Prove that every positive integer can be written as the sum of distinct powers of two.
We will prove this using the method of mathematical induction.

Base case: For n = 1, we can write it as 2^0 = 1. So, the base case holds true.

Inductive step: Assume that every positive integer k (k ≥ 1) can be written as the sum of distinct powers of two. We need to show that k + 1 can also be written as the sum of distinct powers of two.

Case 1: If k is odd, then k can be written as a sum of distinct powers of two with the smallest power being 2^0 (since k is odd). In this case, k + 1 would be even and can be written as a sum of distinct powers of two without including 2^0 (as k+1 is even), which are all distinct from the powers in the representation of k.

Case 2: If k is even, let's consider the binary representation of k. Since k is even, its least significant bit must be 0. Let m be the position of the first 1-bit counting from the right (starting at position 0). Then, k can be represented as:

k = 2^m + ... + 2^p (where p > m)

Now, when we add 1 to k, we get:

k + 1 = 2^m + ... + 2^p + 1

Since adding 1 flips the bits up to and including the m-th bit, the new number has a 0 at position m and a 1 at position (m+1):

k + 1 = 2^(m+1) + ... + 2^q (where q > m+1)

This is again a representation of k+1 as a sum of distinct powers of two.

By the principle of mathematical induction, we have proved that every positive integer can be written as the sum of distinct powers of two.