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Graham's number

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For other uses, see Graham's number (disambiguation).
Graham's number in arrow notation

A concise definition of Graham's number.

Graham's number is a famous large number, defined by Ronald Graham.[1]

Using up-arrow notation, it is defined as:

\begin{eqnarray*} g_0 &=& 4 \\ g_1 &=& 3 \uparrow\uparrow\uparrow\uparrow 3 \\ g_2 &=& 3 \underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{g_1 \text{ arrows}} 3 \\ g_{k + 1} &=& 3 \underbrace{\uparrow\uparrow\uparrow\cdots\uparrow\uparrow\uparrow}_{g_k \text{ arrows}} 3 \\ g_{64} &=& \text{Graham's number} \end{eqnarray*}

Graham's number is celebrated as the largest number ever used in a mathematical proof, although much larger numbers have since claimed this title (such as TREE(3) and SCG(13)). The smallest Bowersism exceeding Graham's number is corporal.

History Edit

Wells

Martin Gardner's famous article introducing the number.

220px-GrahamCube.svg

An example of a cube with 12 planar K4's, with a single monochromatic K4 shown below. If you change the edge on the bottom of this K4 to blue, then the cube will contain no monochromatic planar K4's, thus showing that N* is at least 4.

Graham's number arose out of the following unsolved problem in Ramsey Theory:

Let N* be the smallest dimension n of a hypercube such that if the lines joining all pairs of corners are two-colored for any nN*, a complete graph K4 of one color with coplanar vertices will be forced. Find N*.

Graham published a paper in 1971 proving that the answer exists, providing the upper bound \(F^{7}(12)\), where \(F(n) = 2 \uparrow^{n} 3\) in arrow notation.[2] Sbiis Saibian calls this number "Little Graham". Martin Gardner, when discovering the number's size, found it difficult to explain, and he devised a larger, easier-to-explain number which Graham proved in an unpublished 1977 paper. Martin Gardner wrote about the number in Scientific American and it even made it to Guiness World Records in 1980 as the largest number used in a mathematical proof, although a few years later the title was removed from Guiness World Records.

In 2013, the upper bound was further reduced to N' = 2↑↑2↑↑2↑↑9 using the Hales–Jewett theorem.[3] As of 2014, the best known lower bound for N* is 13.

Gardner's article contained an error, where he claimed that 3↑↑7625597484987 = 3↑(7625597484987↑7625597484987); it is actually 3↑(3↑↑7625597484986).

Comparison Edit

Since g0 is 4 and not 3, Graham's number cannot be expressed efficiently with most major googological functions. It can be approximated with \(3 → 3 → 64 → 2\) in Conway's Chained Arrow Notation or \(\{ 3,65,1,2\}\) in BEAF, with upper bound \(\{3, 66, 1, 2\}\). A rare example of an exact representation is Jonathan Bowers' G functions, where it is.G644 in base 3. 

Tim Chow proved that Graham's number is much larger than the Moser.[4] The proof hinges on the fact that, using Steinhaus-Moser Notation, n in a (k + 2)-gon is less than \(n\underbrace{\uparrow\uparrow\ldots\uparrow\uparrow}_{2k-1}n\). He sent the proof to Susan Stepney on July 7, 1998.[5] Coincidentally, Stepney was sent a similar proof by Todd Cesere several days later.

It has been proven that Graham's number is much less than \(\Sigma(64)\),[6] and later a better upper bound \(\Sigma(22)\) was proven.

In the fast-growing hierarchy, Graham's Number can be shown to be less than f_{\omega+1}(64)

Calculating last digits Edit

The final digits of Graham's number can be computed by taking advantage of the convergence of last digits, because Graham's number is a power tower of threes. Here is a simple algorithm to obtain the last \(x\) digits \(N(x)\) of Graham's number:

  • \(N(0) = 3\)
  • \(N(x) = 3^{N(x-1)} \text{ mod } 10^x\)

For example:

  • \(N(1) = 3^{N(0)} \equiv 3^3 \equiv 27 \equiv 7 \pmod{10}\), so the last digit is 7.
  • \(N(2) = 3^{N(1)} \equiv 3^7 \equiv 2187 \equiv 87 \pmod{100}\), so the last two digits are 87.
  • \(N(3) = 3^{N(2)} \equiv 3^{87} \equiv 323257909929174534292273980721360271853387 \equiv 387 \pmod{1000}\), so the last three digits are 387.
  • etc.

This naive method is not very efficient, since number of digits in the leftmost expression grows exponentially. We can use right-to left binary method instead:

  • Convert the exponent into binary form. E.g. \(87_{10} = 1010111_2\)
  • If last digit of exponent is 1, then multiply base to result and square base.
  • Otherwise, just square base.

Using this, it can be shown that last 20 digits of Graham's number are: \(...04575627262464195387\).[7]

Video Edit

Source: Graham's Number - Numberphile

Graham's Number - Numberphile09:16

Graham's Number - Numberphile


Sources Edit

  1. Graham's Number
  2. Graham & Rothchild 1971 paper
  3. http://arxiv.org/pdf/1304.6910v1.pdf
  4. Proof that G >> M. (This website uses n[m] = n inside an m-gon for Steinhaus-Moser Notation.)
  5. Stepney, Susan. Moser's polygon notation. Retrieved 2013-03-17.
  6. Proof that BB(64) >> G
  7. Last 10000 digits of G

See also Edit

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