Root Pursuit

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:

• √2 = 1·414213562373095048801688724209698078569671875376948073...

It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:

• 2 = 2^1
• 4 = 2^2
• 8 = 2^3
• 16 = 2^4
• 32 = 2^5
• 64 = 2^6
• 128 = 2^7
• 256 = 2^8
• 512 = 2^9
• 1024 = 2^10
• 2048 = 2^11
• 4096 = 2^12
• 8192 = 2^13
• 16384 = 2^14
• 32768 = 2^15
• 65536 = 2^16
• 131072 = 2^17
• 262144 = 2^18
• 524288 = 2^19
• 1048576 = 2^20
[...]

But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?

131072 = 2^17
129140163 = 3^17
1255420347077336152767157884641... = 2^193
1214512980685298442335534165687... = 3^193
2175541218577478036232553294038... = 2^619
2177993962169082260270654106078... = 3^619
7524389324549354450012295667238... = 2^2016
7524012611682575322123383229826... = 3^2016

There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:

32 = 2^5
3125 = 5^5
316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185

The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:

• √10 = 3·1622776601683793319988935444327185337195551393252168268575...

I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:

• 316227... = 2^2728361
• 316227... = 5^2728361
• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 3162277660... = 2^961700165
• 3162277660... = 5^961700165

But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:

32 = 2^5
3125 = 5^5
• 3^2 = 9

316912650057057350374175801344 = 2^98
3155443620884047221646914261131... = 5^98
• 31^2 = 961

3162535207926728411757739792483... = 2^1068
3162020133383977882730040274356... = 5^1068
• 3162^2 = 9998244

3162266908803418110961625404267... = 2^127185
3162288411569894029343799063611... = 5^127185
• 31622^2 = 999950884

• 316227... = 2^2728361
• 316227... = 5^2728361
• 316227^2 = 99999515529

• 3162277... = 2^15917834
• 3162277... = 5^15917834
• 3162277^2 = 9999995824729

• 31622776... = 2^73482154
• 31622776... = 5^73482154
• 31622776^2 = 999999961946176

• 3162277660... = 2^961700165
• 3162277660... = 5^961700165
• 3162277660^2 = 9999999998935075600

The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:

3 162266908803418110961625404267... = 2^127185
3·162277660168379331998893544432... = √10
3 162288411569894029343799063611... = 5^127185

In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:

• 2.24103122055214532500432040411... = √6 (in base 6)

24 = 2^4
213 = 3^4
225522024 = 2^34 in base 6 = 2^22 in base 10
22225525003213 = 3^34 (3^22)
2241525132535231233233555114533... = 2^1303 (2^327)
2240133444421105112410441102423... = 3^1303 (3^327)
2241055222343212030022044325420... = 2^153251 (2^15007)
2241003215453455515322105001310... = 3^153251 (3^15007)
2241032233315203525544525150530... = 2^233204 (2^20164)
2241030204225410320250422435321... = 3^233204 (3^20164)
2241031334114245140003252435303... = 2^2110415 (2^102539)
2241031103430053425141014505442... = 3^2110415 (3^102539)

And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:

• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

55AA4 = 2^M in base 30 = 2^22 in base 10
5NO6CQN69C3Q0E1Q7F = F^M = 15^22
5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606)
5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606)
5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934)
5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934)
[...]
5R4C9 = 3^E in base 30 = 3^14 in base 10
52CE6A3L3A = A^E = 10^14
5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113)
5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113)
5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187)
5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187)
[...]
5OCNCNRAP = 5^I in base 30 = 5^18 in base 10
54NO22GI76 = 6^I (6^18)
5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567)
5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567)
5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786)
5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786)
[...]

So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:

•  √2 = 1·414213562373095048801688724209698078569671875376948073...
• √20 = 4·472135954999579392818347337462552470881236719223051448...

16 = 2^4
125 = 5^3
140737488355328 = 2^47
142108547152020037174224853515625 = 5^46
1413... = 2^243
1414... = 5^242
14141... = 2^6651
14142... = 5^6650
141421... = 2^35389
141420... = 5^35388
4472136... = 2^162574
4472135... = 5^162573
141421359... = 2^3216082
141421352... = 5^3216081
447213595... = 2^172530387
447213595... = 5^172530386
[...]

The Trivial Troot

Here is the square root of 2:

√2 = 1·414213562373095048801688724209698078569671875376948073176679738...

Here is the square root of 20:

√20 = 4·472135954999579392818347337462552470881236719223051448541794491...

And here are the first few triangular numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035...

What links √2 and √20 strongly with the triangular numbers? At first glance, nothing does. The square roots of 2 and 20 are very different from the triangular numbers. Square roots like those are irrational, that is, they can’t be represented as a fraction or ratio of integers. This means that their digits go on for ever, never falling into a regular pattern. So the digits are hard to calculate. The sequence of triangular numbers also goes on for ever, but it’s very easy to calculate. The triangular numbers get their name from the way they can be arranged into simple triangles, like this:

* = 1


*
** = 3


*
**
*** = 6


*
**
***
**** = 10


*
**
***
****
***** = 15

The 1st triangular number is 1, the 2nd is 3 = 1+2, the 3rd is 6 = 1+2+3, the 4th is 10 = 1+2+3+4, and so on. The n-th triangular number = 1+2+3…+n, so the formula for the n-th triangular number is n*(n+1)/2 = (n^2+n)/2. So what’s the 123456789th triangular number? Easy: it’s 7620789436823655 (see A077694 at the OEIS). But what’s the 123456789th digit of √2 or √20? That’s not easy to answer. But here’s something else that is easy to answer. If tri(n) is the n-th triangular number, what are the values of n when tri(n) is one digit longer than tri(n-1)? That is, what are the values of n when tri(n) increases in length by one digit? If you look at the beginning of the sequence, you can see the first three answers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105...

1 is one digit longer than nothing, as it were, and 1 = tri(1); 10 is one digit longer than 6 and 10 = tri(4); 105 is one digit longer than 91 and 105 = tri(14). Here are some more answers, giving triangular numbers on the left, as they increase in length by one digit, and the n of tri(n) on the right:

1 ← 1
10 ← 4
105 ← 14
1035 ← 45
10011 ← 141
100128 ← 447
1000405 ← 1414
10001628 ← 4472
100005153 ← 14142
1000006281 ← 44721
10000020331 ← 141421
100000404505 ← 447214
1000001326005 ← 1414214
10000002437316 ← 4472136
100000012392316 ← 14142136
1000000042485480 ← 44721360
10000000037150046 ← 141421356
100000000000018810 ← 447213595
1000000000179470703 ← 1414213562
10000000002237948990 ← 4472135955
100000000010876002500 ← 14142135624
1000000000022548781025 ← 44721359550
10000000000026940078203 ← 141421356237
100000000000242416922750 ← 447213595500
1000000000000572687476751 ← 1414213562373
10000000000004117080477500 ← 4472135955000
100000000000007771272992046 ← 14142135623731
1000000000000031576491575006 ← 44721359549996
10000000000000140731196136705 ← 141421356237310
100000000000000250760786750861 ← 447213595499958
1000000000000000638090771126060 ← 1414213562373095
10000000000000000479330922588410 ← 4472135954999579
100000000000000000169466805816725 ← 14142135623730950
1000000000000000025572412483843115 ← 44721359549995794
10000000000000000087657358700327265 ← 141421356237309505
100000000000000000097566473134542830 ← 447213595499957939
1000000000000000000987561276980703725 ← 1414213562373095049
10000000000000000003048443380954913921 ← 4472135954999579393
100000000000000000006832246143819194316 ← 14142135623730950488
1000000000000000000014155501020518731556 ← 44721359549995793928

Can you spot the patterns? When tri(n) has an odd number of digits, n approximates the digits of √2; when tri(n) has an even number of digits, n approximates the digits of √20. And what can you call the approximations? Well, in a way they’re triangular roots so I’m calling them troots. Here are the troots for tri(n) with an odd number of digits:

1 → 1
14 → 105
141 → 10011
1414 → 1000405
14142 → 100005153
141421 → 10000020331
1414214 → 1000001326005
14142136 → 100000012392316
141421356 → 10000000037150046
1414213562 → 1000000000179470703
14142135624 → 100000000010876002500
141421356237 → 10000000000026940078203
1414213562373 → 1000000000000572687476751
14142135623731 → 100000000000007771272992046
141421356237310 → 10000000000000140731196136705
1414213562373095 → 1000000000000000638090771126060
14142135623730950 → 100000000000000000169466805816725
141421356237309505 → 10000000000000000087657358700327265
1414213562373095049 → 1000000000000000000987561276980703725
14142135623730950488 → 100000000000000000006832246143819194316
14142135623730950488... = √2 (without the decimal point)

When I first found these patterns, I thought I might have discovered something mathematically profound. I hadn’t. Troots are trivial. I think troots are beautiful too, but a little thought soon showed me how easily and obviously they arise. Remember that the formula for tri(n), the n-th triangular number, is tri(n) = (n^2+n)/2. As you can see above, when tri(n) is increasing in length by one digit, it rises above the next power of 10, which always begins with 1 followed by only 0s. Therefore n^2+n will begin with the digit 2 followed by some 0s, which then becomes 1 followed by some 0s as (n^2+n) is divided by 2. So n for tri(n) increasing-by-one-digit will be the first integer, n, where n^2+n yields a number with 2 as the leading digit followed by more and more 0s.

And that’s why n approximates the digits of √2·0000… and √20·0000…, for tri(n) with an odd and even number of digits, respectively. Similar trootful patterns exist in other bases and for other polygonal numbers, like the square numbers, the pentagonal numbers and so on. The troots are beautiful to see but trivial to explain. All the same, there is a sense in which you can say the mindless sequence of triangular numbers is “calculating” the digits of √2 and √20. It even rounds up the final digits when necessary:

1414214 → 1000001326005
14142136 → 100000012392316
141421356 → 10000000037150046
141421356... = √2
[...]
14142135624 → 100000000010876002500
141421356237 → 10000000000026940078203
141421356237... = √2
[...]
14142135623731 → 100000000000007771272992046
141421356237310 → 10000000000000140731196136705
1414213562373095 → 1000000000000000638090771126060
1414213562373095... = √2
[...]
1414213562373095049 → 1000000000000000000987561276980703725
14142135623730950488 → 100000000000000000006832246143819194316
14142135623730950488... = √2

The Power of Powder

• Racine carrée de 2, c’est 1,414 et des poussières… Et quelles poussières ! Des grains de sable qui empêchent d’écrire racine de 2 comme une fraction. Autrement dit, cette racine n’est pas dans Q. — Rationnel mon Q: 65 exercices de styles, Ludmilla Duchêne et Agnès Leblanc (2010)

• The square root of 2 is 1·414 and dust… And what dust! Grains of sand that stop you writing the root of 2 as a fraction. Put another way, this root isn’t in Q [the set of rational numbers].

Root Rite

A square contains one of the great — perhaps the greatest — intellectual rites of passage. If each side of the square is 1 unit in length, how long are its diagonals? By Pythagoras’ theorem:

a^2 + b^2 = c^2
1^2 + 1^2 = 2, so c = √2

So each diagonal is √2 units long. But what is √2? It’s a new kind of number: an irrational number. That doesn’t mean that it’s illogical or against reason, but that it isn’t exactly equal to any ratio of integers like 3/2 or 17/12. When represented as decimals, the digits of all integer ratios either end or fall, sooner or later, into an endlessly repeating pattern:

3/2 = 1.5

17/12 = 1.416,666,666,666,666…

577/408 = 1.414,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,…

But when √2 is represented as a decimal, its digits go on for ever without any such pattern:

√2 = 1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,462,107…

The intellectual rite of passage comes when you understand why √2 is irrational and behaves like that:

Proof of the irrationality of √2

1. Suppose that there is some ratio, a/b, such that

2. a and b have no factors in common and

3. a^2/b^2 = 2.

4. It follows that a^2 = 2b^2.

5. Therefore a is even and there is some number, c, such that 2c = a.

6. Substituting c in #4, we derive (2c)^2 = 4c^2 = 2b^2.

7. Therefore 2c^2 = b^2 and b is also even.

8. But #7 contradicts #2 and the supposition that a and b have no factors in common.

9. Therefore, by reductio ad absurdum, there is no ratio, a/b, such that a^2/b^2 = 2. Q.E.D.

Given that subtle proof, you might think the digits of an irrational number like √2 would be difficult to calculate. In fact, they’re easy. And one method is so easy that it’s often re-discovered by recreational mathematicians. Suppose that a is an estimate for √2 but it’s too high. Clearly, if 2/a = b, then b will be too low. To get a better estimate, you simply split the difference: a = (a + b) / 2. Then do it again and again:

a = (2/a + a) / 2

If you first set a = 1, the estimates improve like this:

(2/1 + 1) / 2 = 3/2
2 – (3/2)^2 = -0.25
(2/(3/2) + 3/2) / 2 = 17/12
2 – (17/12)^2 = -0.00694…
(2/(17/12) + 17/12) / 2 = 577/408
2 – (577/408)^2 = -0.000006007…
(2/(577/408) + 577/408) / 2 = 665857/470832
2 – (665857/470832)^2 = -0.00000000000451…

In fact, the estimate doubles in accuracy (or better) at each stage (the first digit to differ is underlined):

1.5… = 3/2 (matching digits = 1)
1.4… = √2

1.416… = 17/12 (m=3)
1.414… = √2

1.414,215… = 577/408 (m=6)
1.414,213… = √2

1.414,213,562,374… = 665857/470832 (m=12)
1.414,213,562,373… = √2

1.414,213,562,373,095,048,801,689… = 886731088897/627013566048 (m=24)
1.414,213,562,373,095,048,801,688… = √2

1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,377… (m=48)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376… = √2

1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,6… (m=97)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,5… = √2

1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,8… (m=196)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,5… = √2

1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,43… (m=392)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,40… = √2

1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,408,498,847,160,386,899,970,699,004,815,030,544,027,790,316,454,247,823,068,49
2,936,918,621,580,578,463,111,596,668,713,013,015,618,568,987,237,235,288,509,264,861,249,497,715,42
1,833,420,428,568,606,014,682,472,077,143,585,487,415,565,706,967,765,372,022,648,544,701,585,880,16
2,075,847,492,265,722,600,208,558,446,652,145,839,889,394,437,092,659,180,031,138,824,646,815,708,26
3,010,059,485,870,400,318,648,034,219,489,727,829,064,104,507,263,688,131,373,985,525,611,732,204,02
4,509,122,770,022,694,112,757,362,728,049,574… (m=783)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,408,498,847,160,386,899,970,699,004,815,030,544,027,790,316,454,247,823,068,49
2,936,918,621,580,578,463,111,596,668,713,013,015,618,568,987,237,235,288,509,264,861,249,497,715,42
1,833,420,428,568,606,014,682,472,077,143,585,487,415,565,706,967,765,372,022,648,544,701,585,880,16
2,075,847,492,265,722,600,208,558,446,652,145,839,889,394,437,092,659,180,031,138,824,646,815,708,26
3,010,059,485,870,400,318,648,034,219,489,727,829,064,104,507,263,688,131,373,985,525,611,732,204,02
4,509,122,770,022,694,112,757,362,728,049,573… = √2

Squooh You

Here’s an interesting little puzzle:

Winnie-the-Pooh and Piglet set out to visit one another. They leave their houses at the same time and walk along the same road. But Piglet is absorbed in counting the birds overhead, and Winnie-the-Pooh is composing a new “hum,” so they pass one another without noticing. One minute after the meeting, Winnie-the-Pooh is at Piglet’s house, and 4 minutes after the meeting Piglet is at Winnie-the-Pooh’s. How long has each of them walked? — “A puzzle by S. Sefibekov” viâ Futility Closet

If you’re good at maths, you should see the answer in an intuitive instant. I’m not good at maths, so it took me much longer, because I didn’t understand what was going on. But I can explain the answer like this. Pooh is obviously walking faster than Piglet. Therefore, Pooh and Piglet can’t have met after one minute, because that would mean Pooh takes one minute to walk the distance walked by Piglet in one minute.

So let’s suppose Pooh and Piglet met after two minutes. If Pooh takes one minute to walk the distance walked by Piglet in two minutes, then Pooh is walking twice as fast as Piglet. Does that work? Yes, because Piglet walks Pooh’s distance in four minutes, which is twice as long as Pooh took. Therefore Piglet is walking twice as slowly as Pooh. It’s symmetrical and we can conclude that they did indeed meet after two minutes. Pooh then walks another minute, for three minutes in total, and Piglet walks another four minutes, for six minutes in total.

But guessing is not a good way to find the answer to the puzzle. Let’s try to reason it through properly. Suppose that Pooh and Piglet meet after one unit of time, during which Piglet has walked one unit of distance and Pooh has walked x units of distance, where x > 1. In other words, Pooh is walking x times faster than Piglet. The distances they walk before meeting are therefore in the ratio:

1 : x

Next, note that Pooh will cover the distance Piglet has already walked in 1 unit / x = 1 minute, while Piglet covers the distance Pooh has already walked in x / 1 = 4 minutes. The times they take are therefore in the ratio:

1 / x : x → 1 : x^2 → 1 : 4

And if 1 : x^2 is 1 : 4 (the ratio of the minutes they walk after meeting), then 1 : x (the ratio of the distances they walk before meeting) = 1 : √(x^2) = 1 : √4 = 1 : 2. Pooh is therefore walking 2x faster than Piglet and Piglet is walking 2x slower than Pooh. If Pooh covers Piglet’s distance in 1 minute, Piglet must have taken 2 minutes to walk that distance. And if Piglet covers Pooh’s distance in 4 minutes, Pooh must have taken 2 minutes to walk that distance.

Therefore, when they meet, each of them has been walking for 2 minutes. Pooh therefore walks 2 + 1 = 3 minutes in total and Piglet walks 2 + 4 = 6 minutes in total.

The result can be generalized for all relative speeds. Suppose that Pooh and Piglet meet after m1 minutes and that Pooh then takes m2 minutes to walk the distance Piglet walked in m1 minutes, while Piglet takes m3 minutes to walk the distance Pooh walked in m1 minutes. The time they walk before they meet, m1 minutes, is therefore supplied by this simple equation:

m1 = √(m3 / m2)

And you can call √(m3 / m2), the square root of m3 / m2, the squooh function:

m1 = √(m3 / m2) = squooh(m2,m3)

Now suppose the distance between Pooh’s and Piglet’s houses houses is 12 units of distance and that Piglet always walks at 1 unit a minute. If Pooh walks at the same speed as Piglet, i.e. 1 unit a minute, then:

After they meet, Pooh walks 6 more min = m2, Piglet walks 6 more min = m3.

How long do they walk before they meet?

m1 = m3 / m2 = 1, √1 * 6 = 6

They meet after 6 min.

Now suppose that after they meet, Pooh walks 2 more min, Piglet walks 8 more min.

Therefore, m3 / m2 = 4, √4 * 2 = 2 * 2 = 4 = m1

Therefore they walk for 4 min before they meet and Pooh walks 2x faster than Piglet.

After they meet, Pooh walks 1 more min, Piglet walks 9 more min (m3 / m2 = 9, √9 * 1 = 3)

Therefore they walk for 3 min before they meet and Pooh walks 3x faster than Piglet.

After they meet, Pooh walks 0·6 more min, Piglet walks 9·6 more min (m3 / m2 = 16, √16 * 0·6 = 4 * 0·6 = 2·4)

Therefore they walk for 2·4 min before they meet and Pooh walks 4x faster than Piglet:

After they meet, Pooh walks 0·4 more min, Piglet walks 10 more min (m3 / m2 = 25, √25 * 0·4 = 5 * 0·4 = 2)

Therefore they walk for 2 min before they meet and Pooh walks 5x faster than Piglet.

And so on.

Total Score

The number 23 is always (and trivially) equal to some running total of the digits of its roots in base 2. In other bases, that’s not always true (n.b. numbers inside square brackets represent single digits in that base):

√23 = 23^(1/2) = 100.1100101110111011100111010101110111000001000... in base 2
23 = digsum(100.110010111011101110011101010111011)
23^(1/2) = 11.21011101110011111122022101121121... in base 3
23 = digsum(11.2101110111001111112202)
23^(1/2) = 4.8832850[10]89028... in base 11
23 = digsum(4.883)
23^(1/2) = 4.[14]5[15]53[14]0[12]0[14]5[13]... in base 18
23 = digsum(4.[14]5)
23^(1/2) = 4.[19]29[13][19]4[11][23][19][11][20]... in base 24
23 = digsum(4.[19])
23^(1/2) = 4.[19][22]9[21][17]5[12][10]456... in base 25
23 = digsum(4.[19])

23^(1/3) = 10.11011000000001111010101010011000101000110000001100000010010000101011... in base 2
23 = digsum(10.1101100000000111101010101001100010100011000000110000001001)
23^(1/3) = 2.21121001121111121022212100220... in base 3
23 = digsum(2.2112100112111112102)
23^(1/3) = 2.312000132222212022030003... in base 4
23 = digsum(2.31200013222221)
23^(1/3) = 2.6600365246121403... in base 8
23 = digsum(2.660036)
23^(1/3) = 2.753154453877080... in base 9
23 = digsum(2.75315)
23^(1/3) = 2.93120691571[10]001[10]... in base 11
23 = digsum(2.931206)
23^(1/3) = 2.[12]9[13]0[11]74[11]61[14]2... in base 15
23 = digsum(2.[12]9)
23^(1/3) = 2.[13]807[10][10]98[10]303... in base 16
23 = digsum(2.[13]8)
23^(1/3) = 2.[21]2[10][10][13][11][21][23][15][24][21]... in base 25
23 = digsum(2.[21])
23^(1/3) = 2.[21][24][11][20][24][22][23][25]0[11][11]... in base 26
23 = digsum(2.[21])

23^(1/4) = 10.0011000010011111110100101010011000001001011110001110101... in base 2
23 = digsum(10.001100001001111111010010101001100000100101111)
23^(1/4) = 2.1411772251404570... in base 8
23 = digsum(2.141177)
23^(1/4) = 2.1634161832077814... in base 9
23 = digsum(2.163416)
23^(1/4) = 2.33[15]2[14][13]967[10]6[12]5... in base 17
23 = digsum(2.33[15])
23^(1/4) = 2.6[15][19][11][31][17][10][18][21]30[27]... in base 34
23 = digsum(2.6[15])
23^(1/4) = 2.[12]9[63][18][41][32][37][56][58][60]1[17]... in base 64
23 = digsum(2.[12]9)
23^(1/4) = 2.[21]9[26]6[54][21][20]3[64][86][110]... in base 111
23 = digsum(2.[21])
23^(1/4) = 2.[21][30][66][22][73][19]3[15][51][24]8... in base 112
23 = digsum(2.[21])
23^(1/4) = 2.[21][52][36][111][32][104][66][40][95][33]5... in base 113
23 = digsum(2.[21])
23^(1/4) = 2.[21][74][50][62][27]19[100][70][48][89]... in base 114
23 = digsum(2.[21])
23^(1/4) = 2.[21][96][108]2[101][62][43][18][71][113][37]... in base 115
23 = digsum(2.[21])

23^(1/5) = 1.110111110100011010011101000111111011111011000... in base 2
23 = digsum(1.11011111010001101001110100011111101)
23^(1/5) = 1.313310122131013323323010... in base 4
23 = digsum(1.31331012213101)
23^(1/5) = 1.[10]5714140[10][11][11]61... in base 12
23 = digsum(1.[10]57)
23^(1/5) = 1.[11]45210[12]3974[12]0[11]... in base 13
23 = digsum(1.[11]452)
23^(1/5) = 1.[22][17][15]788[12][20][10][16]5... in base 26
23 = digsum(1.[22])

And in base 10:

23^(1/7) = 1.565065607960239...
23 = digsum(1.56506)

23^(1/11) = 1.32982177397055...
23 = digsum(1.3298)

23^(1/25) = 1.133624213096260543...
23 = digsum(1.13362421)

23^(1/43) = 1.075642836327515...
23 = digsum(1.07564)

23^(1/51) = 1.0634095245502272...
23 = digsum(1.063409)

23^(1/59) = 1.054581462032154...
23 = digsum(1.05458)

23^(1/74) = 1.043282031364111825...
23 = digsum(1.04328203)

23^(1/78) = 1.041017545329593513...
23 = digsum(1.04101754)

23^(1/81) = 1.039468791371841...
23 = digsum(1.03946)

23^(1/85) = 1.037576979258809...
23 = digsum(1.03757)

23^(1/86) = 1.0371320245405187874...
23 = digsum(1.037132024)

23^(1/101) = 1.031531403111493041428...
23 = digsum(1.03153140311)

Self-Raising Power

The square root of 2 is the number that, raised to the power of 2, equals 2. That is, if r^2 = r * r = 2, then r = √2. The cube root of 2 is the number that, raised to the power of 3, equals 2. That is, if r^3 = r * r * r = 2, then r = [3]√2.

But what do you call the number that, raised to the power of itself, equals 2? I suggest “the auto-root of 2”. Here, if r^r = 2, then r = [r]√2. I don’t know a quick way to calculate the auto-root, but you can adapt a well-known algorithm for approximating the square root of a number. The square-root algorithm looks like this:

n = 2
r = 1
for c = 1 to 20
    r = (r + n/r) / 2
next c
print r

r = 1.414213562…

Note the fourth line of the algorithm: r = (r + n/r) / 2. When r is an over-estimate of √2, then 2/r will be an under-estimate (and vice versa). (r + 2/r) / 2 splits the difference and refines the estimate. Using the lines above as the model, the auto-root algorithm looks like this:

n = 2
r = 1
for c = 1 to 20
    r = (r + [r]√n) / 2[*]
next c
print r

r = 1.559610469…


*This is equivalent to r = (r + n^(1/r)) / 2

Here are the first 100 digits of [r]√2 = r in base 10:

1, 5, 5, 9, 6, 1, 0, 4, 6, 9, 4, 6, 2, 3, 6, 9, 3, 4, 9, 9, 7, 0, 3, 8, 8, 7, 6, 8, 7, 6, 5, 0, 0, 2, 9, 9, 3, 2, 8, 4, 8, 8, 3, 5, 1, 1, 8, 4, 3, 0, 9, 1, 4, 2, 4, 7, 1, 9, 5, 9, 4, 5, 6, 9, 4, 1, 3, 9, 7, 3, 0, 3, 4, 5, 4, 9, 5, 9, 0, 5, 8, 7, 1, 0, 5, 4, 1, 3, 4, 4, 4, 6, 9, 1, 2, 8, 3, 9, 7, 3…

And here is [r]n = r for n = 2..20:

autopower(2) = 1.5596104694623693499703887…
autopower(3) = 1.8254550229248300400414692…
autopower(4) = 2
autopower(5) = 2.1293724827601566963803119…
autopower(6) = 2.2318286244090093673920215…
autopower(7) = 2.3164549587856123013255030…
autopower(8) = 2.3884234844993385564187215…
autopower(9) = 2.4509539280155796306228059…
autopower(10) = 2.5061841455887692562929409…
autopower(11) = 2.5556046121008206152514542…
autopower(12) = 2.6002950000539155877172082…
autopower(13) = 2.6410619164843958084118390…
autopower(14) = 2.6785234858912995813011990…
autopower(15) = 2.7131636040042392095764012…
autopower(16) = 2.7453680235674634847098492…
autopower(17) = 2.7754491049442334313328329…
autopower(18) = 2.8036632456580215496843618…
autopower(19) = 2.8302234384970308956026277…
autopower(20) = 2.8553085030012414128332189…

I assume that the auto-root is always an irrational number, except when n is a perfect power of suitable form, i.e. n = p^p for some integer p. For example, autoroot(4) = 2, because 2^2 = 4, autoroot(27) = 3, because 3^3 = 27, and so on.

And here is the graph of autoroot(n) for n = 2..10000:
autoroot

The World as Worm

In “Hymn to Herm”, I wrote about a religion based on √2, or the square root of two, the number that, multiplied by itself, equals 2. In the religion, neophytes learn the mystery and majesty of this momentous number when they try to calculate its exact value. The calculation involves adding and subtracting fractions based on powers of two. The first step is this: 1 x 1 = 1. So that’s too small. Add 1/2^1 = ½ and re-multiply: 1½ x 1½ = 2¼. Too big. So subtract 1/2^2 = ¼, and re-multiply. 1¼ x 1¼ = 1+9/16. Too small. Add 1/8 and re-multiply. 1+3/8 x 1+3/8 = 1+57/64. Too small again. Add 1/16 and re-multiply. And so on.

In effect, what the neophytes are doing is calculate the digits of √2 in binary, or base two. When the multiplication is too small, put a 1; when it’s too big, put a 0. Like this:

1 x 1 = 1 < 2, so √2 ≈ 1·…
1½ x 1½ = 2¼ > 2, so √2 ≈ 1·0…
1¼ x 1¼ = 1+9/16 < 2, so √2 ≈ 1·01…
(1+3/8) x (1+3/8) = 1+57/64 < 2, so √2 ≈ 1·011…
(1+7/16) x (1+7/16) = 2+17/256 > 2, so √2 ≈ 1·0110…
(1+13/32) x (1+13/32) = 1+1001/1024 < 2, so √2 ≈ 1.01101…
(1+27/64) x (1+27/64) = 2+89/4096 > 2, so √2 ≈ 1.011010…
(1+53/128) x (1+53/128) = 1+16377/16384 < 2, so √2 ≈ 1·0110101…
(1+107/256) x (1+107/256) = 2+697/65536 > 2, so √2 ≈ 1·01101010…
(1+213/512) x (1+213/512) = 2+1337/262144 > 2, so √2 ≈ 1·011010100…
(1+425/1024) x (1+425/1024) = 2+2449/1048576 > 2, so √2 ≈ 1·0110101000…
(1+849/2048) x (1+849/2048) = 2+4001/4194304 > 2, so √2 ≈ 1·01101010000…
(1+1697/4096) x (1+1697/4096) = 2+4417/16777216 > 2, so √2 ≈ 1·011010100000…
(1+3393/8192) x (1+3393/8192) = 1+67103361/67108864 < 2, so √2 ≈ 1·0110101000001…

Mathematically naïve neophytes, seeing the process miss 2 by smaller and smaller amounts on either side, might imagine that eventually the exact root will appear and the calculations end. But they would be wrong. They could work a year or a million years: they would never calculate the exact square root of two. There is no ratio of whole numbers, a/b, such that a^2/b^2 = 2. In other words, √2 is an irrational number, or number that can’t be represented as a ratio of integers (please see appendix for the proof).

This discovery, made by Greek mathematicians more than two millennia ago, is both mind-boggling and world-shattering. In fact, it’s mind-boggling in part because it’s world-shattering. √2 shatters the world because the world is too small to contain it: in the words of the Cult of Infinite Hermaphrodites, “Were the sky all parchment, the seas all ink, and gulls all plucked for quills”, the square root of two could not be recorded in full. This is far more certain than tomorrow’s sunrise, because predicting tomorrow’s sunrise depends on fallible scientific reasoning from incomplete knowledge. Proving the irrationality of √2 depends on infallible mathematical reasoning.

At least, it’s as close to infallible as human beings can get. But that’s another part of what is mind-boggling about √2. A finite, feeble human being, with a speck of soon-decaying brain, can prove the existence of things larger than the universe. A few binary digits of √2 are shown above. Here are a few more:

1·
0110101000001001111001100110011111110011101111001100100100001000
1011001011111011000100110110011011101010100101010111110100111110
0011101011011110110000010111010100010010011101110101000010011001
1101101000101111010110010000101100000110011001110011001000101010
1001010111111001000001100000100001110101011100010100010110000111
0101000101100011111111001101111110111001000001111011011001110010 
0001111011101001010100001011110010000111001110001111011010010100 
1111000000001001000011100110110001111011111101000100111011010001 
1010010001000000010111010000111010000101010111100011111010011100 
1010011000001011001110001100000000100011011110000110011011110111 
1001010101100011011110010010001000101101000100001000101100010100 
1000110000010101011110001110010001011110111110001001110001100111 
1000110110101011010100010100011100010111011011111101001110111001 
1001011001010100110001101000011001100011111001111001000010011011 
1110101001011110001001000001111100000110110111001011000001011101 
1101010101001001010000010001001100100000100000011001010010010101 
0000001001110010100101010110110110110001111110100001110111111011 
1110100110100111010000000101100111010111100100100111110000011000 
1000010011001001101101010111100110101010010100010110110010100011 
0111000110011110011010000011011011011111000001000110110110001110 
0000001000001001101110000000001111111100011001000110101001011110 
0110011001010100101111010011111011110111101101000011110101111111 
1110110101000011011111000111111110010100010001000010011000001111 
1011110101000000110001001000001111101111010101010000001110000101 
1000001111111001011110111011110101000101111011111011100001100110 
0011000100000111000101000101110101011111111010111110011101100101 
1010010010011110100101001110110001111111010110010111000100000101 
1111101111111100001011100001111110100111011000111110111100000001 
1111001101011001100111001000001011110010111111100101000000001011 
1000010010001100111100001011110100100101001010101110000001000110 
1011111110011111000111101111011110010100011111010100011001110110 
1001101011111000110000010100101111001100011001111100011111000010 
1001000010111110011101101001001010011011000001010111100011000001 
0000101101011000010011111011010010000111110010010010010011110101 
1011011100011111100000101101110011010010100100000011011000001001 
1101111011101000100100010010100110000011110101001110101010101101 
0000111011101010001100100001111101110100100010011111010001101010 
0111111010010000001100001011111000100000111110110111011010010100 
1110111110110101100011001001100110000100110011011101011100001010 
0001110110101001000001000101110000111101000100110011101000000110 
1000010000100011110101101110001110000011000000111101100100000001 
1011101010011101101000110100011101100110100001000111100101101100 
0101110011010101100101110010110111000000111111110011010101000000 
1100001101000001001010010100001011010110010000000110000100000001 
1110111101101111110001101101111010010001000101001010001010110100 
1111001001001000110001101000100111000110000000001011101101000000 
1010100010110101011010110000010000011111110101011101111001101110 
0000110111010000110001100110110101001000001100011111111001111111 
1111111101010111010101111110010001110001000010011000000011001101 
1011110101011100001001101000010010000101110110100101111010010001 
1011001111100010111100100000010110110111001001110010010110111001 
0111000111010110000010100001111110001000100011110000100010100000 
1010011011100001000000001100110011101101110000101100111001011011 
1101100110001010111011100111000111100100001011100010011010001101 
0011011110100110000001110010111100100010000000100011010001100001 
0011111111111100001000100100010100110100001110011110101010010111 
1010100110011001101101101100100111100011110011100111000111111001 
0100110101100000100100101010110011100001001000001010101110001110 
0101010100001110000011010101010100010001011010001000011000110001 
0111011110001100111101100000001101010000110100000010111111101000 
0101111100101001111011001000101111100101110001110010101110000000 
0111101011110101011101110001101110000010010110110011000010100000 
1110011110000011011101101010100100011100000010001100011010100111 
1111000011111000111100110010001110110011011000101000000111010010 
0010010101101000100111000000101101011010100000100000010001111101 
1011100110001001111101100011101010001010011001001110100001010001 
1001101111000000110100001100000111100010001000101000000001001000 
0100110110010100111101001111100110111011001111010100101100110001 
1101010010001001110101110101001000110001101101011100011000110011 
1100010010000000110010010110101111100101010010011011111101011101 
1001011001100111100010110100110100101100010011011101101010000110 
0111101111011000111001001000000000101001111111101010100011001000 
0001011100110101011001111100001010111010001111010011110011101001 
1101111111100000101111010001101101001101110101110111000100010111 
1000000001010111101101101001010110110111111010101111000110110000 
0101110000100010110010001101010111111110111010111101000001110111 
1111111011001001011011011011100011110111011110001111110000011100 
0010101110111011110011100001101101001001111010111111010110101111 
0100010001100000100010000010100101011000101011011101000000011100 
1010011111110001101101101011110000001011011111101100000110111100 
0110111000001010011011101101101111000110011111111000010110110010 
0111010011100000100001100001101100111010000100110111010101110001 
1011000101100101010010011000011100111101001000010001110001101010 
1010111001101001110110000000000111100101011110010100010001011011 
1100011000001010001111100000101001001111110110001001011010001110 
1011011110010100101111011100011100000010110101101001010001011101 
1010100101001011000001001010010001000000110010101011110010010100 
0011100001111100001111010010011011111101000011110011101111101000 
1010111101100011000001011010010100111010000101110111001010001000 
1010110010100001001111111011010000000110110010011000001010010001 
0101110110000011101110100000110100110101010110001101100000011101 
1101000100010101100111101001011001000011111010101010001001111110 
1011011101110101011110100010000001010010100101110101101101101111 
0100101010001000100111100011110100001001001010111011000111000110 
1000010101001000000011011100001011101001100110010100011110110011 
0111001011011110110110100000010111100010000110010010111110010101 
1011000110111001001001100101000100101011010000000100110000110011
0001100000011101011...

The distribution of 1’s and 0’s seems effectively random, as though the God of Mathematics were endlessly tossing a coin, putting 1 for heads, 0 for tails. Yet √2 is the opposite of a random number. Change a single digit anywhere and it ceases to be √2. Every 1 and every 0 is rigidly determined by “unalterable law”. So are the position and magnitude of the digits of √2 in every other base. Here, for example, is √2 in base 4:

1·
112220021321212133303233030210020230233230103121232222111133
103320322313230011311010213131100212131220233112100230012121
303020222211133210012002013111...

Another word for base-4 is DNA: genes are in fact written in a base-4 code based on the chemicals guanine, adenine, thymine and cytosine, or G, A, T, C for short. If the digits of √2 are truly random, in the statistical sense, then all genomes, actual and potential, occur somewhere along its length: yours, mine, the Emperor Heliogabalus’s, Bilbo Baggins’, the sabre-toothed tiger’s, the dodo’s, and so on. But almost all the “DNA” of √2 in base-4 will be meaningless: although √2 is the opposite of random, it is effectively a typing chimpanzee. Or a typing worm – a type-worm. √2 is like an endless worm that types out its own segments on a typewriter with two keys (for binary numbers) or four keys (for quaternary numbers) or ten keys (for decimal numbers) and so on.

But √2 doesn’t just encode the genomes of individual people, animals and plants: it encodes everything they do throughout their lives. In fact, it encodes the entire universe. And perhaps the universe is √2 or some number like it. Perhaps, in some sense, everything exists within the digits of an irrational number, or a sufficiently large rational number. If so, then √2 has become aware of itself through human beings: the World as Worm has bitten its own tail.

Appendix: Proof of the irrationality of √2

1. Suppose that there is some ratio, a/b, such that

2. a and b have no factors in common and

3. a^2/b^2 = 2.

4. It follows that a^2 = 2b^2.

5. Therefore a is even and there is some number, c, such that 2c = a.

6. Substituting c in #4, we derive (2c)^2 = 4c^2 = 2b^2.

7. Therefore 2c^2 = b^2 and b is also even.

8. But #7 contradicts #2 and the supposition that a and b have no factors in common.

9. Therefore, by reductio ad absurdum, there is no ratio, a/b, such that a^2/b^2 = 2. Q.E.D.