Saturday, April 2, 2011

Graham's Number

Someone asked a question on the Physics Forums asking for an explanation of the largeness of this number, so doing some research this is what I came up with:


It is a number that is impossible to comprehend. Going by the articles on wikipedia covering Graham's number, Knuth's up-arrow notation, Conway's chained arrow notation, and tetration(power tower),

\[\begin{align*}G = g_{64} &= 3 \rightarrow 3 \rightarrow g_{63} \\ ... \\ g_{2} &= 3 \rightarrow 3 \rightarrow g_{1} \\ g_{1} &= 3 \rightarrow 3 \rightarrow 4 \end{align*}\]

\(g_{1}\) is already insanely huge number.

\[\begin{align*} g_{1} &= 3 \uparrow\uparrow\uparrow\uparrow 3 \\ &= 3 \uparrow\uparrow\uparrow 3 \uparrow\uparrow\uparrow 3 \\ &= 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow 3 \uparrow\uparrow 3) \\ &= 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow 3 \uparrow 3)) \\ &= 3 \uparrow\uparrow\uparrow (3 \uparrow\uparrow (3 \uparrow 27)) \end{align*}\]

\(3 \uparrow 27 = 7625597484987\), so \(3 \uparrow\uparrow (3 \uparrow 27)\) is 3 exponentiated by itself 7625597484987 times. Another way to write that is \(^{7625597484987}3\). That still doesn't come close to giving us the value of \(g_{1}\).

We still have to compute \(3 \uparrow\uparrow\uparrow (^{7625597484987}3)\), and all we will have accomplished is to have determined the number of \(\uparrow\) for \(g_{2}\). Considering how impossibly large \(g_{1}\) seems to be when it's Knuth up-arrow notation had a measly 4 \(\uparrow\), then imagine how \(g_{2}\) must dwarf \(g_{1}\) since it's Knuth up-arrow notation will have \(g_{1} \uparrow\). Then that continues on, each step from \(g_{2}\) to \(g_{64}\) being unbelievably larger than the previous step.

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