Sample Mutual Termination Forms: Fill & Download for Free

Which verse in fiction is comparable to SCP levels of power?Honestly this is….. really hard for me to answer without sounding bias towards the SCPverse, in the SCPverse you have beings who adapt to anything instantlySCP-3812 - SCP Foundationbeings who supposedly can destroy all of creationbeings who just can’t be killed no matter what you throw at themSCP-682 - SCP FoundationBeings who can literally eat other beings whos mere shadow can erase entire multiversesItem: SCP-017 Tissue Test Record: Sample “swallowed” by SCP-017 without incident. Termination Test Record: SCP-682 exposed to SCP-017. SCP-682 issues several sounds at extremely high volume, damaging several recording devices. Sound extends across several wavelengths, reported as “the most god-awful roar” by staff. SCP-017 appears to stumble, then return to a far corner of the containment area. SCP-682 attempts to break containment of both SCP-682 and SCP-017. SCP-682 suppressed by Agents, and removed. SCP-682 states, “You foul bags of tissue; you don’t [DATA EXPUNGED]” Notes: It is unclear if SCP-682 somehow damaged SCP-017, or communicated with it. Analysis of the recorded sound is ongoing.Beings who have survived platonic concepts of reality get rewrittenItem: SCP-536 Tissue Test Record: Tissue divided into samples and subject to the individual effects of SCP-536's dials. Notable results follow:Increase in g: Tissue restructures itself into neutron degenerate matter.Decrease in e: Tissue maintains loose integrity as a cloud of ions, regenerates upon reestablishment of normal laws of physics.Decrease in Theta: Tissue disintegrates.Termination Test Record: SCP-682's containment tank inserted into SCP-536. Speed of light, strong nuclear force and fundamental charge dials decreased progressively. 682's containment tank is nearly immediately destroyed, and 682's body begins disintegrating. Due to intense light and radiation, visual is lost. Free neutrons, pions, kaons and more exotic mesons (described in [REDACTED]) are detected. 55s into the experiment, the primary detection equipment fails. Upon bootup of the secondary detection equipment, dials are at minimum levels. 682 is again visible in the chamber, reduced to roughly 1% of its normal size. Analysis suggests 682 has reformed into a previously unknown form of matter, kept together by quantum effects. Asst. Researcher ███████ becomes aggressive and turns the dials randomly and violently before being removed from the premises. 682 recovers its original shape upon restoration of standard physics. Note: I don't blame him. I could swear, at one point, that thing looked.And beings who can eat the very narrative/ erase fictional verses with just a mere thought.SCP-2747 - As below, so aboveAnd you wanna know what’s crazy? These aren’t even all the OP scp’s, yep that’s right their are thousands of more OP scps.In my opinion DC and Marvel, if we were to exclude omnipotent beings(obviously) don’t compare to the SCPverse the strongest in DC that isn’t omnipotent is either Lucifer or CAS(I could be wrong btw I have the 3 stooge knowledge on DC and Marvel)Lucifer and CAS are strong as hell, Lucifer tanked a big bang, killed two cyclops before they gained omnipotence(not sure how that works btw…. but it’s comics so whatever) and is a rival to Elaine iirc, CAS also can pre-adapt to any future threat, and becomes as strong as he needs to be to win. But imo CAS can be countered by SCP-2747.Basically this is what SCP-2747 can do as stated by Calee Chu in this wonderful answer Calee Chu's answer to Who would win, SCP-2747 or Cosmic Armor Superman?it is capable of mutually annihilating narratives.“It is the current hypothesis of the Department of Analytics that SCP-2747 represents evidence of a naturally-occurring anafabula, or anti-narrative: a cluster of interdependent signs, iconography and narremesthat, when included to a sufficient extent within a fictional construct, leads to mutual annihilation. First-hand reconstruction of the anafabula's properties is impossible given its anomalous nature, but second-hand and third-hand descriptions have been generated from Observational Procedure LUCID CHALICE and appended below (see Appendix B). It can effect through layers of metafictional narrative, i.e. a metanarrative containing the anafabula will cease to exist within the narrative, followed by the narrative itself disappearing from our reality.2. The key identifier of the anafabula is that it invariably represents an in-universe antagonist or anathema in all manifestations of SCP-2747, likely due to inherent narreme components indicating its alien, yet centralising, nature.”This also includes narratives outside of the primary narrative stack, of which contain various higher dimensions.Acidverse - SCP FoundationYou know me, I love my Acidverse.So it is my belief that SCP-2747 can take care of CAS and maybe even Lucifer.And like I said before these guys aren’t the only powerful beings SCP verse has, their are still millions of other SCP’s who are just as strong if not stronger then the SCP’s I have mentioned.So who can stack up to the SCPverse in terms of power? Not that many….The only verse I can think of as of right now would be Suggverse which is just wank but an official part of fiction. Other then that I really don’t know who else can even compare to the SCPverse, with the amount of powerful SCP I doubt anyone besides omnipotents can deal with them.PeaceCurrent song:

The answer to this very good question is a yes. But. To make our mutual experience on Quora more enjoyable, a few words on framing technical questions:opening them with Can you introduces a needless and a meaningless personalization. Mary Shmary can, while Joe Shmoe can not. Now what? And how cares? Better choices would be Is it possible to? Does there exist a method to? Does there exist a proof that/of? What is a sample proof that/of? and the likesdifficult is the personalization in disguise - simply drop it, see the aboveintegrals are not solved but evaluated, computed, estimated, investigated for existence and (uniform) convergence among other things. (Systems of) ((non)linear/integral/differential) Equations or well-defined problems are solvedout of a multitude of possible kinds of objects which specific one do you have in mind: Cauchy/Riemann/Stieltjes/Lebesgue/etc., of integrands in single/multiple real/complex variable(s), over bounded or unbounded domains, proper or improper - the relevant menu is deep. Say what you mean, understand what you are sayingThe point is, when your question is technically sound and aligns properly with an underlying mathematicity of it then it is more likely that a qualified individual will be willing to donate a slice of her/his time to answer it.(for the record - I am not a mathematician)One of the reasons why this is a very good question is the width of the mathematical terrain covered by the abstract phenomenon that it touches upon.On a number of occasions a definition of a mathematical object provides no clues to its construction whatsoever - take a definition of a greatest common divisor of two integers, for example.When, on the other hand, a definition of a mathematical object does provide a seemingly explicit recipe for its construction then it may be, and often is, not that convenient tactically.Constructing an object by directly following an explicit recipe embedded into its definition may be referred to, for the sake of brevity, as acting by the book:computing a value of a numeric series via the limit of the corresponding partial sums (see here and here)finding an alternate expression for a functional series via the limit of the corresponding partial sumsobtaining a Laurent series decomposition of a function holomorphic in a given annulus via contour integration in a complex planerecovering a residue of a holomorphic function at a pole via the coefficient [math]c_{-1}[/math] of the corresponding Laurent series expansion about the poleprime-factorizing an integer [math]n[/math] via repetitive division by consecutive primes between [math]2[/math] and [math]\sqrt{n}[/math]computing the value of a definite Riemann integral directly or by its definition or as a limit of the corresponding integral sumand so on.By-the-book practical manipulations are often tedious, to say the least, and sometimes are outright not feasible.So why have such definitions? Among other reasons, because such definitions are convenient in theoretical work and they lend themselves naturally to proofs generation.That is why (not only) mathematicians engage in all sorts of tactical manipulations of previously obtained results until they can beat them into submission and, hopefully, answer a concrete, pointed, practical question at hand.For example, in this Quora answer we have shown the concrete steps of a relatively fancy footwork needed to obtain the Laurent series decomposition of the complex function [math]w(z)=\pi\cot \pi z[/math] about origin via the corresponding Taylor series decomposition of the complex functions [math]\sin z[/math] and [math]\cos z[/math] and the geometric series.You are enquiring about the by-the-book manipulation.Let us pretend that we have successfully read your mind and that you are interested in seeing a sample direct computation of a definite Riemann integral as per its definition.By presenting such a sample computation we prove that the answer to that question is a yes.Sample 1. Compute the value of the following definite Riemann integral [math]I(r),\,|r|\neq 1,\,r\in\mathbb{R}[/math]:[math]\displaystyle I(r) = \int_0^{\pi}\log\left(1 - 2r\cos x + r^2\right)\;\mathrm dx \tag{1}[/math]directly.The integral [math]I(r)[/math] is often attached to the name of a French mathematician S. Poisson (1781–1840). Later on in your mathematical career you may learn about an object [math]P_r(\theta)[/math]:[math]P_r(\theta) = \dfrac{1 - r^2}{1 - 2r\cos\theta + r^2} \tag*{}[/math]which is known as the Poisson Kernel on an open unit disk and in whose denominator you should readily recognize the argument of our integrand.Solution: since it is clearly the case that:[math]0<\left(1 - |r|\right)^2 \leqslant 1 - 2r\cos x + r^2 \tag*{}[/math]it follows that, with the [math]|r|\neq 1[/math] constraint in hand, our integrand is a function that is continuous on the given interval and that, thus, the integral in question exists.By the book, partition the interval [math][0,\pi][/math] into [math]n[/math] sub segments of uniform length (Fig. 1):with the purpose of constructing the corresponding integral sum:[math]\displaystyle \sigma_n = \dfrac{\pi}{n}\;\sum_{k=1}^n \log\left(1 - 2r\cos\dfrac{\pi k}{n} + r^2\right) \tag{2}[/math]whose limit as [math]n[/math] grows without bound is [math]I(r)[/math]:[math]\displaystyle I(r) = \lim_{n\to\infty}\sigma_n \tag*{}[/math]As we can see, the battle to get a handle on the sum in (2) now unfolds.Playing the devil’s advocate: is that battle important? interesting? Does it constitute Mathematics?First off, we capitalize on the sum-to-product property of logarithms:[math]\displaystyle \sigma_n = \dfrac{\pi}{n}\;\log\prod_{k=1}^n \left(1 - 2r\cos\dfrac{\pi k}{n} + r^2\right) \tag{3}[/math]shifting the pressure toward getting a handle on the product [math]p_n[/math] such that:[math]\displaystyle p_n = \prod_{k=1}^n \left(1 - 2r\cos\dfrac{\pi k}{n} + r^2\right) \tag{4}[/math]which we introduce for brevity:[math]\sigma_n = \dfrac{\pi}{n}\;\log p_n \tag*{}[/math]and because with time you will (or should) develop a nose or an eye for reusable results that can stand on its own and be useful elsewhere.Find the parallel here with a proper software development - that is how (reusable) shared libraries, written in C, come into existence.We attack [math]p_n[/math] by chipping away from it little piece by little piece.When [math]k = n[/math] then the trigonometric function of cosine, namely [math]\cos\pi[/math], folds into a negative unity, leaving is with a nice factor of just [math]\left(1+2r+r^2\right)=(1+r)^2[/math]:[math]\displaystyle p_n = (1+r)^2\prod_{k=1}^{n-1} \left(1 - 2r\cos\dfrac{\pi k}{n} + r^2\right) \tag{5}[/math]noting that the upper bound of our product is now [math]n-1[/math].Whenever we see this very peculiar signature of the trigonometric function of cosine, that is whenever [math]k[/math] varies as it does while [math]n[/math] remains fixed, we can not help it but to recall the roots of unity.In this Quora answer we have already demonstrated that the solutions of the equation in [math]z[/math] such that [math]z\in\mathbb{C}[/math] and [math]n\in\mathbb{N}[/math]:[math]z^n = 1 \tag*{}[/math]are given by the complex numbers known as roots of unity [math]\omega_n(k)[/math]:[math]\omega_n(k) = \text{exp}\left(\dfrac{2\pi k}{n}\right), \;k\in\mathbb{Z} \tag{6}[/math]which can be rewritten in their trigonometric equivalent as follows:[math]\omega_n(k) = \cos\dfrac{2\pi k}{n} + i\;\sin\dfrac{2\pi k}{n} \tag{7}[/math]The distinct roots of unity are obtained when the variable [math]k[/math] runs through the set of least non negative residues modulo [math]n[/math]:[math]k = 0, 1, 2, 3, \ldots , n - 1 \tag{8}[/math]and that is almost what we have in our [math]p_n[/math]. Why almost?Because the signature of our trigonometric function of cosine lacks the [math]2[/math] present in (6, 7).That problem is straightforwardly cured by replacing [math]n[/math] with [math]2n[/math] in (7):[math]\omega_{2n}(k) = \cos\dfrac{\pi k}{n} + i\;\sin\dfrac{\pi k}{n} \tag{9}[/math]Aha.Now we are getting somewhere. We can see the light at the end of the tunnel.Walking away from the [math]z^n = 1[/math] equation, we reason that we have a polynomial [math]P(z) = z^{2n} - 1[/math] which, by the fundamental theorem of algebra, can be represented as a product of terms linear in [math]z[/math]:[math]\displaystyle P(z) = z^{2n} - 1 = \prod_{k=0}^{2n-1}\left(z - \omega_{2n}(k)\right) \tag{10}[/math]What does it mean geometrically?Because [math]2n[/math] is necessarily even, geometrically it means two things:that not only the points in a complex plane representing the roots of unity are sprinkled uniformly over the circumference of an origin-centered unit circle, but such points in the upper half plane are the mirror images of the points in the lower half plane with respect to the real axis and vice versaand that the point corresponding to the root of unity [math]k = n[/math] will always sit on the real axis at negative unity (Fig. 2):In Figure 2 above we see the case when [math]n = 5[/math], [math]2n=10[/math] and we, thus, have shown the points that are the mirror images of each other in the real axis in the same (orange) color and we have shown the points that sit on the real axis, [math]\omega_{2n}(0)[/math] and [math]\omega_{2n}(n)[/math], in a different same, violet, color.From the product in (10) we, then, chip away the factor that corresponds to [math]k = 0[/math], namely [math](z - 1)[/math], and we roll the result starting from [math]k=1[/math]:[math]\displaystyle z^{2n} - 1 =(z-1)\cdot \prod_{k=1}^{2n-1}\left(z - \omega_{2n}(k)\right) \tag{11}[/math]From the product in (11) we, then, chip away the factor that corresponds to [math]k = n[/math], namely [math](z + 1)[/math], and we still terminate the result with [math]k=2n-1[/math]:[math]\displaystyle z^{2n} - 1 =\left(z^2-1\right)\cdot \prod_{k=1}^{2n-1}\left(z - \omega_{2n}(k)\right) \tag{12}[/math]But wait a second.The notation in (12) sucks wind because it hides the fact that the factor corresponding to [math]k = n[/math] has been dealt with already.As such, break the wholesome product into a product of two products - roll the first product for [math]k[/math] between [math]1[/math] and [math]n-1[/math], geometrically covering the upper half plane, and roll the second product for k between [math]n+1[/math] and [math]2n-1[/math], geometrically covering the lower half plane:[math]\displaystyle z^{2n} - 1 =\left(z^2-1\right)\cdot \prod_{k=1}^{n-1}\left(z - \omega_{2n}(k)\right)\cdot \prod_{k=n+1}^{2n-1}\left(z - \omega_{2n}(k)\right) \tag{13}[/math]Now capitalize on the fact that the roots of unity that are left standing in the products in (13) are what is known as the conjugate images of each other:[math]\omega_{2n - k}(k) = \overline{\omega_{2n}(k)} \tag{14}[/math]Refresher: a conjugate image of a complex number [math]z = x+iy[/math] is [math]\overline z = x - iy[/math].With (14) the product in (13) can be rewritten as:[math]\displaystyle z^{2n} - 1 =\left(z^2-1\right)\cdot \prod_{k=1}^{n-1}\left(z - \omega_{2n}(k)\right)\cdot \left(z - \overline{\omega_{2n}(k)}\right) \tag{15}[/math]At this point, as we multiply out the factors in parentheses in (15), a straightforward arithmetic calculus of complex numbers takes over:[math]\left(z - \omega_{2n}(k)\right)\cdot \left(z - \overline{\omega_{2n}(k)}\right) = \tag*{}[/math][math]z^2 - z\left(\omega_{2n}(k)+ \overline{\omega_{2n}(k)} \right) + \omega_{2n}(k)\cdot\overline{\omega_{2n}(k)} \tag*{}[/math]One by one, according to our conjugate image definition above:[math]\omega_{2n}(k)\cdot\overline{\omega_{2n}(k)} = 1 \tag*{}[/math]and, with (9):[math]\omega_{2n}(k)+ \overline{\omega_{2n}(k)} = 2\cos\dfrac{\pi k}{n} \tag*{}[/math]because in this case the imaginary parts of the roots of unity vanish.Thus:[math]\displaystyle z^{2n} - 1 = \left(z^2-1\right)\cdot \prod_{k=1}^{n-1}\left(z^2 - 2z\cos\dfrac{\pi k}{n} +1\right) \tag{16}[/math]Substitute [math]z = r[/math] into (16):[math]\displaystyle r^{2n} - 1 = \left(r^2-1\right)\cdot \prod_{k=1}^{n-1}\left(r^2 - 2r\cos\dfrac{\pi k}{n} +1\right) \tag{17}[/math]and inject the LHS of (17) into the RHS of (5):[math]p_n = \dfrac{(r+1)^2}{r^2 - 1}\cdot\left(r^{2n} - 1\right) \tag*{}[/math]which is to say that:[math]p_n = \dfrac{r+1}{r-1}\cdot\left(r^{2n} - 1\right) \tag{18}[/math]Put the RHS of (18) into the RHS of (3):[math]\displaystyle \sigma_n = \dfrac{\pi}{n}\;\log\left(\dfrac{r+1}{r-1}\cdot\left(r^{2n} - 1\right)\right) \tag{19}[/math]and take the limit of both sides of (19) as [math]n[/math] grows without bound:[math]\displaystyle \lim_{n\to\infty}\sigma_n = \lim_{n\to\infty}\dfrac{\pi}{n}\;\log\left(\dfrac{r+1}{r-1}\cdot\left(r^{2n} - 1\right)\right) \tag*{}[/math]We see that we must consider two cases:when [math]|r|< 1[/math] andwhen [math]|r|>1[/math]Consider the first case when [math]|r|<1[/math]. Then, clearly:[math]\displaystyle \lim_{n\to\infty}|r|^{2n} = 0 \tag*{}[/math]Therefore:[math]\displaystyle \lim_{n\to\infty}\sigma_n = \lim_{n\to\infty}\dfrac{\pi}{n}\;\log \dfrac{r+1}{1-r} = 0 \tag*{}[/math]Therefore, when [math]|r|<1[/math] then it is the case that:[math]\displaystyle I(|r|<1) = \int_0^{\pi}\log\left(1 - 2r\cos x + r^2\right)\;\mathrm dx = 0 \tag{20}[/math]Consider the second case when [math]|r|>1[/math].Here we need to engage in some creative arithmetic.We can, for example, divide and multiply the argument of the logarithm by [math]r^{2n}[/math]:[math]\dfrac{r+1}{r-1}\cdot\left(r^{2n} - 1\right) = \dfrac{r+1}{r-1}\cdot\dfrac{r^{2n} - 1}{r^{2n}}\cdot r^{2n} \tag*{}[/math]so that:[math]\sigma_n = \dfrac{\pi}{n}\;\log\left(\dfrac{r+1}{r-1}\cdot\dfrac{r^{2n} - 1}{r^{2n}}\right) +2\pi\log|r| \tag*{}[/math]which means that:[math]\displaystyle \lim_{n\to\infty}\sigma_n = 2\pi\log|r| \tag*{}[/math]which means that when [math]|r|>1[/math] it is the case that:[math]\displaystyle I(|r|>1) = \int_0^{\pi}\log\left(1 - 2r\cos x + r^2\right)\;\mathrm dx = 2\pi\log|r| \tag{21}[/math]You should compute [math]I(r)[/math] by other means for comparison purposes.Sample 2. Compute the value of the following improper integral:[math]\displaystyle \int_0^{+\infty}\dfrac{\sin^2 x}{x^2} \; \mathrm dx \tag*{}[/math]directly.Solution: see this Quora answer.Extra for experts.Observe that in Sample 1 to compute a real integral directly we had to step out on [math]\mathbb{R}[/math], into [math]\mathbb{C}[/math], and take a tiny swipe at elementary number theory while at it.The above experiment tells us at least three things.Mathematics is internally consistent - distinct solution paths lead to the same result.We spared you virtually no pain to demonstrate what a by-the-book manipulation may look like. The source of the said pain is the fact thatThere are no shortcuts or no gifts in mathematics - if an intermediate result is needed then that result will never fall out of the sky one day. Someone must generate that result. Guess who that someone is?

Append this answer to the end of the great answer already given by Alan Bustany:… and depending on the purpose and the specific application of the need to render base-ten digits in higher bases it is you who will be designing your own extended translation alphabet.Here is a real life (whatever that means) example of me doing exactly that in a professional capacity of a computer programmer. What follows is an exact reproduction of a real business need with some non technical details omitted for obvious reasons.Problem/Requirement: before a fixed length string of nine base-ten digits is transmitted over a network connection somewhere else, it is to be maximally compressed in length in such a way that it can be uncompressed into its original on the receiving side.Solution: came in 3 seconds or less - the required length compression is achieved exactly by converting the input base-ten number into its base-sixty-four equivalent (and [math]64[/math] turns out to be the sweet spot for this problem).It took all of 15 minutes to implement this idea soup to nuts.Optional (elementary number) theory: any positive integer [math]a[/math] can be rendered in a given integral base [math]B>1[/math] in a unique way as follows:[math]a = a_nB^n+a_{n-1}B^{n-1}+\ldots a_2B^2+a_1B+a_0 \tag{1}[/math]or, in a compressed form:[math]\displaystyle a = \sum_{k=0}^na_kB^k \tag{2}[/math]where the set of the coefficients [math]\{a_k\}[/math] forms what is known as the alphabet of the given numeral system in the base [math]B[/math].Such coefficients are associated in the minds of most people with just one symbol - and that is the crux of this matter because such a symbol or a small picture or a small hieroglyph can be just about anything we want, as long as we all agree on the meaning of each such a symbol and insure that each such symbol is unique.To such a hieroglyph these days we refer to as a base-[math]B[/math] digit or simply a digit.Historically, the set of Arabic hieroglyphs rendered as [math]0[/math], [math]1[/math], [math]2[/math], [math]3[/math] and so on won over the competitors and now makes up the alphabet of a base-ten numeral system, in which a sample number such as [math]9459[/math], as per (1), is depicted as follows (since [math]B=10>1[/math]):[math]9459 = 9\cdot 10^3 + 4\cdot 10^2 + 5\cdot 10^1 + 9\cdot 10^0 \tag{3}[/math]The numeral systems in which the objects of its domain are recorded as shown in (1) and (3) are called positional because the actual value of each underlying digit in such systems depends heavily on the exact geometry, the exact location or the exact position of such a digit within the string that represents a given number.For example, in our sample number above the leftmost digit [math]9[/math] carries the weight of one thousand, while its rightmost digit [math]9[/math] carries the weight of just one.It should now be clear that if we take our input base-ten positive integer [math]a[/math] (that is nine digits long, say) and attempt to record it in base-sixty-four:[math]a = a_n64^n+a_{n-1}64^{n-1}+\ldots a_264^2+a_164+a_0 \tag{4}[/math]then we will face the challenge of recovering the numbers [math]\{a_k\}[/math] each of which corresponds to or relates to an element from the alphabet of the numeral base-sixty-four system.How can we do that?Ah … this is where you come in.Geometrically we can process the RHS of the equation shown in (4) in two mutually opposite directions - either left-to-right (Fig. 1):or right-to-left, correct?If we were to process the RHS of the equation shown in (4) left-to-right working on the highest possible integral powers of [math]64[/math] first, then we would clearly need the access to the perfect integral powers of [math]64[/math]. We shall not pursue that approach here.Instead.We shall process the RHS of the equation shown in (4) right-to-left working on the lowest integral powers of [math]64[/math] first, by using the Divide and Conquer problem-solving approach on a micro scale.First, we shall compute the integral remainder [math]r[/math] of the integer-division of the input number [math]a[/math] by the base [math]B[/math]: according to the elementary number theoretic result known as The Division Algorithm Theorem, separately, each summand on the RHS of (1) located to the left of the coefficient [math]a_0[/math] is evenly integer-divisible by the number [math]B[/math].Therefore, by the same theorem, our input number [math]a[/math] can be represented as follows:[math]a = q\cdot B + r,\;0\leqslant r < B \tag{5}[/math]and in our notation the remainder [math]r[/math] will be none other than the coefficient [math]a_0[/math]: [math]r = a_0[/math].For example, for our sample input number [math]a=9459[/math], the remainder [math]r[/math] of the first integer division of [math]a[/math] by [math]B=64[/math] will be [math]51 = a_0[/math]:[math]9459 = 147\cdot 64 + 51,\;0\leqslant 51 < 64 \tag*{}[/math]Because I operate in the C/C++/UNIX/Linux world, we will use the modular arithmetic-related operator % that carries out the above maneuver in one fell swoop:r = 9459 % 64 Now, as we stare at the blessed number [math]51[/math], we wonder - what on Earth are we to do with it?Ah, the answer to that question is exactly the answer to your question - we use that remainder as the index into the array of symbols (small hieroglyphs/pictures) that represent the base-sixty-four alphabet (Fig. 2):And exactly what such a base-sixty-four alphabet looks like?Well, we are the designers of such an alphabet - we can make it anything we want so long as the definition of the alphabet of a positional numeral system is honored.As such, we can, but we do not have to, roll our base-sixty-four alphabet with the digits zero through nine first:alphabet[] = "0123456789"; We can then extend such an initial version of our alphabet with [math]26[/math] upper case English alphabet letters ‘A’ through ‘Z’ for a total of [math]36[/math] units:alphabet[] = "0123456789ABC...XYZ"; We can further extend our alphabet with [math]26[/math] lower case English alphabet letters ‘a’ through ‘z’ for a total of [math]62[/math] units:alphabet[] = "0123456789ABC...XYZabc...xyz"; and we are almost home free - we need two more units.How do the underscore character _ and the dash character - grab ya?Don’t like those? No problem - how about ~ (the tilde) and ! (the exclamation mark)?Don’t like those? No problem - how about [ and ], the square brackets?Don’t like those? No problem - how about < and >, the angle brackets (wink, wink)?Don’t like those? No problem - how about ( and ), the parens?Don’t like those? No problem - how about ^ and &, the cap and the ampersand?and so on - we hope that you are getting the overall drift: for our alphabet we can use just about any printable symbols that fit into eight bits. For demonstration purposes let us agree that we settle on the star * and the pound sign #.Lastly, we make our large problem smaller by integer-dividing the input number [math]a[/math] by the base [math]B[/math]:a = a / B; and applying the same computational maneuver to the new (smaller) input number [math]a[/math].Why would the new input number [math]a[/math] be smaller? Because in the context of the relation shown in (5) the number [math]a[/math] is the quotient [math]q[/math].We, thus, keep taking the remainder of the division of the smaller and smaller numbers [math]a[/math] by the base [math]B[/math] and translating these remainders into the alphabet of the base-sixty-four numeral system until one such number [math]a[/math] vanishes, at which point our conversion process terminates.Why such a conversion process must terminate? Because of the Well-Ordering Principle, according to which every non empty set of non negative integers contains a least element.For demonstration purposes we are being more explicit and verbose than we really have to and in order to avoid the essence-obfuscating clutter we omit any error-handling:#include <stdio.h> #include <stdlib.h>  static char const Alphabet[] = "0123456789"  "ABCDEFGHIJKLMNOPQRSTUVWXYZ"  "abcdefghijklmnopqrstuvwxyz"  "*#";  static char* toBase64( int a, char* buf, int bufLen ) {  int r;  char* b64a = buf + bufLen;   *b64a = '\0';  b64a--;  do  {  r = a % 64; /* Compute the remainder */  *b64a = Alphabet[ r ]; /* Pick up the base-64 digit */  a = a / 64; /* Make the problem smaller */  b64a--; /* Move to the next output position */  }  while ( a );   b64a++;  return b64a; }  extern int main( int argc, char* argv[] ) {  int a = atoi( argv[ 1 ] );  char* b64a = NULL;  char buf[ 64 + 1 ] = { 0 };   b64a = toBase64( a, buf, sizeof( buf ) - 1 );  printf( "b64a = %s\n", b64a );   return 0; } Assuming that we save the above code in the b64.c disk file, compile it as follows:gcc -g -o b64 b64.c and test it as follows:./b64 999999999 b64a = xcid# and so on.Experiment at will:./b64 987654321 b64a = wtcYn or./b64 123456789 b64a = 7MyqL We see that instead of [math]9[/math] bytes, our base-[math]64[/math] string is only [math]5[/math] bytes long - and that should answer your question.If we wanted to add some silly obfuscation to the compressed string, we could cook up our alphabet in any which way by mixing its existing contents in any wild order:Alphabet[] = "AxQbzTJC7p521..."; as long as the same code is symmetrically used to compress and uncompress our numbers, that would make no difference whatsoever.A such, the next section is optional.Once we receive a compressed string at some destination point, we uncompress it into the original number symmetrically with respect to the code used by the toBase64() function - by using exactly the same alphabet and by reversing the process encoded in the toBase64() function: multiply the said remainder by the appropriate power of [math]64[/math] and add the resultant summand (s) to the running total (a).In the code below we have added the indexOf() and fromBase64() functions that implement such a reverse process:#include <stdio.h> #include <stdlib.h> #include <string.h>  static char const Alphabet[] = "0123456789"  "ABCDEFGHIJKLMNOPQRSTUVWXYZ"  "abcdefghijklmnopqrstuvwxyz"  "*#";  static char* toBase64( int a, char* buf, int bufLen ) {  int r;  char* b64a = buf + bufLen;   *b64a = '\0';  b64a--;  do  {  r = a % 64;  *b64a = Alphabet[ r ];  a = a / 64;  b64a--;  }  while ( a );   b64a++;  return b64a; }  static int indexOf( char b64digit ) {  int ndx = 0;   for ( ndx = 0; ndx < sizeof( Alphabet ) - 1; ndx++ )  {  if ( b64digit == Alphabet[ ndx ] )  {  break;  }  }   return ndx; }  static int fromBase64( const char* b64a, int b64aLen ) {  int ndx;  int a = 0;  int power = 1;  int s = 0;  const char* here = b64a + b64aLen - 1;   while ( here >= b64a )  {  ndx = indexOf( *here );  s = ndx * power;  a = a + s;  power = power * 64;  here--;  }   return a; }  extern int main( int argc, char* argv[] ) {  int orig;  int a = atoi( argv[ 1 ] );  char* b64a = NULL;  char buf[ 64 + 1 ] = { 0 };   b64a = toBase64( a, buf, sizeof( buf ) - 1 );  printf( "b64a = %s\n", b64a );  printf( "sending %s over ...\n", b64a );  printf( "received %s, uncompressing ...\n", b64a );  orig = fromBase64( b64a, strlen( b64a ) );  printf( "The original number is: %d\n", orig );   return 0; } Build the new code like so:gcc -g -o b64 b64.c and test it like so:./b64 999999999 b64a = xcid# sending xcid# over ... received xcid#, uncompressing ... The original number is: 999999999 or like so:./b64full 123454321 b64a = 7MyDn sending 7MyDn over ... received 7MyDn, uncompressing ... The original number is: 123454321 etc.Feel free to instrument the above code with more informative printf()s that will deepen your understanding of how are digits in bases higher than 10 represented.