Very large number(s)
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# Very large number(s)

[From: ] [author: ] [Date: 11-05-22] [Hit: ]
such as, for example, Conway chained arrow notation.To quote from wikipedia, “Conway chained arrow notation, created by mathematician John Horton Conway,......
Ok, this is a question for all you math geniuses out there.

I've been doing some research on mathematics, particularly into the area of very large numbers. I stumbled across some interesting things, such as, for example, Conway chained arrow notation.

To quote from wikipedia, “Conway chained arrow notation, created by mathematician John Horton Conway, is a means of expressing certain extremely large numbers. It is simply a finite sequence of positive integers separated by rightward arrows, e.g. 2→3→4→5→6.
As with most combinatorial symbologies, the definition is recursive. In this case the notation eventually resolves to being the leftmost number raised to some (usually enormous) integer power.”

So my question here is, would this number: 1→6→1→8 be MUCH larger than, let's say for example, Graham's Number?

Thanks for taking the time to read my question, I look forward to hearing the answer soon! (I think the answer is “Yes”, but I wanted to confirm it with some people smarter than me!)

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No it wouldn't be.

From what I understand, the arrows are just raising to the power, therefore:

1-->6-->1-->8

would be the equivalent of

1^6^1^8

and 1 raised to any power is just 1.

Edit: I read a bit further and realize I don't understand quite how Conway chained arrow notation works (working on it) but, it would still be 1, because:

"notation eventually resolves to being the leftmost number raised to some (usually enormous) integer power" -from Wikipedia

meaning you would get

1^(BIG NUMBER)

Which is still just 1.

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i think no.

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yes..
1
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