Proof about the sum of integers in the form of 2^n The question is: Prove that every positive integer can be uniquely expressed as a sum of different numbers, where each number is of the form 2^n for some non-negative integer n.
My proof: 

We see that it works from $1, 2, \dots, 10$ -- assume it works from
  $1, 2, \dots, k$. 
Then let $2^n$ be the biggest power of 2 smaller than $(k+1)$. 
Then we have that $(k+1) - 2^n \leq k$, then it is representable as a
  sum of powers of 2. 
Then, appending $2^n$ to $((k+1)-2^n)+2^n = k+1$.

Can I get some feedback, on what I am missing? Maybe some intuition?
 A: First, as an intuition, this question is actually exploring the binary system of representing positive integers (integral for computers) -- essentially we are looking to prove that every positive integer can be written in binary in a unique way. 
Remember how our decimal numeral system works: we add powers of ten to each other to represent any positive integer e.g. $1234 = 1*10^3 + 2*10^2 + 3*10^1 + 4*10^0$. We are being asked to prove that a similar thing can be done for binary. 
As you seem to be hinting at, you can prove the existence of a binary representation for every integer using strong induction. NOTE that we will need to introduce something else to show that such a representation is unique. 
So let's begin showing the existence of such a binary representation. Our inductive hypothesis is that every positive integer can be expressed as a sum of different numbers, where each number is of the form $2^n$ for some non-negative integer $n$.
Our base case is 1 (note the question asks for a proof about positive integers, not nonnegative integers, a very minor technicality). Well, it's quite obvious that we can express 1 as $1 = 2^0$. Thus our base case is proven. 
Now for our induction step. Assume that the statement above holds true up to some $N$, such that for all nonnegative integers $0 \leq x \leq N$, the statement is true (this is the strong induction part). In other words, for every $0 \leq x \leq N$, we have that $x$ can be written as the sum of different numbers, where each number is of the form $2^n$ for some non-negative integer $n$.
Now the onus is on us to prove that it is also true for the number $N+1$. Let $2^k$ be the largest power of 2 such that $2^k \leq N+1$. 
Consider the number $(N+1) - 2^k$. Now since we know that $2^k \geq 1$ for nonnegative $k$, we can see that $(N+1) - 2^k \leq (N+1) - 1 = N$. Our strong inductive hypothesis then indicates to us that since $(N+1) - 2^k \leq N$, we can write $(N+1) - 2^k$ as the sum of different numbers each in the form $2^n$. This immediately implies that we can then write $N+1$ as $[(N+1) - 2^k] + 2^k$, and since $(N+1) - 2^k$ can be written as the sum of different numbers each in the form $2^n$, and since $2^k$ is also in the form $2^n$, we see that $(N+1)$ is also the sum of different numbers each in the form $2^n$.
Now since $(N+1) - 2^k$ is assumed, by inductive hypothesis, to already be written as the sum of different powers of 2, and we are simply adding $2^k$, we now need to show that $2^k$ does not show up in the expression $(N+1) - 2^k$ as a sum of distinct powers of 2. 
To do this, assume, by way of contradiction, that $(N+1) - 2^k = S + 2^k$, where $S$ is a sum of distinct powers of 2. 
Since $S \geq 0$, we see that $2^k \leq (N+1) - 2^k$. This implies that $2^k + 2^k \leq (N+1)$, or that $2^{k+1} \leq N+1$. This is a contradiction: remember that earlier we assumed that $2^k$ must be the largest power of 2 such that $2^k < N+1$. This contradiction shows that our assumption must be false, and that $2^k$ is not repeated twice. 
So far we have shown the existence of a representation of positive integers as the sum of distinct powers of 2. To show uniqueness, let us consider some number $n$, with 2 distinct binary representations:
$$
n= a_ma_{m-1}...a_0 \\
n= b_kb_{k-1}...b_0 \\
a_ma_{m-1}...a_0 \neq b_kb_{k-1}...b_0
$$
Note i'm using shorthand here, where $a_ma_{m-1}...a_0$ actually represents a sum of distinct powers of 2, with coefficients $a_i \in \{0,1\}$, i.e.
$$
n = \sum_{i=0}^m a_i2^i = \sum_{i=0}^k b_i2^i,\hspace{2cm} a_i,b_i \in \{0,1\}
$$ 
since there are supposed to be 2 distinct binary representations here, let $x$ be the index of the first coefficients become unequal, i.e. $a_x \neq b_x$. 
This means that $a_{x-1}...a_0 =  b_{x-1}...b_0$. Thus we can subtract these from $n$ to show that $n' = a_ma_{m-1}...a_l0...0 = b_kb_{k-1}...b_l0...0$.
Since $a_l \neq b_l$, one of them is 0 while the other is 1. Thus one representation of $n'$ (WLOG, assume the $a$ representation) is divisible by $2^{l+1}$ since it ends in $l+1$ zeroes. (To see this, simply note that $\sum_{i=0}^m a_i2^i = \sum_{i=l+2}^m a_i2^i$ and note that you can factor out $2^{l+1}$) while the other representation ($b$ representation here) cannot. This contradicts the notion that the $a$ and $b$ representations are representations of the same number, thus the two strings must be equal, and this proves uniqueness. 
Note I have exploited the base 2 representation for the uniqueness proof. it stands to reason that any positive integer basis should both exist and have unique representations, and indeed you can prove this more generally, as it is done here: https://proofwiki.org/wiki/Basis_Representation_Theorem
