Aleph-Math

Ok, So I am back, and now working on this page again. - Oromero9
“You enter a realm where physical or digital skills don’t matter, Here, knowledge is power.”
-Aleph-Math description for upcoming tower. (No not DC those are overused)
==What is Aleph-Math?==
Aleph-Math is a class 26 difficulty. Even if this difficulty is nerfed, there are no words to describe how hard this difficulty is. Example: comparing low Aleph-Math to high Aleph-Math Is like comparing high N/A to low KLJY. And since the gap is so big, it is logic-defying and utterly impossible to conquer or even get to this difficulty.
Let me give you a description of the pain you will feel while doing Aleph-Math:
So much PAIN!!! WHY THE HECK would you do this to me! THE PAIN IS UNBEARABLE! IF I FAIL, WELL, LET's JUST SAY YOU'RE IN TROUBLE.
Even a SINGLE STUD of this difficulty would collapse into a SUPERMASSIVE BLACK HOLE that would rip apart the fabric of space-time itself, and then eating the EARTH, then the SOLAR SYSTEM, then the GALAXY, then the UNIVERSE, then the MULTIVERSE, and then creating a DEATH WAVE OF HEAT AND LIGHT BURNING ANYONE AND EVERYONE IN ITS PATH, and then cooling to a giant piece of gas the size of our UNIVERSE killing anything that goes near it. The red gas would be the only thing left of our ENTIRE MULTIVERSE.
The weight of every multiverse in the OMNIVERSE are on the tip of my shoulders, and I need to balance them while doing RIDICULOUSLY hard obstacles like a >EXPAN^^^^EXPAN stud wraps SIMULTANEOUSLY!
====Catches:====
*It can absorb any powers from Unclassified difficulties and any other difficulty below, it wants to chose.
*You have the strength and intelligence of a 10 month old baby, unable to walk or talk or even think.
==Explaining Aleph-Math:==
I feel like Aleph-Math should be explained why it’s the first Class 26 difficultly and Class 26 Mythical, so I am making a separate heading to explain how’s its not Class <10 or something like that.
What makes Aleph-Math a mythical difficulty, is that it’s not skill that requires you to complete the difficulty, it’s intelligence. The key to completing this difficulty is to understand and solve the problems. You can understand or comprehend a Class 10 difficulty, and can even do the same to a class 22 difficultly. You may not be able to imagine it, but you can understand the words themselves and what they mean.
This is why it’s the lowest Class 26. After Aleph-Math, is The Super Absolutely Infinity Meltdown, which has “Clover times Umbrella” stud wraparounds, which is so stupidly crazy, it doesn’t even make sense. This is what’s makes Aleph-M different, Because the way to complete the obstacles is to comprehend the obstacles. And that’s only just understanding it, we need to solve each obstacle you encounter along the way.
But wait, if the obstacles are trying to comprehend and solve the problems, then what about Epsilon Naught? You can understand it. It’s difficulty level is now higher. Hmm. Thus we can continue. Abomination, no problem. DEADLY TUBULAR DEATH, piece of cake. ABSOLUTE INFINITY, just a small crack in the sidewalk, a walk in the park. Penultimate starts to become a challenge, be we can absorb difficulties, using :‍, we can surpass its properties, and finally defeat The Final Difficulty.
But we can still continue. MetaToo-Hard is nothing compared to us. Meta-Destructive, Class 32, Class 40, Class 58, making it to TooTooHard Wiki. We don’t need to stop anymore. TooTooTooHard Wiki, and beyond, all in out grasp.
But where is it taking us…
==Aleph-Math Terms:==
You may need to read these number definitions to understand this difficulty later on. (Although you can’t even comprehend most of this page, so just read this. IDK)
===TORSHAVN (Ꙭ)===
How long it will take with of a fraction of a Planck Lengths to reconstruct the entire town of Torshavn, Capital of the Faroe Islands.
(Fun Fact: This term is slightly Inspired from Torshing, I basically just changed the last letters randomly, and I got the name of a town. So yeah.)
===YANO-TORSHAVN (ꙮ)===
TORSHAVN timed by YAONTING.
===YOTTASILL ( ҈ )===
YANO-TORSHAVN to the power of itself 3 times multiplied by TORSHAVN and half of YAONTING.
==Obstacles in Aleph-Math:==
===Obstacles in Aleph-Math Terms:===
Jesus Christ. You really wanna try this? You’re gonna need a IQ of 345672389578132985238953423491453912746813515623675673248657934275934248675836477589431526745865816548753765842376534826523846534284652836458326582376485631563248567382654738265836823478326582675893276539287589326543298675793267594236759 IQ, so get ready.
====Low Aleph-Math====
How to solve this:
*When you had simplified to a triangle, i simplified again to a line. place 2 dots on the line and check the odds that the dots contain the center. when i did that, i noticed that the odds are basically just 1/2 regardless of where the first point lands, no matter where it lands, it just needs to be on the other side of the center. so i was able to divide the line into 2 and solve it by finding out if they were on opposite sides. this is a massively simplified version of the 3D question, but it helped for visualizing in 2D. i increased the size to a circle and divided it into 2 and quickly found that you can fix the first dot to one point on the circle by rotating the circle regardless of where it lands. from here, i divided the circle in 2 and placed the first dot in the middle of the line on one of the sections of the circle. from here, it goes similar to the video. the second dot has a 50% chance to end up on the other side and a 50% chance to end up on the closer side. on average, it will be right in the middle, and I realized that the section that the 3rd dot has to land on to contain the center is between the first 2 but on the opposite side so the odds are 25%. with this, i move on to the actual question and fix a point on the sphere and i found that wherever the next dot is, you could fix a plane that crosses through there and the first dot (and the center). this helped massively simplify because that meant that the second dot could be fixed to a line and the odds were, on average, that it would land in the center. I took that and moved on and found that the whether the third dot lands on one side or the other of this plane, it can be inverted without changing anything (before the fourth dot is placed). So, it would always land on one half of the sphere and where is the average point there? the center of that hemisphere. the fourth dot must land within the spherical triangle that the first 3 dots created which ultimately, if the second and third dots were at their average points, the fourth dot has a 1/8 chance to land on the other side. this means that, on average generically, the fourth dot will create a tetrahedron which contains the center 12.5% of the time.] suppose that we focus on one student and disregard who they are cheating off of there is a 50% chance that just the person to their left is cheating off of their test and a 50% chance that just the person to their right is cheating off of their test, that means a 25% chance that both are cheating off their test, and a 25% chance that neither are cheating their test, this means that there is a 75% chance that any given person has someone else cheating off of their test. so, 1/4 people should be left without their test being cheated off of. also i did consider it, but the states of the people beside any student doesn't matter, and you have to do this 10 times again, if you fail once.
====Mid Aleph-Math====
God, what are you? Albert Einsten’s Dad?
*Four points are chosen at random on the surface of a sphere. What is the probability that the center of the sphere lies inside the tetrahedron whose vertices are at the four points? (It is understood that each point is independently chosen relative to a uniform distribution on the sphere.) The problem has a geometric immediacy that makes it tantalizing: the tetrahedron so formed is readily visualized and no great mathematical background is necessary to understand the question being asked. Further, it is almost impossible to resist the urge to generalize the problem. Some of the variants that spring to mind quickly are: (1) Suppose n+1 points are chosen at random from the surface of a ball in Rn. What is the probability that the center of the ball lies inside the simplex in Rn whose vertices are the n+1 points (i.e. the convex hull of the n+1 points)? (2) Four points are chosen at random from within a ball in R3 (or n+1 points from an n-ball in Rn). What is the probability that the center of the ball lies within the convex hull of the points? (3) Four points are chosen at random from the surface of some other object in R3 (or n+1 points from the surface of some object in Rn). What is the probability that a fixed interior point of the object lies inside the convex hull of the four (respectively, n+1) points? (4) More vaguely, assume the action is centered about the origin in Rn, and that n+1 points are chosen ``at random'' in Rn. What is the probability that the convex hull of the n+1 points contains the origin? The list can easily be extended, but as question (4) demonstrates we have already reached the point where the questions need to be more carefully posed. Despite the fact that the original Putnam question is so easily understood, the solution is (not surprisingly) not arrived at with equal ease. This sentiment is supported by the fact that 123 of the top 203 scorers on the Putnam exam submitted no solution at all to problem A-6, and a relatively low number of 9 of the top scorers received a full 10 points for the problem. This difficulty in answering such an easily grasped problem just makes it more intriguing, of course, and suggests that the problem and its generalizations are worth investigating. In this paper we will develop a surprisingly simple answer to questions (1) and (2). In addition, our result answers rather general forms of questions (3)and (4). In [3], Klosinski, Alexanderson and Larson offer the following solution to A-6. Assume the sphere is centered at the origin, and that the first point P0 is located at the north pole of the sphere, with the three remaining points then located at random locations on the sphere. We can assume that these remaining points are chosen in a two-step process: first a diameter Pi1Pi2 (i Î {1,2,3}) is fixed and then one of the two end-points {Pi1,Pi2} is selected as a vertex of the tetrahedron. Figure 1 below illustrates a typical orientation of the choices. The eight possible tetrahedra P0P1j1P2j2P3j3 (with each ji being 1 or 2) are equally likely. Further, we can assume that the result is an honest tetrahedron and that the origin does not lie on any face. (Recall that the plane through three noncollinear points P1, P2 and P3 consists of all affine combinations where a1+a2+a3 = 1. With probability one, neither the fourth vertex nor the origin lies in the plane through any three vertices.) Figure 1: Typical choice of vertices. In particular, the four vertex vectors must be linearly dependent, so there exists a 4-tuple (w,x,y,z) for which/and for which w,x,y and z are all non-zero. Then since the eight equations have the solutions Each point in the tetrahedron with vertices P0, P1j1, P2j2,P3j3 can be uniquely represented as a convex combination (where each bi ³ 0 and b0 + b1 + b2 +b3 = 1), so the origin is contained in the tetrahedron P0P1j1P2j2P3j3 if and only if the 4-tuple solving the associated vector equation consists of four coordinates of the same sign. Since only one of the above eight solutions has this property, only one of the eight equally likely tetrahedra contains the origin, and hence the probability that the origin is contained in the randomly chosen tetrahedron is 1/8. 2 First Generalization So far, so good. This solution generalizes in the obvious way and gives us the answer of 1/2n to question (1) in the above list. But what of question (2)? The above approach seems inadequate in this case, since points can now be chosen anywhere along the randomly chosen diameters. Let us employ one of the standard procedures when faced with a difficult problem: that of changing the problem to something easier. We will attempt first to answer question (2) in R2. Specifically, if three points are chosen at random from the unit disk B2, what is the probability that the triangle thus formed contains the origin? Let us further simplify the problem by assuming that we are choosing three points at random with respect to a probability measure P on B2 which is rotationally invariant ; that is, measures of subsets of B2 are unchanged under rotational translations. We will also continue to assume the appropriate degree of non-degeneracy of the measure (more on this in the next section). Since we are assuming rotational invariance, we can assume that the first point P1 is fixed between 0 and 1 on the positive x-axis. With probability one, the second point P2 of the triangle is not located at the origin, and we can form the ray starting at the origin and passing through P2. Let q be the angle between the positive x-axis and this ray. The question can now be posed as a conditional probability problem: given q, what is the probability that the third point P3 defines a triangle which contains the origin? Integrating this probability over all possible q's will then give us the answer we seek. In order to simplify our work, let us agree upon some notation. Given a point P in B2 -{(0,0)}, let Q(P) denote the angle from the positive x-axis to the ray beginning at the origin and passing through P (see Figure 2). Thus, q1 £ Q(P) £ q2 will indicate that P lies in the sector of B2 defined by the angles q1 and q2. Let P(capture) denote the probability that the origin is captured within the triangle formed by the three points P1, P2 and P3. Thus, the first task is to calculate P(capture | Q(P2) = q), for each q Î [0,2p]. Figure 2: Illustration of Q(P2) for a typical P2. Suppose first that 0 £ q £ p. It is not difficult to see that a necessary and sufficient condition for capture is that p £ Q(P3) £ p+ q. That is, the ray from the origin to P3 must pass through S1 (the boundary of B2) at a point between p units and p+ q units, as measured from the positive x-axis. Since the length of this arc is q, this conditional probability is q/2p, i.e. P(capture | Q(P2) = q) = q/2p. Similarly, if p £ q £ 2p, P(capture | Q(P2) = q) = 1 - q/2p. We can now approximate our solution with an appropriate Riemann sum. Let { q0, ¼, qn } be a partition of [0,2p]. Then In the limit of finer and finer partitions, we obtain Examination of this argument shows that we have answered more than we set out to, since the fact that P is a probability measure on B2 is really irrelevant. As long as P is a probability measure on R2 which is rotationally invariant and suitably non-degenerate, the result is the same. We are already aware of one consequence of this: if P is a uniformly distributed probability measure on S1, the method of Klosinski, Alexanderson and Larson tells us that with probability 1/4 the origin will be contained in a randomly chosen triangle. Anyway, I doubt you're even reading this so here's some gibberish to extend this paragraph and artificially increase the difficulty. First, We can also begin to make sense of question (4) by noting that if P is the usual Gaussian probability measure on all of R2, the probability that three randomly chosen points captures the origin is again 1/4. A related problem in geometric probability, whose many variants are dealt with in [1], [2], [4], [5] and [6], is to find the probability that three points chosen at random from a region in the plane will form an acute triangle. One version can be easily answered here. Since the origin is captured by three points chosen at random from the unit circle if and only if the three points form an acute triangle, the probability that an acute triangle is formed by three points chosen at random from S1 is also 1/4. The results above suggest that under rather general circumstances n+1 points chosen randomly from a region in Rn which is symmetric with respect to the origin will capture the origin with probability 1/2n. Our main result gives conditions which guarantee the validity of this conclusion, and thus provides answers to questions (2), (3) and (4). 3 Second Generalization We begin with a theorem, in which we finally describe the amount of non-degeneracy of the measures that we require. We also discard rotational-invariance for a weaker condition. Theorem 1 Let Rn be endowed with a probability measure m which is symmetric with respect to the origin and such that when n+1 points are chosen independently with respect to m, with probability one their convex hull is a simplex. Then the probability that the origin is contained in the simplex generated by n+1 such random points is 1/2n. Recall that a measure m is symmetric if m(-S) = m(S) for any measurable set S. Also, n+1 points x1,x2,¼,xn+1 from Rn are vertices of a simplex if and only if none of these points lies on a hyperplane containing the other n points. Thus simplexes in R2 are triangles and simplexes in R3 are tetrahedra. As examples of how the theorem can be applied, we could let m be a uniformly distributed probability measure on a rectangle centered at the origin in R2, or on the boundary of such a rectangle. In either case, if three points are chosen at random with respect to the measure, the probability that the origin is contained in the triangle thus formed is 1/4. In R3, the original Putnam answer of 1/8 applies to four points chosen at random from a cube, or from the surface of a cube, as well as from the sphere. More generally, let D be any domain in Rn with finite volume and which is symmetric with respect to the origin. Then the probability that the origin is in the convex hull of n+1 points chosen uniformly and independently from D is 1/2n. Note that, as in Figure 3, it is not necessary for D to contain the origin, or even for Dto be connected. Figure 3: A disconnected domain D to which the theorem applies. We will now proceed with the proof of the theorem. Proof: Begin by defining a new probability space W = Rn ×¼×Rn (n+1 factors) with the product measure m×¼×m. Let For each set of indices I Ì {1, ¼, n+1} let where for w = (x1,x2,¼,xn+1) we define wI = (y1,y2,¼,yn+1) with Thus A = AÆ, where Æ is the empty set of indices. Our proof now rests on four observations. (i) AI = AIc where Ic is the complement of I in {1,¼, n+1} (ii) AI and AJ are essentially disjoint if J is neither I nor Ic (iii) W = È{AI | I Ì {1, ¼, n+1}} (iv) The sets AI all have equal measure. To prove (i), suppose w Î AI, with w = (x1, x2, ¼, xn+1). Then wI Î A, i.e. with 0 £ bi £ 1 for each i and åbi = 1. But then and since it follows that wIc Î A, so w Î AIc. Of course, the same argument shows that AIc Ì AI as well, so AI = AIc. Next, let x1, ¼, xn+1 be chosen independently with respect to m from Rn, and let w = (x1, ¼, xn+1). Because of the hypothesis that with probability one none of the random points xi is in the hyperplane spanned by the other n points, the origin has a representation as an affine combination of x1,x2,¼,xn+1 that is almost surely unique, say where åai = 1. Thus, except possibly for w in a set of measure zero in W, the set I = {i | ai £ 0} for the above representation is uniquely determined by w, and picks out the set AI = AIc of which w is a member. Thus observation (ii) is proved. Observation (iii) follows from noting that for any n+1 randomly chosen points x1,x2,¼,xn+1 from Rn, the zero vector can be expressed as an affine combination of these points, and the set AI = AIc which contains w = (x1,x2,¼,xn+1) is essentially uniquely determined by this expression. Finally, (iv) follows from the hypothesis that m is invariant under reflection through the origin, so that for any index set I the mapping w® wI on W that changes the signs of the coordinates in I preserves the product measure. This mapping sends AI onto A, so m(AI) = m(A) for all I. The index set Á = {I | I Ì {1,¼,n+1}} has 2n+1 elements, but AI = AIc so our claims show that W can be written as an essentially disjoint union of 2n subsets each with the same measure. Thus each of these subsets has measure 1/2n. In particular A has measure 1/2n, completing the proof of the theorem. ¨ After discovering the above proof, we learned that our result for points chosen from the surface of the unit sphere in Rn could be deduced from a result due to J. G. Wendel and found in [8]. Wendel showed that if N points are chosen at random from the surface of the unit sphere in Rn, the probability that all N points lie in the same hemisphere is given by If we let N = n+1, then the probability that the origin is contained in the convex hull of n+1 points chosen at random from the unit sphere in Rn is 1 - pn,n+1, which as the reader can verify is 1/2n. A referee pointed out another interestingly odd result related to our investigation that can be found on page 124 in [7]: If five points are chosen at random from a ball in R3, the probability that one of them is contained in the tetrahedron generated by the other four is 9/143, and do this 1 trillion times a nanosecond.
====High Aleph-Math====
ok, I like your funny words, “magic man.”
*
a1	®
P	1	®
P	2	®
P	3	,
	®
P	0	,	®
P	11	,	®
P	21	and	®
P	31	
	®
0	= w	®
P	0	®
P	11	®
P	21	®
P	31	
	®
P	i1	= -	®
P	i2	,
	®
0	= w	®
P	0	®
P	1j1	®
P	2j2	®
P	3j3	
(w,x,y,z),(w,x,y,-z),(w,x,-y,z),(w,-x,y,z),
(w,x,-y,-z),(w,-x,-y,z),(w,-x,y,-z),(w,-x,-y,-z).
b0	®
P	0	®
P	1j1	®
P	2j2	®
P	3j3	
	®
0	= w	®
P	0	®
P	1j1	®
P	2j2	®
P	3j3	
P(capture)
=	n-1
å
i = 0	qi £ Q(P2) £ qi+1)	Dqi
2p	.
P(capture)
=	ó
õ	2p
0	Q(P2) = q)	dq
2p	
=	ó
õ	p
0		q
2p		dq
2p	ó
õ	2p
p		æ
ç
è	1 -	q
2p	ö
÷
ø		dq
2p	
= 1/4.
the origin ofRn is in the convex hull of x1, ¼, xn+1}
wI Î A},
yi =	ì
í
î		
if i Î I
xi
if i Î Ic.
	å	biyi =	®
0	,
	å	bi(-yi) =	®
0	,
ì
í
î		
if i Î Ic
xi
if i Î I
	®
0	=	å	aixi,
pn,N = 2-N+1	n-1
å
k = 0		æ
ç
è				ö
÷
ø	.
N-1
k
*
====Advanced Aleph-Math====
Dont you love speaking enchanted table?
*P0 ^ P31 ^ P11 ^ P-0 ^ P21 ^ 𝜋 x 2
*
====Skyline/Peak Aleph-Math====
my man, all of Harvard needs your help breh
*𝜋 x 2,000 ^ P200 ^ θ (P200) ^ P1 ^ 𝜋 x 2,000
===Obstacles in Average Class 24 Terms:===
For those who have the brains of a small bottomless pit, I’ll put it in JJT Terms or more specifically, Class 24 Terms, so here you go. Brain ded people.
====Low Aleph-Math (0.01-0.1% of Power)====
How to access The World Wide Web:
*First do a ∞*100 stud jump. For each -Planck Second, Take a -EXPAN^EXPAN^EXPAN of a single quark until you have taken out every quark in every other multiverse, with Negative CROUNTILLION Walk-speed and Jump-speed. Now wait a CROUNTILLION years for the entire omniverse to reappear, once that happens, do the thing all over again, until quantum Mechanics can’t regenerate the omniverse out of nothing. Take all of the quarks you took out in the first step, rearrange the quarks into a barrel-verse, put one every BARRELFILL eons, and repeat once again for a Drazer times, after all of that, head back to earth, wait for every timeline of earth to finish or end. Then and after that, you can access the World Wide Web. If you fail this even once, you’ll be sent to Ancient-Brazil, and forced to do this all over again.
====Mid Aleph-Math (0.1%-1% of Power)====
Ok, did you do all of that? If you did, then lets move onto the next step: How to access Roblox.
*False-Collide Instant Ban-brick Wall-Walk all the way to Roblox headquarters a TORSHINGxTORSHING times, while with -YAONTING^YAONTING Jump-speed and Walk-speed. Next, reconstruct every possible alternative version of the entire Multiverse, with being blind, deaf, can’t smell or touch. After that, keep repeating until a OBAMA yottaseconds has passed. Then live through the entire history of the universe, remake roblox from absolutely nothing, copy everything that happened in the history of roblox, and if we mess up even just a Negative 0.0000000001% of a Planck second, you’ll have to do everything again, including every Low Aleph-Math Obstacle. Then download roblox with only 0.NaN% of a byte being downloaded every NaN^NaN^NaN^NaN^NaN years. Search “The Aleph-Game” while having everything in the site glitched and unreadable. Next, click on the game to play it, and move onto the next step.
====High Aleph-Math (1%-25% of Power)====
Holy Cow of the Moly, Oh my Lord, Thank the gods for your intelligence I mean, skill. Now, How to access the game.
*Once clicking on the play button, do a confirmation of every account in every parallel universe a NaN times, after that, give your account to everyone on the solar system, hope that no-one ever uses it for a OBAMA years, then finally enter the game, but you are a insta-kill and ban brick, with a ban of EXPAN decades, and you cant move your mouse, have speed and jump of -NaN^NaN, while trying to control your character from your mind, on opposite sides of the Omniverse, While having the brain power of a —TORSHING of a small broken pebble. Also, you need to get a Girlfriend, (Which Is Impossible), And you need to work together from this point. Then we can finally move onto the next step.
====Advanced Aleph-Math (25%-50% of Power)====
Class 24 terms here we go. How to do a stud.
*Now back onto the game-play itself, once you have finally loaded in, which would take EXPAN^OBAMA^200 years for each pixel to load, you would be instantly getting deleted and reprogram the entire game back every -10^1000^NaN^NaN*1 Millicentillion Planck Nano-seconds while trying the run and play the game by yourself in roblox studio with a Stone Age computer on 0.1% Battery, giving you only 39 fe to seconds to do all of that. Now, use your girlfriend and put her in the game shredder, in the top left corner of the screen, (Which is the size of a 0.0005 sub pixel on a 1024K TV.) After that, you have to wish on every gods life and hope that the remains of your girlfriend turn into a Key, and has a chance of -YANO-TORSHAVN of beings pickable and is definitely a insta-perm-ban brick.
====Skyline/Peak Aleph-Math (50-100% of Power)====
Aaaaaa I can’t handle le this!
*You are nearly done! Although, you haven’t even done the obby itself. You now have to do every possible and impossible obby made on all versions of roblox across every Omniverse to beyond The Outside. Phew, ok. It is time to start the REAL challenge… WIP.
====Beyond Aleph-Math (100-101% of Power)====
Literally just move onto the next difficulty, lol. The Super Absolutely Infinity Meltdown.
*The Super Absolutely Infinity Meltdown’s Obstacles
*Y e s
==Joke Obstacles:==
A bit of a Funni moment.
*Solve what 77 + 33 is. (Low A-M)
*Calculate how big and fat your mother is. (Peak A-M)
*Do 6 divided by 2 times (1 + 2). (Mid A-M)
*Find the distance to the guy who asked. (Peak A-M)
==Towers in this Difficulty:==
===Fictional Towers===
*Great Citadel of Aleph-ing (Fictional, so far)
*N/A
*N/A
*N/A
*N/A
*N/A
===Real Towers===
*N/A
*N/A
*N/A
*N/A
*N/A
==Continuing on...==
===Normal Way:===
Congratulations, You somehow beat this. This isn’t surprising, since you’ve managed to defeat every other difficulty, and also this is one of the easiest Class 24 difficulties. Move onto any Class 24 difficulty, first, The Super Absolutely Infinity Meltdown, I am sure Hell would DEFINITELY break loose there.
===Aleph Sequence Way:===
You somehow managed to beat every other Aleph-Sequence difficulty, Be happy. Your “Aleph” adventure has ended. For now.
==Icon Representation:==
===Aleph-Math Icon 1:===
Aleph-Math’s Icon is represented with Aleph-Null’s original icon, but most of it is black instead of grey and in glitchy/glowy text, mixed with a representation of the Hardest question known to mankind.
===Aleph-Math Icon 2:===
Basically the same as Icon 1, but it is slightly brightened and the “zero” is barely warped.
==Decal ID:==
If you want the decal, here you go:
Icon 1: https://www.roblox.com/library/6802377573/Aleph-Math-Difficulty
Icon 2: https://www.roblox.com/library/7898388469/Aleph-Math-Aleph-Icon-1-1
==Music:==
Add any type of Music that you believe suits this Difficulty!
Aleph-0 (By Leaf & Optie)
==Trivia:==
*This is The Hardest Difficulty in the Aleph Sequence, so far.
*This page is the 2nd longest in every Class 26 Difficulty.
*Wow, 7th most popular page, But why are so many popular pages now Difficulty Charts?
**Oh no now it’s 6th aaaa
***Now Aleph-Hold Cube is 6th
****Let’s get it to 5th
*****Aleph gang movement
*This Difficulty is part of The Aleph-Movement, including Aleph-Hold Cube.
*This is the only Aleph themed difficulty in Class 26. MEGADEATH-aleph-void.exe was the second, but it is now deleted.
*This is currently the easiest difficulty in Class 26. Used to be MEGADEATH-aleph-void.exe, but it was deleted.
*The math problem behind the Aleph-Null is from a YouTube video by 3Blue1Brown titled: The hardest problem on the hardest test. Go watch it, it’s really good.
==Gallery:==
Aleph-Math.jpg|Aleph-Math Icon 1
Aleph-Math 2.jpg|Aleph-Math Icon 2