# 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.

Note by X Zero
4 years, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 4 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 $$2.$$

- 4 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. :)

- 4 years, 6 months ago

Thank you very much Riju Roy .

- 4 years, 6 months ago