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Thought Experiment Take a piece of paper one thousandth of an inch thick Now fold it in half and then in half again Do this 50 times (I know that this.

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Presentation on theme: "Thought Experiment Take a piece of paper one thousandth of an inch thick Now fold it in half and then in half again Do this 50 times (I know that this."— Presentation transcript:

1 Thought Experiment Take a piece of paper one thousandth of an inch thick Now fold it in half and then in half again Do this 50 times (I know that this is not practicable) How thick do you think it will be ?

2 Each time we fold the paper, it doubles in thickness So after the first fold it is 2x as thick After the second fold it is 2x2 ie 4 times as thick After 50 folds it is 2x2x2.......... 2x2 times as thick That is 2 50 Divide by a thousand to get the thickness in inches Divide by 63,360 to get the thickness in miles And we get......

3 2 50 = 1125899906842624 Dividing by a thousand = 1125899906842 inches Dividing by 63360 = 17,769,885 miles

4 Carol recently gave a talk on the metal Bismuth. One of its interesting facts was it's half-life This is currently calculated at :- 1.9 x 10 19 years

5 If we assume that there are 64 generations between now and the start of the Christian Era Then you will have approximately 1.8 x 10 19 ancestors

6 There is a story about an Indian temple which contains a large room with three posts in it surrounded by 64 golden disks. Brahmin priests, acting out the command of an ancient prophecy, have been moving these disks, in accordance with the immutable rules of the Brahma, since that time. The puzzle is therefore also known as the Tower of Brahma puzzle. According to the legend, when the last move of the puzzle will be completed, the world will end

7 if the priests were able to move disks at a rate of one per second, using the smallest number of moves, it would take them 2 64 −1 seconds or roughly 585 billion years or 18,446,744,073,709,551,615 turns to finish ie 1.8 x 10 19 moves The world will certainly have ended !!!

8 Just a little larger are the 4.3 x 10 19 different starting positions for a Rubik's Cube Any ideas of the smallest number of moves required to solve the cube from any starting position ? (Known as God's Number)

9 Any cube can be solved in 22 moves at most, but there is strong evidence that only 20 moves are required. Hence God's Number is either 20 or 22

10 Consider this 200 digit number 914743972814745128948036774162014302835642105 034332853395613272769334542296093046464719250 945181147710162588965929074414263498975565041 455709602039255036791052451991423388060824942 54050610000000000000 You now have 70 seconds to extract its 13th root assuming that you wish to claim the world record

11 The answer is 2,407,899,893,032,210

12 Compare this to :- Age of the universe 13.798×10 9 years or 4.354×10 17 seconds Number of elementary particles in the observable universe between 10 80 and10 97

13 But these numbers are mere beginners Googol Googolplex Skewe's number Gijswijt's sequence Ramsey Numbers Graham's Number

14 Googol Originally named by Edward Kasner a Googol = 10 100 ie 1000000000000000000000000000000000 0000000000000000000000000000000000 0000000000000000000000000000000 (The search engine Google copied the named but spelled it incorrectly)

15 Any idea what's special about this number ? 2 57885161 -1 this is approximately 10 10000000

16 It's the largest prime number currently know It was discovered in 2013 and is some 17.5 million digits long It is also the 48th Mersenne Prime ie of the type 2 n -1 It's the largest known number that cannot be expressed in terms of smaller numbers

17 Googolplex There isn't enough room or time to write out this number If you can write 2 digits/second, you would need 1.5 x 10 92 years, which is some 10 82 times greater than the age of the universe 10 100 or 10 googol

18 Skewes' Number Gauss generated a formula to give the number of Primes up to any given number. It is known that it generally over-estimates the true number. Stanley Skewes showed that Gauss's formula would under- estimate when we got above the number that bears his name

19 This result, however depends upon the Riemann Hypothesis being true If the Riemann Hypothesis is ever shown to be false, then Skewes number increases to Which is 10 1010 34 10 1010 963

20 Gijswijt's sequence 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 1............. where does 4, 5, etc appear ?

21 The first time 3 appears is in the ninth position. 4 appears for the first time in the 221 st position. The first 5 occurs in position 10 100000000000000000000000 ie 10 10 23

22 The other numbers will eventually appear, although with no sense of urgency

23 Ramsey Numbers You are going to have a party. How many people do you need to invite to be sure that at least 3 people will be mutual aquaintances, or at least 3 people will be mutual strangers ?

24

25 However, if we need to know the minimum number of guests such that either 5 people are mutually aquainted or mutually strangers, there are some 2 903 or 6.7 x 10 271 cases to try out using brute computer force alone ! We do know that between 43 and 49 guests would be required

26 Graham's Number If we now consider Ramsey's numbers but venture into multi-dimensional space the numbers get very large. So large, that we need a whole new way to represent them

27 Once again, say we have some points, but now they are the corners of an n dimensional hypercube. They are still all connected by blue and red lines. For any 4 points, there are 6 lines connecting them. Can we find 4 points that all lie on one plane, and the 6 lines connecting them are all the same colour? Graham's Number gives an upper limit of the points needed for this to be certain

28 Knuth's up-arrow notation Invented by Donald Knuth in 1976 It uses a series of up arrows and looks like :- g 64 =3↑↑↑↑ g 63 times ↑↑↑↑3 If each digit could occupy a Planck length 10 -35 m, then you could get 10 35 x 10 35 x 10 35 digits ( ie 10 105 ) in a cubic meter, there would still not be enough space in the universe to write it down

29 We do know the last 500 digits of his number - the last digit is a 7 Also the lower limit is 13


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