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 January 31st, 2011, 10:09 AM #1 Member   Joined: Jul 2010 Posts: 42 Thanks: 0 Binary Proof Any number "n" can be expressed as either the sum of powers of 2 or a single power of 2. $\mathbb{R}_{i,odd} = 2^{b}B + 2^{0}|_{b>=1} \\ \mathbb{R}_{i,even} = 2^{b}B$ Both $\mathbb{R}_{i,odd}$ and $\mathbb{R}_{i,even}$ are valid to the assertion. With n as any number, we have... $n=2C + 2^{0}D|_{D=1 or 0}$ Case 1: $\\\\ _{\mathbb{R}_{i,odd}} {\wedge^{n}__{2^{a}}}$ $n = 2^{a}(\mathbb{R}_{i,odd}) \\n = 2^{a}(2^{b}B + 2^{0}) \\n = 2^{a+b}B + 2^{a}$ By definition of $\mathbb{R}_{i,odd}$ where $b=>=1$ $\\B \subseteq \mathbb{R}_{odd},$ $n = 2^{a+b}(\mathbb{R}_{i,odd}) + 2^{a} \\ = 2^{a+b}(2^{b}B+2^{0}) + 2^{a} \\= 2^{a+2b}B + 2^{a+b} + 2^{a} \\...$ Case 2: $\\ _{2^{a}}{\wedge^{n}}__{1^{a}}$ $n= 2^{a}$ (satisfies assertion) Case 1 continues the pattern as you substitute in for $B$. The pattern follows that for each substitution into $B$, another sequence of powers of $2$s will be transformed into the equation. The pattern follows this ultimate equation: $= 2^{a+cb} + 2^{a+(c-1)b} + 2^{a+(c-2)b} + ... + 2^{a+(c-(c-1))b} + 2^{a}$ Thus proving that for any given valid entry for any variable, n can always be expressed as the additive sequence of a power of $2$s.
 January 31st, 2011, 12:49 PM #2 Global Moderator   Joined: May 2007 Posts: 6,511 Thanks: 585 Re: Binary Proof It is true - any number can be expressed in base 2.
 February 3rd, 2011, 11:27 AM #3 Member   Joined: Jul 2010 Posts: 42 Thanks: 0 Re: Binary Proof My apologies, I'm well aware that any number can be expressed in the base of 2, or any base for that matter. I'm interested in whether or not the proof works and makes sense.

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# how to proof binary numbers

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