14.1.4. Question Number 9

In question 9, you asked to prove that every positive integer n can be expressed as sum of distinct powers of two basically this is a fundamental fact that we learn that you take any positive integer, it has a binary representation and a binary representation of that number is nothing but sum of distinct powers of two.

In this section, we start with a fundamental concept: every positive integer can be expressed as a sum of distinct powers of two. This links directly to how numbers are represented in a binary system. For example, the decimal number 5 is represented in binary as 101, which is equivalent to 1×2² + 0×2¹ + 1×2⁰, or 4 + 0 + 1. This representation confirms that 5 can indeed be expressed as a sum of distinct powers of two (specifically 2² and 2⁰).

Think of each unique power of two as a different flavor of ice cream. When you want to create your own sundae (a number), you can only pick each flavor (power of two) once. The combination of flavors you choose represents the unique number you're creating (like 5 from the earlier example).

So we will prove it by induction because this is a universally quantified statement my base case will be n equal to 1. If my integer n is 1, then I can represent this as 20.

The proof begins with an inductive approach. The base case is chosen as n = 1, the smallest positive integer. Here, 1 can be represented as 2^0. This establishes the foundation of our proof, showing that the statement holds true for the simplest case.

Imagine you are building a tower with blocks. The first block you place (representing the number 1) is a power of two (2^0). It’s the most basic building block, just as 20 is for representing the number 1 in a power of two format.

Assume the statement is true for n = k, that means you give me any integer k where k is arbitrary, it can be expressed as the sum of distinct powers of 2 or it has a distinct binary representation.

The next step is to assume that the statement is true for some arbitrary positive integer k. This means that we can express k as a sum of distinct powers of two. This assumption is crucial, as it will be used to prove that the statement must also hold for k + 1.

Think of it like climbing stairs: if you can step up k stairs without falling (the k case), then you should be able to step up one more stair (k + 1) using the same logic. The idea is that if it worked for k, it will work for the next step.

So, how do I proceed here? So I use proof by cases. Case one, if k is even; now if k is even and since k has a binary representation a unique binary representation namely k is expressible as sum of distinct powers of two.

The proof then branches into two cases based on whether k is even or odd. If k is even, it can be expressed as a unique sum of distinct powers of 2. The logic here is that since k does not have 2^0 in its representation, we can add this power (2^0) to create k + 1, which is also a sum of distinct powers of two.

Imagine having a chocolate cake (k) which you have made using multiple distinct ingredients (powers of two). If you are adding one more ingredient (2^0) that you haven't used, your new cake (k + 1) will still be made from distinct ingredients and thus fit the rules of our cake recipe (the binary representation).

Case two is when k is odd. Now, if k is odd then it follows that k + 1 will be even.

In the second case, if k is odd (meaning it includes the term 2^0), then k + 1 will be even. We take half of k (let's call it l). Since l is less than or equal to k, we know from our inductive hypothesis that l can be expressed as a sum of distinct powers of two. Then, to convert l back up to k + 1, we just need to double each power, still keeping all distinct.

This can be compared to a group of friends who each have different special skills. If one friend (k) decides to team up with others while ensuring all skills are distinct, they can create a stronger team (k + 1) with new combinations of skills but still no duplicates. The idea is that the distinct nature of their skills means new combinations still fit the rules.

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