Graham's Number

          This weeks post will be on the Graham's number, and what that number means. Grahams number is known to be an immensely large number in the mathematical community, and hopefully this post will help shows its scale, and purpose. First this post will cover what the number refers to, and then it will talk about how absurdly big graham's number really is.

          Graham's Number is a number named after a mathematician by the name of Ron Graham. Graham's number is ridiculously big, and it refers to a geometry problem with respect to higher planes of spatial dimensions. Starting with spatial dimensions, the zero dimensional space refers to a singular point in space. The point has no dimensions, and is a singularity. Moving up in dimensions, the first spatial dimension refers to a connection between two zero dimension singularities via an infinite number of points, or a line segment. The second spatial dimension refers to connecting two lines to created a two dimensional plane. This plane could be pictured as a square, and if all the vertices were connected with lines including the diagonals, there would be six lines. If the lines were colored either red or blue there are six possible configurations. In terms of binomial expansion, there are four vertices, and two colors, so the notation would be four choose two, which are six. That's great, but who cares? The point of the problem is to look for certain avoidable configurations of colors and lines. The conditions that trying to be avoided are four points, that are flat, with six line connections, and all those line connections are the same color. Essentially the pain tis avoid the square that was mentioned earlier being all the same color. Obviously, this is avoidable in two dimensions, because as long as one line is a different color, then the condition is avoidable. However, as the number of spatial dimensions increases, the condition becomes avoidable. The question is which spatial dimensions is the configuration avoidable. Well spatial dimensions 2 through 12 are all avoidable, but all spatial dimensions between 13 and Graham's number of spatial dimensions are unknown. Graham's number of spatial dimensions makes the criteria unavoidable, but if there are smaller numbers of spatial dimensions that are unavoidable is still unknown. 

       So how big is Graham's Number? Graham's number is best explained in a notation called arrow notation. An example of arrow notation would be three arrow three or 3 ^ 3. This means 3 to the third power, or 27. Three double arrow (3^^3) means three arrow, three arrow three, or 3^(3^3), that’s the same as three raised to the twenty-seven, or 7.6 trillion. This is creating a tower of threes, or three rose to the power three raised to the power three. Three triples arrow (3^^^3) means that there is a tower of threes 7.6 billion threes long. The point to take away from this is (3^3) = 27, (3^^3) = 7.6 billion, and (3^^^3) = roughly 1.26 X 10 ^ 3,638,334,640,024. Meaning arrow notation gets big numbers very fast. Three four arrow three (3^^^^3) is a ridiculously large number. Take that crazy number, and that is how many arrow are in the next step, or (3(3^^^^3 number of arrows) 3). Repeat this process again by taking that even bigger number and making it the number of arrows in the next step. Repeat this process 64 times and then you have yourself Graham's number. Unsurprisingly this number once held the world record for the biggest mathematical number used in a proof.

    Graham's number is so large that there are not names for number to describe the number of digits in Graham's Number. Mathematicians understand enough properties of the powers of three to figure the last 500 digits, but it is thought that the first digit will never be known. Now how does a number like this even get figured out? What process lead to this number? As mentioned earlier this number has a meaning, it is the upper bound of number of spatial dimensions that follows a certain criteria. Ron's insight did not allow him to solve the problem, but rather a process that would limit the bound from infinity. This is similar to the post on the twin prime conjecture in that in both situations the method used was not optimized, bath rather the start. However unlike the twin prime conjecture there have not been any recent break through limiting the bounds, but hopefully there will be.

         Sources https://www.youtube.com/watch?v=GuigptwlVHo

                       https://www.youtube.com/watch?v=HX8bihEe3nA

                       http://mathworld.wolfram.com/GrahamsNumber.html

The Twin Prime Conjecture

         This weeks post will be covering a progression of the twin prime gap conjecture. The problem is faulty simple, and is only recently had a major breakthrough. Although there have been breakthroughs the problem remains unsolved. Firstly the problem itself will be covered. Secondly the order of progressions of the problem will be covered in this post.

         The twin prime conjecture is a fairly straightforward conjecture. It has been proven that there are an infinite number of prime numbers. The means that there is an infinite number of numbers that have no divisors other then one and itself. The twin prime conjecture focuses on the gap between prime numbers. There is one exception, but for the most part the smallest gap between two numbers is two. This is because every other number is even, so every other number is divisible by two. The only exception to that rule is, two, the only even prime number. If two prime numbers have a gap of two between them, then they are called twin primes. Primes with a gap of four are called cousin primes, and gaps of six are called sexy primes. The conjecture says that there is an infinite number of town primes. Meaning there is an infinite number of primes with only a gap of two between them. The origins of this problem are unknown. Some individuals say that the work comes from Euclid a couple thousand years ago. This could be very possible, but it was only formally written down a few hundred years ago, so it is at least that old. 

         There have been a number of breakthroughs in recent years towards this problem, but the first big breakthrough, that got the wheels turning happened within the last five years. From the University of New Hampshire a man by the name of Yitang Zhang released a paper putting a limit on the gaps. Before Yitang there was no approach to the problem, what he managed to do was say that there is an invite number of primes with gap N, N is between one and seventy million. To be clear he did not prove that the gap between primes are seventy million or less, but rather that there is an infinite number of primes with gap of seventy million. This breakthrough allowed for the first time a method of whittling down this seventy million to two. Yitang's result was the breakthrough, but it was not optimized. This means that though his result was seventy million, it was also understood that by tweaking the argument the bounds could shrink. This lead to a competition to see who could tweak the argument in such a way, to get the smallest bound. These developments lead to the gap being shrunk to 4,680, which is a big improvement, but not quite the end goal. Though the twin prime conjecture has not been proven, there have been more recent results that whittle down the gap number even more. Before the work of Yitang, there was an approach to solve the problem by mathematicians Goldston, Pintz, and Yildirim. Their work was important, but it required outside proofs, meaning that this works, but only if this other problem is solved. By sorting through these hiccups of the work of Goldstone, pints, and Yildirim, James Mayard, and Terrence Tao were able to whittle it down even further. Their method was completely different to Yitang, and both developed it simultaneously, and separately. Thus their work was named the Mayard-Tao method, and it brought the gap down to 256. 

         Even though this newly found method offers new insight into the problem, it has a flaw. The method used has a limit to it. If the method is completely optimized, it sill can not be used to prove the twin prime conjecture. Theoretically the Mayard-Tao method can only bring the gap down to there being an infinite number of primes with a gap of six. So even if the this method is completely optimized it is only a victory for the sexy primes. The win prime conjecture still needs a new approach, or new idea, to be solved.




Sources https://www.youtube.com/watch?v=vkMXdShDdtY
                  https://www.youtube.com/watch?v=QKHKD8bRAro
                  http://mathworld.wolfram.com/TwinPrimeConjecture.html

Gauss' Problem

            This week's post will be covering a problem solved by Gauss, and how it can be applied to another problem. To start things off I want you to add all the digits, of all the numbers from one to a million. Notice how it’s not the numbers, but the digits of those numbers that are being added up. So its not twenty-five plus twenty-six, it is two plus five plus two plus six. Think about this, and I will get to it at the end of the post. First the history of Gauss will be covered in their post. Then his similar problem will be solved. Finally I will connect the two and give the solution to the first problem I presented.

            Johannes Carl Friedrich Gauss was born in southern Germany in the late 1700's. Though he is not as famous as Newton, Gauss is still considered to be a great contributor to world of mathematics. His works included developments in number theory, such as advances in algebra, and statistics. He we also know for some of with work in the world of physics, as many mathematicians where in his time. He died in the mid eighteen hundreds.

         While in his early years Gauss was known to be a child prodigy. He seemed to have a natural ability with mathematics, and often times astonished people. One day while still in his youth Gauss went to school as he normally would. However, the professor was especially grumpy that day. Many variations of the story say he was old, and cranky, and would beat kids with his cane, making him seem more menacing then he probably was. The grumpy professor walked into the one room classroom, and since he did not feel like putting up with the children, he told them to add all the numbers from one to a hundred. At this time there was not pens and paper, but instead the children had blackboard slates. Now seconds after the professor had challenged the boys, Gauss runs up and puts his slate face down on the teacher’s desk. Thus submitting his answer. At this point in time Gauss is not the oldest boy in the room, and many of the other older boys look down on him. They were said to be thinking, poor little gauss, he has gone with a guess. Well time goes on, and all the other boys start working out the problem, using addition, multiplication, subtractions, and division, anything to try to get done first. Once the boys finished the problem they did like Gauss had previously done, putting their slate face down on the teachers desk, stacking them up. Once the time was up, the professor walked up to the desk and flipped all of the slates over. Now all the slates were in order from first to last, with the answers facing upward. Sitting on top of the pile was Gauss' answer. Gauss' answer was the correct answer, which is 5050. 

         So how did Gauss do it? Gauss, then explained, that the problem needs to be looked at in a different way. The most brute force way would be to add one plus two plus three, all the way to one hundred. Instead of doing it like that Gauss looked at in a different way. If all the number one through hundred are laid out one a number line, then Gauss' method can be seen. If you take one and add it to one hundred the answer is one hundred and one. The answer is the same if you add two and ninety nine. The same would apply to all the pair of numbers that can be created, all the way down to fifty and fifty-one. All Gauss had to do was take one hundred and one, and multiple by the number of pairs. In this case that would be fifty, so the answer would be fifty times one hundred and one. Fifty times one hundred and one is 5050.

          Going back to the problem at the start of post. Using the same process, start by writing out all the numbers from one to a million. Now notice that if you add to one to 999,998 it equals 999,999. This process can be done again and again. This process is the same pairing process gauss used. So then i would take the sum of all the digits of 999,999 and multiple that by the number of pairs. If zero is included that would make 500,000 pars. So 54 times a half a million would be twenty seven million. Not forgetting the one in one million would make the solution to the original problem 27,000,001.

Sources http://planetmath.org/gausscarlfriedrich

                  https://www.youtube.com/watch?v=Dd81F6-Ar_0

The Eight Queen Problem

          In this week's installment of mathematical insight, the topic will focus on the eight-queen problem. The post will start off with a small history, and backstory to the problem. Followed with the problem itself paired with some information on the scale of the numbers used in the problem. 

          The problem was originally published by Max Bezzel in the mid eighteen hundreds. Bezzel was a both a mathematician, and a chess enthusiast. It took two years for solutions to appear. Franz Nauk released them. Nauk then went to continue the problem into a more generalized state of understanding. The problem was modified from having eight queens to having "N" number of queens. After changing the number of queens Nauk modified the dimensions of the board. Instead of the traditional eight by eight board, it became "N" by "N" boards. "N" is not only the length, and width, but also the number of queens on the board. Mathematicians such as Gauss and Gunther have since contributed to the problem. 

         Looking primarily at the standard value of eight for "N", this meaning the board is a standard eight by eight chessboard, and there are eight queens on the board. The problem asks, if possible can all "N" number of queens be on the board and not be threatened by another queen? Color of the queen does not matter in this case, and if it can be done, how many possible ways can it be done? A queen in chess is considered to be the single most powerful piece, as it is the most versatile piece to move around the board. It can move both horizontally, vertically, and across diagonals. This makes positioning eight of them without threatening another difficult on a confined space. There are solutions to the standard eight by eight board. Starting with the total possible ways that eight queens can be positioned on a eight by eight board is the first part. There are sixty-four tiles on a standard chessboard. That means there are sixty-four factorial possible tile positions. If eight of the positions are taken up with queens that leaves fifty-six factorial tile positions. Since the queens are interchangeable among their given positions eight factorial also comes into play. So to find the number of ways that eight queens can be positioned on a chessboard, the total, sixty-four factorial, would be divided by 56 factorial multiplied by eight factorial. A factorial refers to that number multiplied by every previous whole number before that number. For example six factorial would be six, times five, times four, times three, times two, times one. A factorial has the nice property of showing how many ways you can arrange a number of times. For example if you wanted to know how many ways to arrange three items, the answer would be three factorial, or six. That makes numbers such as sixty-four factorial an immensely large number. Same with respect to fifty-six factorial. The net calculation on the number of ways to arrange the eight queens is roughly around four billion. This includes all permutations, but without the restriction of non-intersection between queens. Factoring that in reduces the number down to 92, continuing the sifting of possible outcomes, by removing redundancies due to reflections and rotations, the number goes down to twelve. There are twelve possible solutions to the eight-queen problem.

          To put factorials into scale of how big these numbers are, fifty-two factorial is less then the number of possible tiles, sixty-four factorial, and the number on non queen tiles fifty six factorial. Fifty-two factorial is the number of possible ways a deck of cards can be arranged. This is chosen because there is rather nice statistic about fifty-two factorial, and it is smaller then some of the numbers used in this problem. To understand fifty-two factorial, first stand on the equator. Start playing solitaire with a deck of cards, assume every new game is a new deck, and you win every game. Every billion years take a step forward, which should allow for a couple trillion games every step. Every time you walk around the earth remove a drop from the Pacific Ocean. Once the ocean is empty start stacking pieces of paper on the bottom. Once you have reached the sun, roughly one hundred million miles away. Notice not all decks have been played yet, so remove the papers one at a time, and put the water back in the ocean drop by drop, and go to and from the sun a thousand more times. Now about a third of 52 factorial decks have been played. That is how big fifty-two factorial is.

Sources http://www.datagenetics.com/blog/august42012/

                https://www.youtube.com/watch?v=jPcBU0Z2Hj8