2's Problem

Please help me to prove that "Every natural number can be uniquely represented in sum of power's of 2". Example 5=2^2 + 2^0.

#HelpMe! #MathProblem

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Riju Roy - 7 years, 6 months ago

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 $2.$

Michael Tang - 7 years, 6 months ago

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. :)

Ivan Koswara - 7 years, 6 months ago

Thank you very much Riju Roy .

X zero - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$