Positive integers expressable as sums of powers of 2

Question

I need to prove that any positive integer is expressable as $$x=2^{j_0}+2^{j_1}+2^{j_2}+...+2^{j_m}$$ where $m\ge 0$ and $0\le j_0\lt j_1\lt j_2\lt ... \lt j_m. $

I think I get the gist of the proof; what I mean is I think I intuitively understand what is happening here, but looking for verification, either a more simplistic argument or proper rigor.

I start by saying a positive integer is expressable as an even or odd integer. So, a positive integer, $x=2k$ or $x=2k+1$ for some positive integer $k$. If $x$ is odd, then we can write $1=2^0$. Now, an integer $k$ is also expressable as an even or odd integer. Thus we now have $2^2=4$ possibilities; if $x$ is even; $$x=2k=2(2k_1)=2^2k_1$$ $$x=2k=2(2k_1+1)=2^2k_1+2^1$$ if $x$ is odd; $$x=2k+1=2(2k_1)+1=2^2k_1+2^0$$ $$x=2k+1=2(2k_1+1)+1=2^2k_1+2^1+2^0$$ it is clear that $k>k_1$ since $k=2k_1$. Now we repeat the process with $k_1=2k_2$ or $k_1=2k_2+1$ and eventually, $k_n=2k_{n+1}$ or $k_n=2k_{n+1}+1$. Since $x$ is a positive integer, $k_i$ is positive for all $i=1,2,...,n$ and thus the larger the $i$, the smaller the $k_i$. This process eventually terminates since $k_i>0$ and thus $k_n=1$

Now how do I simplify this argument, assuming it's correct. If it is not correct,, how do I repair it or make it more rigorous?

Answer 1

We can do it by (strong) induction. Let $P(x)$ be the assertion that $x$ is a sum of $0$ or more distinct powers of $2$. The number $0$ is a sum of $0$ or more distinct powers of $2$. So $P(0)$ holds.

Suppose that $P(k)$ is true for all $k\lt x$. We show that $P(x)$ is true. Let $2^p$ be the largest power of $2$ which is $\le x$. Then $x-2^p \lt 2^{p-1}$. By the induction hypothesis, $x-2^p$ is expressible as a sum of $0$ or more distinct powers of $2$. All these powers of $2$ are $\lt 2^p$, since $x-2^p\lt 2^p$. It follows that $x=2^p$ plus a sum of $0$ or more distinct powers of $2$ that are less than $2^p$. So $x$ is a sum of distinct powers of $2$.

Answer 2

Much more generally, you can prove this:

Let $(b_k)_{k=1}^{\infty}$ be a sequence of integers such that each $b_k \ge 2$ and let $B_0 = 1$ and $B_k = \prod_{j=1}^k b_j$ for $k \ge 1$. Then every positive integer $n$ can be represented uniquely in the form $n = \sum_{k=0}^{D(n)} d_kB_k$ where $D(n)$ is a positive integer that depends on $n$ and the $d_k$ are integers such that $0 \le d_k < b_{k+1}$.

If all the $b_k$ are equal to $b$, this gives the standard representation in base $b$.

If $b_k = k+1$, this the "factorial" representation.

There is a converse to this:

If $B_k$ is an increasing sequence of positive integers with $B_1 \ge 2$ and $\frac{B_{k+1}}{B_k} \ge 2$ then every positive integer $n$ can be represented in the form $n = \sum_{k=0}^{D(n)} d_kB_k$ with $d_k$ integers such that $0 \le d_k < \frac{B_{k+1}}{B_k}$ and the representation is unique if and only if $B_k $ divides $B_{k+1}$ for all $k$.

Answer 3

This is just the expression of $x$ as a binary (base $2$) number, isn't it?

Answer 4

As you say, $x=2k$ or $x=2k+1$. Now, $k<x$ and so, by induction, $k$ can be expressed as a sum of distinct powers of $2$. Then $2k$ can also be so expressed. If $x=2k$, we're done. If $x=2k+1$, then $x=2k+2^0$, and we're done, because $2^0$ does not appear in the expression of $2k$.

Answer 5

For a variation, we can use contradiction.

Assume there is a set of positive integers than cannot be expressed as the sum of distinct powers of two. Since the positive integers are well-ordered, there must be a smallest one. Call that one $x$.

Now, subtract from $x$ the largest power of two that is still smaller than $x$. Call this number $y$.

$y = x-2^N$

If $y$ is expressible as a sum of distinct powers, then $x = y + 2^N$, contradicting our assumption. We are not in danger of $2^N$ already appearing in $y$ since $y$ must be smaller than $2^N$. If this wasn't the case, $2^N$ would be smaller than $x/2$ and thus not be the largest power of two less than $x$). Therefore, $y$ cannot be expressed as a sum of distinct powers of two.

But, $y$ is smaller than $x$, contradicting our assumption that $x$ was the smallest example. Our only possible escape is to set $y=0$, since $0$ isn't a positive integer, thus avoiding the "smallest example" part contradiction. However, this means that $x=2^N$, which obviously contradicts the "not a sum of powers of two" assumption.

Since our assumption leads to nothing but contradictions, it must be false. All numbers are expressible as the sum of distinct powers of two.

Answer 6

Proof by Contradiction. Assume $n$ can be written as the sum of distinct powers of 2 in two different ways. Find the smallest powers of $2$ where the two ways disagree, say $2^a$ and $2^b$. Assume, WLOG than $a < b$. Subtract all of powers of $2$ that are smaller than $2^a$ From both ways, producing a (possibly smaller) number that is the sum of distinct powers of $2$ in two different ways. Divide both numbers by $2^a$ producing a (possibly smaller) number that is the sum of distinct powers of $2$ in two different ways. The way that had $2^a$ as a power now has $2^0 = 1$ as a power and is therefore odd, but the other way is even. Thus they cannot be equal, which is a contradiction.

Answer 7

Theorem: any non-negative integer is a sum of 0 or more distinct powers of 2.

Proof: By the well-odering principle. Assume there is a set, $C$, of positive integers that can not be expressed as a sum of 0 or more powers of 2. Since $C$ is a subset of the positive integers, by the well-ordering principle, there must be a smallest element, $m$, in C.

It's clear that 0 is the sum of 0 powers of 2, so $m>0$.

Now, let's subtract from $m$ the largest power of two, $2^n$, that is still smaller than $m$ itself. Let's call it $j$:

$j = m-2^n$

Now let's use cases analysis. There are 2 cases:0

1. $j$ is a sum of 0 or more distinct powers of 2.;
2. $j$ is not a sum of 0 or more distinct powers of 2.

In the first case, by manipulation:

$m = j + 2^n$

In this case, $m$ is expressed as the sum of 2 raised to distinct powers. Since $2^n$ is the largest power of 2 in $m$ so $2^n>j$. It makes it a contradiction and, therefore, $C$ must be empty.

In the second case, $j$ is smaller than $m$. It's a contradiction since $m$ should be the smallest counterexample in $C$.

QED