how big is graham's number and can you count that far?

Answer 1

Let's assume that you can count pretty quickly. Say 10 numbers per second. Let's assume that you start counting when you are born and count at a rate of 10 numbers per second until you die 100 years later. that means you have counted 10*60*60*24*365.25*100=3,155,760,000 numbers. That is pretty good. Let's call this number N. And note that 3^3=27, 3^(3^3)=3^27=7625597484987>2000*N. So, you could count to 3^(3^3) if you had 2000 lifetimes to do so.

Now let's look at graham's number:

3^3=27
3^^3=3^(3^3)
3^^^3=3^^(3^^3), where 3^^n=3^(n^n)
3^^^^3=3^^^(3^^^3), where 3^^^n=3^^(n^^n)

3^^3>2000N
3^^^3>>N (this means a hell of a lot bigger)
3^^^^3>>>>>>>>>>>>>N

but 3^^^^3=g_1

graham's number is g_64 where g_n is defined recursively by
g_1=3^^^^3 and g_n=3^^^. . .^^^3 where there are g_(n-1) ^ between the 3s.

So, if you had all the computers in the world counting from the Big Bang until now, you would probably have made it to about g_3 or maybe g_4 (if scientists are very wrong about how old the universe is).

Answer 2

The World Champion largest number, listed in the latest Guinness Book of Records, is an upper bound, derived by R. L. Graham, from a problem in a part of combinatorics called Ramsey theory.
Graham's number cannot be expressed using the conventional notation of powers, and powers of powers. If all the material in the universe were turned into pen and ink it would not be enough to write the number down. Consequently, this special notation, devised by Donald Knuth, is necessary.

3^3 means '3 cubed', as it often does in computer printouts.

3^^3 means 3^(3^3), or 3^27, which is already quite large: 3^27 = 7,625,597,484,987, but is still easily written, especially as a tower of 3 numbers: 333.

3^^^3 = 3^^(3^^3), however, is 3^^7,625,597,484,987 = 3^(7,625,597,484,987^7,625,597,484,987), which makes a tower of exponents 7,625,597,484,987 layers high.

3^^^^3 = 3^^^(3^^^3), of course. Even the tower of exponents is now unimaginably large in our usual notation, but Graham's number only starts here.

Consider the number 3^^^...^^^3 in which there are 3^^^^3 arrows. A largish number!

Next construct the number 3^^^...^^^3 where the number of arrows is the previous 3^^^...^^^3 number.

An incredible, ungraspable number! Yet we are only two steps away from the original ginormous 3^^^^3. Now continue this process, making the number of arrows in 3^^^...^^^3 equal to the number at the previous step, until you are 63 steps, yes, sixty-three, steps from 3^^^^3. That is Graham's number.

There is a twist in the tail of this true fairy story. Remember that Graham's number is an upper bound, just like Skewes' number. What is likely to be the actual answer to Graham's problem? Gardner quotes the opinions of the experts in Ramsey theory, who suspect that the answer is: 6

Answer 3

Graham's number is so big that it has more digits than the universe has particles. There is absolutely no way anyone could count that far since there aren't enough particles in the universe to simply represent the number.

Answer 4

you've used this particular representation to symbolize an exceedingly tremendous huge type yet is fairly all you've performed is shift the region from utilising undemanding index notation to at least one utilising the notation which has for use for pretty tremendous numbers. although you come back up with you may continuously flow to the subsequent degree, it quite is the equivalent of including one. Graham's huge type continues to be the optimum that has been utilized in a mathematical situation. i imagine it replaced into an top certain in a combinatorics situation. curiously although, a chap in the Horizon software about infinity thinks there's a optimum huge type and including a million takes you again to 0. i visit't faux to appreciate the mathematics in contact right here!

Answer 5

it is too far to count.
the last 10 digits are 2,464,195,387.
check out this page though:
http://en.wikipedia.org/wiki/Graham's_number

Answer 6

http://www-users.cs.york.ac.uk/susan/cyc/g/graham.htm


no comment

Answer 7

15 and yes, i think

Answer 8

i dont get it

Answer 9

www.users.cs.york.ac.uk/~susan/cyc/g/graham.htm

Answer 10

what...