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Author Topic: 68k 64-bit multiplication  (Read 14176 times)

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Offline AngelFish

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Re: 68k 64-bit multiplication
« Reply #15 on: August 07, 2011, 09:13:12 pm »
Base 0 isn't defined, but base 1 is by 0↦0↺ ∀ {ai+ 110∈ ℝ}
« Last Edit: August 07, 2011, 09:19:18 pm by Qwerty.55 »
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∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

Offline christop

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Re: 68k 64-bit multiplication
« Reply #16 on: August 07, 2011, 10:09:55 pm »
Quote from: calcdude84se on August 07, 2011, 08:42:35 pm
Quote from: Qwerty.55 on August 07, 2011, 03:54:58 pm
With a calculator handy :P

The basic fact you end up using is that any number x can be represented in base z as ∑aizi
With the exception of when z = 0 or |z|=1, of course ;)
Awww, no nullary number system? :)

With base 1, do you use as many digits as will add up to the value? For example, is 7 decimal equal to 1111111 unary? I also don't see how fractions can be represented in unary.
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Christopher Williams

Offline AngelFish

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Re: 68k 64-bit multiplication
« Reply #17 on: August 07, 2011, 10:15:04 pm »
Unary is one type of Base 1 system. However, there are other types involving symbol collision which are currently in dispute on IRC :P
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∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ
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