still encompass the same parameters… right? Even numbers smaller than 1 keep getting smaller. This is why I SOMETIMES hate math. But mostly I love it. I makes some cool stuff

Ok, I don't completely get this but doesn't the infinite hotel problem say the opposite, that inf^2 is 2nd degree of infinity? E.g. no matter how many times you multiply by 2, you still just get a really large number - infinity - but if you square this infinity, you have infinitely many infinities?

Put x as 3 😂

square root of infinity^2 is just infinity and suare root of 2^infinity is 2^(infinity/2) since infinity divided by 2 is still infinity, square root of 2^infinity > square root of infinity^2 and 2^infinity > infinity^2

Look, if you treat that Infinity as N(R), 2^Infinity is like N(A = {x such that x is a subset on R}) and Infinity^2 is N(B = {(x,y) such that x and y belong to R}. Let's call C = {Q such that Q belongs to B and Qx = Qy} and D = B - C. Then, C = {(x,x) such that x belongs to R}, N(C) = N(R) = N({x such that x belongs to A and N(X) = 1}). Also, D = {(x,y) such that x and y belongs to R and x != y} and N(D) = N(R cartesian R) - N(R) = N({{Qx, Qy, Qx-Qy} such that Q belongs to D}) = N({x such that x belongs to A, N(x) = 3, x = {a, b, a-b}, a and b belongs to R and a != b}). That way, N(B) = N({x such that x belongs to A, (N(x) = 1 or (N(x) = 3 and x = {a, b, a-b}))}). Thus, A have N({x such that x belongs to A, N(x) = 2 or N(x) > 3 or (N(x) = 3 and x = {a, b, c} such that c != a-b and a, b, c belongs to R)}) more elements that B.

You don't have to understand this with logarithms let me clear this short way Infinity is a thing so can't be multiplication divided If you take this infinity as a large number then it can be minus and divided but not multiple and add because nothing can be higher than large number no it can be lower 2^infinity doesn't exist same for infinity^2 But you can do this infinity-n and z≥0/infinity :) You can ask your class teacher does 2^infinity exist

I think infinity^2 larger, because: 2^infinity is infinitely big set of numbers, while infinity^2 is infinitely big set of infinite sets of numbers.

Answer is 2 power infinity

I think it's same because if X = 2 (♾️=♾️)