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This wouldn’t have passed as a rigorous proof in my math class 😂

This video is wrong. The answer is either “this question is nonsense” or “they are both equal to aleph naught”. If you interpret it to be asking “is 2 multiplied by itself aleph naught number of times greater than aleph naught multiplied by itself”, they’re equal. Consider Hilbert’s hotel, the hypothetical hotel with an infinite number of rooms, each assigned a natural number, all of them empty. The desk clerk receives a shipment of 2^(aleph naught) mints and his boss tells him to put one on each pillow in the hotel. So he takes one mint and puts it on the pillows in room 2. On his second trip, he takes twice as many mints as the last trip, and put them on the pillows in rooms 3 and 4. On his third trip, he takes a twice as many again and so on. Every trip t, he takes 2^(t-1) mints and places them in each room in order until he gets to room 2^t. At some point on his trip, he does the math and realizes his first mistake. The sum of the infinite series 2^(t-1) only equals (2^t)-1 so he’ll have one mint left over. But then he remembers his second mistake: he forgot to put a mint in room 1. So on one of his trips, he takes an extra mint and drops it off in room 1 before resuming his rounds. After an infinite number of trips, every mint has been distributed. Later that night, aleph naught buses pull up out front with aleph naught passengers in each bus. The hotel clerks assigns everyone on the first bus to room 2^s, where s is their seat number. Then he assigns everyone on the second bus to room number 3^s. For the third bus he assigns them to room 5^s, for the fourth bus he assigns 7^s and so on. For each bus b, he assigns each passenger to bth prime number raised to their seat number s. Everyone on every bus got a room, and as it happens there were infinitely many rooms left over, one for every number that is not a power of a prime number. So as to not waste rooms, he asks the guest in the occupied room with the lowest number (room 2) to move into room 1. Then he asks the guest in the occupied room with the second lowest number (room 3) to move into room 2, and so on, until every room is occupied. In other words, each of the (aleph naught)^2 guests booked into a room and found one of the 2^(aleph naught) mints waiting on their pillow, with no empty rooms or unclaimed mints.

Are you serious?? Do you really mean Infinity + 1 > Infinity + 0 ??😂😂

ok what the flipify flop dude

Can't you divide 2ˣ by x² to compare, and then use lim x→ ∞ 2ˣ/x², and then use l'hopital's rule to get (2ˣ*ln2)/2x and then use l'hopitals rule again to get lim x→ ∞ ln(2)²2ˣ/2 = lim x→ ∞ ln(2)²2ˣ⁻¹. This means the value is ln(2)²2^(∞-1) = very large number, or ∞. So yea, 2^∞ is bigger than ∞²

Basically it’s the middle of A and B

2^inf > inf^2, i guess inf ^2 is limited on multiplying itself twice but 2 is an infinite number if we use logic a number is basically limited because it provides what it stands for 2 is 2 pieces period. But the ^ inf makes it sound like a paradox

Imagine X=1

Before i watch video, my guess is ♾^2 because your either Squaring Infinite by infinite. Or 2^♾ which is basically ♾×2 ♾×♾ > ♾×2 I was wrong 😭

Multiplying infinity by anything is like multiplying 0 by anything; you’ll just get the same thing

If we remove the unlimited limit (think of it as a rule) the two numbers are equal to each other.

X = 2 They can be equal

neither, infinity is infinity

2^x grows faster than x², so 2^infinite > infinite²

Even 1.0001^inf would be larger than inf^2 or inf^10000

you're comparing infinities and you're not even talking about cardinalities or cantor

2 power infinity is only larger when the infinity of the both will be same like 10,10 or 29,29 etc. but 24, 45 can be if both infinity is not same

They both equal to infinity. There is no concept of less infinity or greater infinity

Ambos son infinitos, es decir son lo mismo. Solo que 2^infinito tiende más rápido a este

If the two infiniteis are equal than exponential is much greater

This is only valid if both the infinities are defined the same way

the only way to actually realize infinity is as something that keeps growing/increasing without limit thus being infinite in this sense the bigger inf is not the actual ammount but wich is faster at growing/increasing, 2^inf will continuesly double in size nonstop while the inf^2 will keep growing at the same pace but significantly more than that of a normal infinity thus by that logic 2^inf will start slower but since inf will continue forever eventually 2^inf will just outspeed the grow of inf^2 since inf^2 speed doesnt increase

My guess is that infinity^2 starts off way higher but halfway to infinity, 2^infinity catches up and dominates...

Why dosent he have a water bottle

He can just ho home and get water there

Just compute the limit of the ratio and you find 2^x grow bigger

n = x^2 + y^2 my solution: n = 2^2 + 3^2 = 4 + 9 = 13

i guessed that 2∞ would be "bigger." but really theyre the same cause 2∞=∞, just lkke ∞ 2=∞

I feel like infinity squared grows faster, because in order for 2^infinity to get anywhere, it must start at the number 2, and multiplication must happen a certain number of times. But since infinity^2 starts with infinity, it’s already been at the point where 2^infinity is going to from the first step, even without squaring it.

Let's take x=3 and substitute it to infinity. We get:- 2^3 = 8 & 3^2 = 9 In this case 3^3 is greater than 2^3 hence its also possible that 2^infinity can be larger. Please don't do this extreme level of mathematics just by childs method. In fact sum of all numbers from 1 to infinity is -1/12, shocking right. Yes it's true. Maths is not a piece of cake

By your logic, infinity+1 > infinity If you draw the graph of x, and x+1, x+1 will always have the greater output

A.(A+1) So its 3x4 equals 12

You can only do this if x is a real number Infinity is imaginary

"If you want to make a number bigger by powering it focus on making the power greater not the number" as simple as that

Lmao infinity is just a concept So even if you square infinity or exponentially raise infinity, the answer is infinity.

I feel like infinity squared is still bigger. No matter how many twos there are is always just a multiple or two and thus finite, but infinite is infinite and regardless of how many, it’s still infinite

Take the limit of the ratio between the two. When you use lhopitals rule the x^2 term goes away after two derivatives while 2^x remains in the expression.

me : nah its equal oh

but if the result is always infinity in both cases, then 2 to the power of infinity = infinity to the power of 2.

I think that same becouse infinity x 2 = infinity and 2^infinity = infinity

Can we use that irl?

You cannot compare between infinites. (Unless it's infinite compared to a fixed value)

But if say x is a natural number, substitute x with 3 and all this proving goes down

Bro aren't they're the same? 💀💀💀

Bruh both cannot exist. 😂 cuz our universe is finite.

2×2×2×2×2×2.... = inf × inf × inf × inf.... Inf × inf < inf × inf × inf... Inf is well inf?

what if infinity power infinity happens

This video has a big problem, it’s has the assumption that conventional notions of size apply to infinity. The affirmation “ some infinity’s are bigger than others” is technically incorrect, the correct one is “some infinity’s have a bigger cardinality than others”. Being cardinality the amount of terms of a set. For every set of the image of 2^x we have a equivalent term in the set of the image of x^2, meaning that both have the same cardinality, or in other words “same size”

Good rationale, wrong conclusion.

They are equal.