It follows from that for any
\[ n \] we can find
\[k\] such that \[2^k\le n<2^{k+1}\].We will proceed with induction.For \[n=1\] we see \[1=2^0\].This is the only way to express it.Suppose it is true for all \[i< n\].We show for \[n\].We show for \[n\].Now as before we can find \[k\] as per the equation \[2^k\le n<2^{k+1}\].So we see this \[k\] is unique.Now consider \[n-2^k\] and since this is \[<n\] it has a unique representation.So every positive integer has one such representation.btw this is actually proving that each number in decimal system has a unique representation in the binary system.

Also, since \(n - 2^k < 2^{k+1} - 2^k = 2^k,\) the representation of \(n - 2^k\) doesn't contain \(2^k.\) Then it follows that each number can be represented as a sum of unique powers of \(2.\)

I almost wanted to claim that the statement is wrong as it's currently worded (since \(5 = 2^0 + 2^0 + 2^0 + 2^0 + 2^0\) too), but you already put it here. :)

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TopNewestIt follows from that for any

\[ n \] we can find \[k\] such that \[2^k\le n<2^{k+1}\].We will proceed with induction.For \[n=1\] we see \[1=2^0\].This is the only way to express it.Suppose it is true for all \[i< n\].We show for \[n\].We show for \[n\].Now as before we can find \[k\] as per the equation \[2^k\le n<2^{k+1}\].So we see this \[k\] is unique.Now consider \[n-2^k\] and since this is \[<n\] it has a unique representation.So every positive integer has one such representation.btw this is actually proving that each number in decimal system has a unique representation in the binary system.

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Also, since \(n - 2^k < 2^{k+1} - 2^k = 2^k,\) the representation of \(n - 2^k\) doesn't contain \(2^k.\) Then it follows that each number can be represented as a sum of

uniquepowers of \(2.\)Log in to reply

I almost wanted to claim that the statement is wrong as it's currently worded (since \(5 = 2^0 + 2^0 + 2^0 + 2^0 + 2^0\) too), but you already put it here. :)

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Thank you very much Riju Roy .

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