# Googolplex

A googolplex is the number 10googol, i.e. 10(10100). The reciprocal of the googolplex is called googolminex.[1]

## History

In 1938, Edward Kasner's nine year old nephew, Milton Sirotta, coined the term googol, which is 10100, then proposed the further term googolplex to be "one, followed by writing zeroes until you get tired". Kasner decided to adopt a more formal definition "because different people get tired at different times and it would never do to have Carnera be a better mathematician than Dr. Einstein, simply because he had more endurance and could write for longer".[2] It thus became standardized to 10(10100).

## Size

In the PBS science program Cosmos: A Personal Voyage, Episode 9: "The Lives of the Stars", astronomer and television personality Carl Sagan estimated that writing a googolplex in standard form (i.e., "10,000,000,000...") would be physically impossible, since doing so would require more space than is available in the known universe.

A typical book can be printed with 106 zeros (around 400 pages with 50 lines per page and 50 zeros per line).[3] Therefore it requires 1094 such books to print all zeros of googolplex. If each book has a size of 210 mm × 297 mm × 13 mm, the total volume of all the books is 8.1×1090 m3, which is many orders of magnitude larger than the visible universe, which has a volume of 'only' 4×1080 m3.[4]

With only about 1080 [5] elementary particles in the observable universe, even if only one elementary particle is used to represent each digit, there are not enough particles to represent a googolplex.

Printing digits of a googolplex in one long line would be unreasonable, even in one-point font (0.353 mm per digit). Writing a googolplex in that font would take about 3.53×1097 meters. The observable universe is estimated to be 8.80×1026 metres, or 93 billion light-years, in diameter;[6] the required line to write the necessary zeroes is therefore 4.0×1070 times as long as the observable universe.

Writing the number takes too long: if a person can write two digits per second, then writing a googolplex would take around about 1.51×1092 years, which is about 1.1×1082 times the age of the universe.[7]

A Planck space has a volume of a Planck length cubed, which is the smallest measurable volume, at approximately 4.22×10−105 m3 = 4.22×10−99 cm3.[8] Therefore 2.4 cm3 contain about a googol Planck spaces. Only about 4×1080 cubic metres exist in the observable universe,[4] giving about 9.5×10184 Planck spaces in the entire observable universe; therefore, the number googolplex is about $10^{(googol-184)}$ times larger than even the number of the smallest measurable spaces in the observable universe. However, if one Planck space is used to represent each digit of a googolplex, a mere 2.4 cm3 of volume gives us enough Planck spaces for that task.

## Scale

### In pure mathematics

In pure mathematics, there are several notational methods for representing large numbers by which the magnitude of a googolplex could be represented, such as tetration, Knuth's up-arrow notation, Steinhaus-Moser notation, or Conway chained arrow notation.

### In the physical universe

One googol is presumed to be greater than the number of hydrogen atoms in the observable universe, which has been variously estimated to be between 1079 and 1081.[9] A googol is also greater than the number of Planck times elapsed since the Big Bang, which is estimated at about 8×1060.[10] Thus in the physical world it is difficult to give examples of numbers that compare to the vastly greater googolplex. In analyzing quantum states and black holes, physicist Don Page writes that "determining experimentally whether or not information is lost down black holes of solar mass ... would require more than 1076.96 measurements to give a rough determination of the final density matrix after a black hole evaporates".[11] The end of the Universe via Big Freeze without proton decay is expected to be around 10(1075) years into the future, which is still short of a googolplex.

In a separate article, Page shows that the number of states in a black hole with a mass roughly equivalent to the Andromeda Galaxy is in the range of a googolplex.[7]

If the entire volume of the observable universe (assumed [4] to be 4×1080 m3) were packed solid with fine dust particles about 1.5 micrometres in size, then the number of different ways of ordering these particles (that is, assigning the number 1 to one particle, then the number 2 to another particle, and so on until all particles are numbered) would be approximately one googolplex.[12]

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## References

12. ^ We assume that each particle has a size of $1.5 {\rm\ {\mu}m}$, this means (if we assume that it is a cube) its volume is $V_{\rm{particle}} = (1.5 {\rm\ {\mu}m})^3 = 3.38 \times 10^{-18}{\rm\ m}^{3}$. If the observable universe, which has a volume of $V_{\rm{universe}} = 4 \times 10^{80}{\rm\ m}^{3}$,[4] is filled with such particles, the total number $n$ of such particles is $n=\frac{V_{\rm{universe}}}{V_{\rm{particle}}}=\frac{4 \times 10^{80}{\rm\ m}^{3}}{3.38 \times 10^{-18}{\rm\ m}^{3}}=1.18\times 10^{98}$. The number of different ways in which the particles can be ordered is equal to the factorial $n!$ of the total number of particles. It can be calculated using Stirling's approximation as: $n! \simeq \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n$$= 10^{\left ( \log_{10}\left ( \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n \right ) \right )}$$= 10^{\left ( \frac{1}{2}\times\log_{10}\left ( {2\pi n} \right ) + n\times\log_{10}\left ( \frac{n}{e} \right ) \right )}$$\simeq 10^{\left ( n\times\log_{10}\left ( \frac{n}{e} \right ) \right )}$$= 10^{\left ( 1.18\times 10^{98}\times\log_{10}\left ( \frac{1.18\times 10^{98}}{2.718...} \right ) \right )}$$= 10^{\left ( 1.18\times 10^{98}\times97.6 \right )} = 10^{\left ( 1.15\times10^{100} \right )}$$\simeq 10^{\left ( 10^{100} \right )}$$= 1{\rm\ googolplex}$