In my article “Composing Powers of Two Using The Laws of Exponents” I showed how to combine powers of two using the standard laws of exponents. There are two other rules I use when combining powers of two; I call them the *add duplicate power of two rule* and the *subtract half power of two rule*. These are nonstandard rules, applying only to powers of two. Although these are special cases of the existing multiplication and division rules, I’ve found value in recognizing them in addition and subtraction form. I’ll state these rules and show examples of their usage.

## Add Duplicate Power of Two Rule

Here’s the rule I call the *add duplicate power of two rule*:

2^{a} + 2^{a} = 2^{a+1}

(2^{a+1} is the power of two following 2^{a}; that is, *double* 2^{a}.)

For example, 2^{4} + 2^{4} = 2^{5}, and 2^{-2} + 2^{-2} = 2^{-1}.

You don’t normally think of the laws of exponents as applying across addition, but in this case they do. Adding something to itself is the same as doubling it, which is multiplying it by 2 — a power of two. This makes this rule a form of the *product of powers of two rule* in disguise. Here’s the math:

2^{a} + 2^{a} = 2^{a} · 2 = 2^{a} · 2^{1} = 2^{a+1}

### Example of Usage

One example where this comes up is when summing binary numbers represented as powers of two. For example, to add 101 and 1100, you add 2^{2} + 2^{0} to 2^{3} + 2^{2}, getting 2^{3} + 2^{2} + 2^{2} + 2^{0}. You want to combine those to eliminate duplicate powers of two. We know by the *add duplicate power of two rule* that 2^{2} + 2^{2} = 2^{3}. Now with two copies of 2^{3}, we have 2^{3} + 2^{3} = 2^{4}. This gives the final sum 2^{4} + 2^{0} = 10001 in binary.

## Subtract Half Power of Two Rule

Here’s the rule I call the *subtract half power of two rule*:

2^{a} – 2^{a-1} = 2^{a-1}

(2^{a-1} is the power of two preceding 2^{a}; that is, *half* of 2^{a}.)

For example, 2^{4} – 2^{3} = 2^{3}, and 2^{-2} – 2^{-3} = 2^{-3}.

Again, you don’t normally think of the laws of exponents as applying across subtraction, but it works here. Subtracting half of something is the same as halving it, which is multiplying by 1/2 or dividing by 2 — both powers of two. This makes this rule a form of the *product of powers of two rule* or *quotient of powers of two rule* (respectively) in disguise. Here’s the math:

2^{a} – 2^{a-1} = 2^{a} – 2^{a}/2 = (2·2^{a} – 2^{a})/2 = (2^{a}(2-1))/2 = 2^{a}/2 = 2^{a-1} .

That was the long way of doing it, because I wanted to show the halving explicitly; here’s a more direct way:

2^{a} – 2^{a-1} = 2^{a-1}(2-1) = 2^{a-1}

### Example of Usage

I used this rule when trying to solve this infinite geometric series:

1/2 − 1/4 + 1/8 − 1/16 + 1/32 – 1/64 + …

Wikipedia solves it by first factoring out 1/2:

1/2 − 1/4 + 1/8 − 1/16 + 1/32 – 1/64 + …

= 1/2 (1 – 1/2 + 1/4 – 1/8 + …)

It then uses this formula to replace the summation:

Using *r* = -1/2, the series simplifies to

I solved it differently. Instead of factoring out 1/2, I started by replacing adjacent terms of the series using the *subtract half power of two rule*:

1/2 − 1/4 = 1/4, 1/8 − 1/16 = 1/16, 1/32 – 1/64 = 1/64, …

That left me with the series 1/4 + 1/16 + 1/64 + 1/256 + 1/1024, which is written

I then used this formula to replace the summation:

With r = 1/4, this simplifies to

I like my proof because it uses a positive *r*, which is easier to conceptualize; you can think of the series as a binary number (0.01).