Graham's Number 1 The Construction of Graham's Number by Ian Tame Background The primary goal of this article is to formally define and attempt to con- struct Graham's number, which appeared in the 1980 Guinness Book of World Records as being the largest number used in a formal mathematical proof (Gardner [1]). To undertake this task, we first consider the represen- tation of large numbers - in the sense of numbers that are of a significantly higher order than numbers used in everyday life, and to contrast these num- bers against the magnitudes of very large physical astronomical quantities. Explicit reference is made to the infamous googolplex and extensive use is made of the powerful Knuth up-arrow operators, which are a standardized notation for very large numbers, and are used to define Graham's number. Standardized Notation for very large numbers We proceed by introducing a special notation which will enable us to use a standardized system for writing very large numbers; and to compare and sort large numbers in increasing order, the Knuth up-arrow notation. First recall the fundamental relationship of multiplication with natural num- bers, whereby multiplication by a natural number is defined as iterated (re- peated) addition: a × b = a + a + ::: + a : | {z } b copies of a The next fundamental relationship is exponentiation (taking powers) for a natural power, defined as iterated multiplication: ab = a × a × ::: × a : | {z } b copies of a Grasping this structure of nested iterated operations is essential to under- standing how to interpret and build-up your own expressions of large num- bers within the Knuth system. You will soon appreciate that the versatility of this notation to represent incredibly large numbers lies in the fact that it 2 Graham's Number is based on iterated exponentiation; which is one of the fastest growing of all conventional functions! Now we introduce the Knuth up-arrow notation. First the single up-arrow operator """, which simply denotes exponentiation, a " b = ab: (Recall this can be thought of as repeated multiplication). Example 1. 5 " 3 = 53 = 5 × 5 × 5 = 125: Example 2. 10 " 4 = 104 = 10 × 10 × 10 × 10 = 10; 000: Now we extend this concept to denote iterated exponentiation by defining the Knuth double up-arrow operator """". a . a "" b = a " (a " (::: " a)) = aa | {z } | {z } b copies of a b copies of a IMPORTANT: expressions with the double up-arrow operator must be eva- luated from right to left (ie they are right-associative). Also, observe this right associative nested exponentiation is often called a power tower. Example 3. 22 4 2 "" 4 = 2 " (2 " (2 " 2)) = 22 = 22 = 216 = 65; 536: Example 4. In this example we make reference and comparison to volume defined in terms of Planck units. According to string theory, lengths smaller than the Planck length do not make any physical sense. To get some comprehen- sion of how this compares to an atomic scale, the radius of a proton is 1020 times larger than the Planck length. Graham's Number 3 5 "" 3 = 5 " (5 " 5) 5 = 5(5 ) = 53125 = (10log 5)3125 = 103125×log5 = 102184:28 = 100:28 × 102184 ≈ 1:91 × 102184 > 8:5 × 10184(the number of Planck volumes in observable universe)1 In the last example, the ability of the Knuth up-arrow operators to generate very large numbers has been starkly highlighted. With just the double up- arrow operator we have been able to generate a number that is unimaginably larger than even the number of Planck volumes in the known universe. But we have only just begun! The notation is now extended to define a triple up-arrow operator which is used to denote iterated application of the double up-arrow operator, as follows: a """ b = a "" (a "" (::: "" a)) | {z } b copies of a Again remember the right to left order of evaluation rule. Example 5. 3 3 """ 3 = 3 "" (3 "" 3) = 3 "" 33 = 3 " (3 " (::: " 3)) | {z } 3 33 copies of 3 3 But 33 = 327 = 7; 625; 597; 484; 987, so 3 """ 3 is given by . 3 . 3 " (3 " (::: " 3)) = 33 : | {z } | {z } 7,625,597,484,987 7,625,597,484,987 copies of 3 copies of 3 1A Planck volume is 4:22419×10−105 m3 which is based on a cube with Planck length 1:6162 × 10−35 m. 4 Graham's Number The number 3 """ 3 and the results from Example 5 will feature later in the description of Graham's Number. To appreciate the scale of magnitude of 3 """ 3, we now proceed to demonstrate that 3 """ 3 has a magnitude significantly greater than 10 "" (1010). First we need to prove the following two Lemmas. n Lemma 1: 33 > 10n (n 2 Z+) n 33 ≥ 33n since exponentiation dominates linear functions > 3nlog310 n = 3log310 = 10n: Lemma 2: 3 "" (2n) > 10 "" n fn 2 Z+g The proof is by mathematical induction. Induction Basis: n = 1 3 "" 2 = 3 " 3 = 33 = 27 > 10 = 10 "" 1: Hence establishing the Induction Basis. Induction Hypothesis. Assume result is true from some n = k 3 "" (2k) > 10 "" k: (1) Now consider the result for n = k + 1. 3 "" (2(k + 1)) = 3 "" (2k + 2) 3""(2k) = 33 10""k > 33 from (1) > 1010""k from Lemma 1. since 10 "" k is a positive integer = 10 "" (k + 1): Hence the result is true for n = k + 1. Induction Step. So it follows the result is true for n = 1 by the Induction Basis, and hence for n = 2; 3; 4; ::: for all n 2 Z+ by the Principle of Mathematical Induction. We can now prove the main theorem for 3 """ 3: Graham's Number 5 Theorem: 3 """ 3 > 10 "" (1010) 3 """ 3 = 3 "" (3 "" 3) 3 = 3 "" (33 ) = 3 "" (327) = 3 "" ((10log3)27) = 3 "" (1027log3) > 3 "" (1012:882) > 3 "" (1012) > 10 "" (5 × 1011) from Lemma 2. > 10 "" (1010): Observe from the last result that 3 """ 3 is significantly greater than . 10 . 10 "" (1010) = 1010 ; | {z } 1010 copies of 10 which obviously dwarfs in magnitude the googolplex, which is defined as 100 102 1010 1010 = 1010 < 1010 : | {z } 4 copies of 10 In general, the nth up-arrow operator expands into a right-associative series of (n − 1) up-arrow operators according to a "" ::: " b = a " ::: " (a " ::: " (::: " ::: " a )) | {z } | {z } | {z } | {z } n n-1 n-1 n-1 | {z } b copies of a Refer to Example 6. below, in order to see how this works in specific cases. Example 6. a) 3 """" 5 = 3 """ (3 """ (3 """ (3 """ 3) )) b) 5 """ 3 = 5 "" (5 "" 5) c) 2 "" 4 = 2 " (2 " (2 " 2) ) d) 7 "" 2 = 7 " 7 So with the generalised Knuth up-arrow notational framework established, we are in a position to formally define Graham's number, G. 6 Graham's Number Graham's Number G is calculated in 64 steps using 64 layers. 8 G = 3 "" ::::::::::::::: " 3 = g > 64 > | {z } > 3 "" ::::::::: " 3 = g > 63 <> | {z } 64 layers ······ > > 3 "" ::: " 3 = g2 > | {z } > :> 3 """" 3 = g1 | {z } The value of layer 1 is g1, the value of layer 2 is g2, and so on, up to layer 64 with value g64. The number of up-arrows in each layer is determined by the value of the layer below. So that the number of up-arrows in layer n = gn−1. That is layer n is of the form, "3 followed by gn−1 up-arrow operators follo- wed by 3". So to begin to understand the magnitude of Graham's Number, we have to start with the calculation of g1 (the lowest layer), and then the value of g2, and so on, eventually up to the value for g64. So let us begin with step 1, the computation of g1. By definition: g1 = 3 """" 3 = 3 """ (3 """ 3) (2) . 3 . = 3 """ 33 (3) | {z } 7,625,597,484,987 copies of 3 Observe the number of 3's in the expression (3) for g1 is 3 """ 3; that is 3 exponentiated by itself 7,625,597,484,987 times, which is an incomprehensi- bly large number. And remember this is just the number of 3's in the power stack of g1. (Recall we have shown in Theorem 3. that this number is grea- ter than 10 "" (1010), and hence unimaginably greater than the number of Plank volumes in the observable universe). Yet we have not even come close to determining the actual value for g1. To attempt this we will make use of a convenient shorthand notation for representing the Knuth double up-arrow operator, viz: nb = b "" n: Refer to Example 7. below, in order to see how this works with specific cases. Graham's Number 7 Example 7. a) 3 "" 3 = 33 3 b) 3 """ 3 = 3 "" (3 "" 3) = 3 "" 33 = 33 Note such nested left power towers are evaluated from the top down. A useful generalisation of the result in Example 7.b (verifiable by mathema- tical induction) of the left power tower derivation is: b . .. b """ n = b "" (b "" (::: "" b)) = b b : (4) | {z } | {z } n copies of b n copies of b Then when we apply the result in Example 7.b to the second occurrence of 3 """ 3 to the expression for g1 in (2), we yield the relationship g1 = 3 """ (3 """ 3) 3 = 3 """ 33 (5) So that applying (4) to (5) generates the following more compact left power stack form for g1: 3 .. .3 g1 = 3 : (6) | {z } 3 33 copies of 3 3 3 But now recall that 33 designates a right power tower with 33 (= 33 = 7,625,597,484,987 copies of 3).