Strong Induction Proof: Every natural number = sum of distinct powers of 2 The question I'm looking at, is to show that every positive integer $n$ can be written as a sum of distinct powers of two. 
I can see that you can form any number based on the highest $2^t$ that is less than the number, plus some combination of $2^j<n$'s. And that you can make the number odd, by adding $2^0$ at the end. 
I just don't know how to create the formula for the proof. I'm trying to figure out my base case, and then my inductive formula to figure out $k+1$, and I've got nothing.
 A: Perhaps people will find this method easier...
$Proof$. We will use strong induction to prove that $\forall n \in \mathbb{Z}^+$ we can express $n$ as a sum of distinct powers of $2$.
Base Case. For $n=1$ note that $2^0=1$ and thus for $n=1$ our proposition holds.
Inductive Hypothesis. Assume $m\in\mathbb{Z}^+$ with $1\le m \le k$ and that our proposition holds for $m$.
Inductive Step. We consider two cases.
Case 1. If $k+1$ is even then observe that $\frac{k+1}{2}$ must be an integer. Now since $1\le \frac{k+1}{2} \le k$ we know by our inductive hypothesis that $\frac{k+1}{2}$ is the sum of distinct powers of $2$. But then multiplying $\frac{k+1}{2}$ by $2$ gives
$$\frac{k+1}{2}\cdot2=k+1$$
and since (by the distributive property of multiplication over addition) each distinct power of $2$ in the sum $\frac{k+1}{2}$ is multiplied by a factor of $2$, each power of $2$ is increased by $1$ and thus remains distinct.
Case 2. If $k+1$ is odd then we know $k$ is even. Furthermore we know by our inductive hypothesis that $k$ may expressed as the sum of distinct powers of $2$. But if $k$ is even we know $k$ does not contain a $2^0=1$ in its sum of distinct powers of $2$. 
Remark: To observe that $2^0=1$ does not exist in the sum of distinct powers of $2$ that makeup $k$, note that $k$, by the definition of an even integer, may be expressed as $2m$ for some $m\in\mathbb{Z}^+$ (since $k$ is positive we limit $m$ to the positive integers) and since we can view multiplication as repeated addition we have $2m = 2 + 2 + \cdots + 2 + 2$ where we are taking the sum of $m$ twos. If the sum were to contain a $2^0=1$ we clearly would need to express our integer as $2m+1$ which would make it odd not even.
Thus we see that 
$$k+1=k + 2^0$$
and if $k+1$ is odd we may express $k+1$ as the sum of distinct powers of $2$.
It follows by strong mathematical induction that $\forall n \in \mathbb{Z}^+$ we can express $n$ as a sum of distinct powers of $2$. $\Box$
A: Finding such a representation is equivalent to expressing $n$ in binary. We can do induction as follows: Let $2^h$ be the highest power of $2$ less than or equal to $n.$ Then we must have $n-2^{h}< 2^{h+1} - 2^{h}=2^{h}(2-1)=2^{h}.$ Hence the greatest power, say $2^{g},$ of $2$ such that $2^{g}\le n-2^{h}$ must satisfy $g<h.$ By strong induction on $h$ we can assume that $n-2^{h} = \sum_{i=0}^{h-1}c_i2^i$ where each $c_i$ is either $0$ or $1.$ But then we're done, since  this implies $n= \sum_{i=0}^{h-1}c_i2^i + 2^h$ is a sum of distinct powers of $2.$
To be explicit, let's include the induction hypothesis. Namely, for $h=0$ every positive $n\le 2^h=2^0=1$ can be expressed as a power of $2,$ since the only possibility is $1=2^0.$ Thus, we assume that for a given $h>0$ and any $n\le2^h,$ that such a representation exists.
A: The statement is obviously true for $n=0$.
Assume that we are given an $n\geq1$ and that it is true for all $m$ with $0\leq m<n$.
When $n=2m$ then $m<n$ and therefore $m=\sum_k 2^{p_k}$ with finitely many $p_k$, all of them different. It follows that $n=\sum_k 2^{p_k+1}$ with all $p_k+1$ different.
When $n=2m+1$ with an $m$ as before then $n=2^0+\sum_k 2^{p_k+1}$ with all $p_k+1$ different and different from $0$.
A: Well, technically, since all the numbers up to 2^X can be represented by powers of 2, that makes it 2^X2-1, which is 1 less than 2^X2, etc.
A: Not an inductive proof, but I thought I'll add it anyway.

$\require{cancel} $
Consider the generating function
$$\prod_{k = 1}^\infty (1 + x^{2^k}) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)... \tag{1}$$
In $(1)$, $(1 + x^{2^k})$ can be interpreted as saying that $2^k$ can be either used zero times or one time in a partition of $n \in \mathbb{Z}$. Hence, the coefficient of $x^n$ in the expansion of the product finds all partitions of $n$ consisting of only distinct powers of $2$.
Note from $a + b = \frac{a^2 - b^2}{a - b}$ that
$$1 + x^{2^k} = \frac{1 - x^{2^{k + 1}}}{1 - x^{2^k}}$$
Hence, we can rewrite $(1)$ as follows:
$$(1) = \frac{\cancel{1 - x^2}}{1 - x} \cdot \frac{\cancel{1 - x^4}}{\cancel{1 - x^2}} \cdot \frac{\cancel{1 - x^8}}{\cancel{1 - x^4}} \cdot ... = \frac{1}{1 - x} \tag{2}$$
We know, from geometric series, that
$$(2) = \frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \; ... \tag{3}$$
$$\implies (1) = (3)$$
Note that the coefficient of $x^n$ in $(3)$ can also be interpreted to denote the number of partitions of $n$ subject to some conditions.
From $(1) = (3)$ and their combinatorial interpretations, we can deduce that every natural number can be represented by the sum of distinct powers of $2$.
$\blacksquare$
A: The subtle thing is the exact wording of the proposition.
If you want to prove:
$P(n) = $ $n$ can be expressed as a distinct sum of powers of $2$.
Initial case: $n = 1$ $P(1)$ is provable.
Induction step: $P(k) \implies P(k+1)$
then this is impossible.  $P(k) \not \implies P(k+1)$ and there is simply no way to show it does.
But if instead you try to prove.
$Q(n) = $ of all $j;  0 < j < 2^n$, $j $ can be written as a distinct product of powers of $2$
then $Q(k) \implies Q(k+1)$ is pretty easy to prove.
(if $0 < j \le 2^{n+1}$ then either $j = m$ or $j = m + 2^n$ where $0 < m < 2^n$ so $m$ can be distinctly written as $\sum 2^i$.  As $m < 2^n$ none of its powers of $2$ are to the $n$ power.  So  $m + 2^n = 2^n + \sum 2^i $ is a sum with distinct powers of $2$.  So $j$, whethe $j = m$ or $j = m + 2^n$ can be written as a sum of distinct powers.)
...
Another way to view this, albeit, maybe not as rigorous.  Is that induction need not be a direct $P(n) \implies P(n+1)$ progression.  And $P(n) \implies P(f(n))$ where $\cup_{i \in \mathbb N} F_i= \mathbb N$ where $F_0 = \{1\text{ or } 0\}$ and $F_{i+1} = \{f(a)|a \in \cup_{k \le i}F_i\}$, qualifies as induction as well.
Here $f(n) = n + 2^k$ where $k$ is the smallest power where $2^k > n$.
(although the bit $\cup_{i \in \mathbb N} F_i= \mathbb N$ most likely will boil down to a $Q(n) \implies Q(n+1)$.  e.g.  How do we know $\{0,1\}, \{0,1, 2, 3\},\{0, 1, 2, 3,4,5,6,7\} ....$ will span $\mathbb N$?  Probably by induction.)
