Quote:
Originally Posted by sanctus
now let's be mathematicians, because I agree with Donk when you look at his example it is obvious/intuitive, but a "true mathematician" would like to prove it...or a real geek ;-)
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Though I’m not sure I fit this description, having little drive to prove the obvious proposition that any finite positive integer can be represented as a numeral, when I can’t imagine a useful consequence of failing to prove or disproving it, I’m a sucker for a direct challenge, so here goes…
Define:
1. A numeral base b is a string of zero or more digits base b, where b is a integer and b>1
2. A digit base b D is a glyph (character) mapping to an integer v_D, 0<=v_D<b
2a. For the sake of this proof, a numeral may begin with the digit mapping to 0 (the digit “0”), but may not begin with the digit mapping to b-1.
Assume the proposition to be disproven:
3. There exist one or more finite integers that can’t be represented as a base b numeral.
4. Because integers are ordered (consequence of the
Peano postulates), there must then exist a smallest of these integer, m.
5. Because all integers have successors S(n)=n+1 (Peano postulates), an integer p that can be represented as a base b numeral exists such that m=S(p)=p+1.
6. Because p can be represented as a base b numeral, it must consist of
S_L A S_R
Where S_L is any string of zero or more digits base b, A a digit mapping to integer v_A<b-1, and S_R a string of zero or more of the digit B mapping to the integer v_B=b-1
7. The successor of p, m=p+1, then, must consist of
S_L B S_Z
Where B is a digit mapping to integer v_A+1, and S_Z is a string of zero or more of the digit “0” such that length(S_Z) = length(S_R)
8. For it to not be representable as a base B numeral, S_L, B, or S_Z must not exist. S_L exist by definition. S_Z exists if the digit “0” exists, so exists. Because every integer has a successor, if v_A exists, v_A+1 must exist.
9. So for B not to exist, there must be an v_A+1 < b that does not map to a digit, which contradicts [2].
10. Therefore B exists, and m=p+1 can be represented as a numeral base b, which contradicts [4], so [4] is false. If [4] is false, [3] is false, and the proposition to be disproven is disproven.
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