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How Many Digits in Graham's Number?

Date: 11/11/2005 at 21:15:24
From: Alex
Subject: How many digits in Graham's number

I have heard that Graham's number is the largest number with 
mathematical use.  I have seen it expressed in arrow notation but 
that does not give me a sense of how large it is.  Is there a way to 
express the number of digits it contains?  It's hard to imagine a
large number without knowing the number of digits it has.



Date: 11/12/2005 at 10:32:38
From: Doctor Vogler
Subject: Re: How many digits in Graham's number

Hi Alex,

Thanks for writing to Dr. Math.  This number is so insanely large it's
unbelievable.  The number of digits it has is so insanely large it's
unbelievable.  Writing out the number of digits it has would take more
atoms that can be found on the planet, and that wouldn't even be
remotely close!

The number of digits that a number has is roughly the logarithm of the
number (precisely, it is the log to base ten, rounded down, plus one).
But when you're talking arrow notation, recall that

  log (x^y) = y * log x,

which means that

  log (x^x^x^x^x^x) = (x^x^x^x^x) * log x,

so you only lose one x.  That is,

  log(x^^n) = x^^(n-1) * log x.

But then x^^^2 = x^^x, so its log is still huge.  And then

  3^^3 = 3^3^3 = 7625597484987,

so

  3^^^3 = 3^^3^^3 = 3^^7625597484987,

which means to power up 3^3^3^3^3^....^3 until there are over seven
trillion threes stacked up.  This is already _way_ more than the
number of atoms in the universe.  If you take the log of this number,
then you get

  3^^7625597484986 * log 3,

which doesn't seem much smaller.  In fact, it's a lot smaller, but
it's still so huge that it's unbelievable, so you can't really tell
that it's a lot smaller.  But this is still not nearly Graham's
number.  This is just 3^^^3, and constructing Graham's number _begins_
with the number

  3^^^^3 = 3^^^3^^^3 = 3^^^(3^^7625597484987),

which means 3^^3^^3^^...^^3 with as many 3's in that stack as that
unbelievably huge number that I was just speaking of.  And this is a
whole lot bigger than the previous number, because if you only had
three of them, then you get that unbelievably huge number we already
discussed, and we have more of them than you can imagine.  And that's
just g_1.  Then g_2 is

  3^^^^^^...^^^^^^^^3,

where the number of arrows is g_1.  I can't even begin to describe
this, because just three or four arrows makes such a gigantic number
that it's unbelievable, so having g_1 arrows is...  Well, suffice it
to say that g_2 is a whole heck of a lot bigger than g_1 (in fact, the
number of digits in g_2 is a whole heck of a lot bigger than g_1),
which was already a whole heck of a lot bigger than anything you could
possibly imagine.  And Graham's number is g_64.  I hope that I have
conveyed the idea that both Graham's number and its log (the number of
digits) are so gigantic as to be beyond believing.  Nothing in the
universe could possibly be so large; only in mathematics can we even
speak of such numbers.  So you shouldn't expect to be able to imagine
such a large number, but you should recognize that it is so big as to
be beyond big.

I am reminded of a math teacher who said that some finite numbers are
so big that they seem bigger than infinity.  After all, infinity is
easy to imagine; you just count up to five or ten and then put three
dots.  But billions and trillions are hard to imagine.  A googol is
worse.  Graham's number is off the scale, by far.

See also

  Graham's Number
    http://mathworld.wolfram.com/GrahamsNumber.html 

  Arrow Notation
    http://mathworld.wolfram.com/ArrowNotation.html 

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Discrete Math
College Number Theory
Elementary Large Numbers

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