3.2.1 Converting Integers

One of the topics I’ve found rather difficult teaching to students is the concept of changing bases. It seems that students have the concept of a base-10 system so ingrained in their mind (almost always unbeknownst to them) that it is difficult considering other base systems. Hopefully this section will be a good introduction to the process of changing bases and doing basic operations in other number systems. First, let’s observe how we look at numbers in the usual base-10 fashion.

Everyone knows that 1254 means that you have one-thousand, two-hundred, and fifty-four of something, but expressing this in an unusual manner we can say:

$$1294 = 1 \cdot 1000 + 2 \cdot 100 + 5 \cdot 10 + 4 \cdot 1 = 1 \cdot 10^3 + 2 \cdot 10^2 + 9 \cdot 10^1 + 4 \cdot 10^0$$

From this we can see where this concept of “base-10” comes from, we are adding combinations of these powers of tens (depending on what $0 - 9$ digit we multiply by). So, you can express any integer $n$ in base-10 as:

$$n = a_m \cdot 10^m + a_{m-1}10^{m-1} + a_{m-2} \cdot 10^{m-2} + \dots a_1 \cdot 10^1 + a_0 \cdot 10^0$$

Where all $a_m$’s are integers ranging from $0 - 9$.

The fact that we are summing these various powers of 10 is completely an arbitrary one. We can easily change this to some other integer (like 6 for example) and develop a base-6 number system. Let’s see what it would look like:

$$n = a_m \cdot 6^m + a_{m-1}6^{m-1} + a_{m-2} \cdot 6^{m-2} + \dots a_1 \cdot 6^1 + a_0 \cdot 6^0$$

Where all $a_m$’s are integers ranging from $0 - 5$.

So to use an example, let look at what the number $123_6$ (where the subscript denotes we are in base-6) would look like in our usual base-10 system:

$$123_6 = 1 \cdot 6^2 + 2 \cdot 6^1 + 3 \cdot 6^0 = 1 \cdot 36 + 2 \cdot 6 + 3 \cdot 1 = 36 + 12 + 3 = 51_{10}$$

From this we have found the way to convert any base-n whole number to base-10!

Let’s look at another example:

$$3321_4 = 3 \cdot 4^3 + 3 \cdot 4^2 + 2 \cdot 4^1 + 1 \cdot 4^0 = 3 \cdot 64 + 3 \cdot 16 + 2 \cdot 4 + 1 \cdot 1 = 192 + 48 + 8 + 1 = 249_{10}$$

So now that we know how to convert from base-n to base-10, let’s look at the process on how to convert the opposite direction:

1. Find the highest power of $n$ which divides the base-10 number (let’s say it is the $m$-th power).
2. Figure out how many times it divides it and that will be your $(m + 1)$-th digit in base-n.
3. Take the remainder and figure out how many times one less than the highest power divides it (so see how many times $n^{m-1}$ divides it). That is your $(m)$-th digit.
4. Take the remainder, and continue process.

I know that this might seem complicated, but let’s look at an example we have already done in the “forward” direction to illustrate how to go “backwards.” Convert $51_{10}$ to base-6:

1. Well we know $6^2 = 36$ and $6^3 = 216$, so the highest power which divides 51 is $6^2$.
2. 36 goes into 51 one time, so our 3rd digit is 1.
3. The remainder when dividing 51 by 36 is 15.
4. Now we see how many times $6^1$ goes into 15 (which is 2 times, so our 2nd digits is 2).
5. The remainder when dividing 15 by 6 is 3.
6. $6^0 = 1$ divides 3 obviously 3 times, so our 1st digit is 3
7. So after conversion, $51_{10} = 123_6$, which corresponds to what we expected.

As a warning, some digits might be zero when you do the base conversion. Let’s look at an example of this: Convert $18_{10}$ to base-4:

$4^2 = 16$ and $4^3 = 64$, so $4^2 = 16$ goes into 18 once with a remainder of 2: Third Digit is 1 Now $4^1 = 4$ doesn’t go into 2: Second Digit is 0 $4^0$ goes into 2 twice: First Digit is 2 Answer: $102_4$

This seems like a lot of steps in making a base conversion, but after substantial practice, it will become second nature. Here are some practice problems with just converting bases from base-n to base-10 and reverse.

Problem Set 3.2.1