3.2.4 Changing Between Bases: Special Case

When changing between two bases $m$ and $n$, the standard procedure is to convert the number from base-m to base-10 then convert that into base-n. However, there are special cases when the middle conversion into base-10 is unnecessary: when $n$ is an integral power of $m$ (say $n = m^a$, $a$ an integer) or vice versa. The procedure is relatively simple, take the digits of $m$ in groups of $a$ and convert each group into base-n. For example, if we are converting $1001001_2$ into base-4, you would take 1001001 in groups of two (since $2^2 = 4$) and converting each group into base-4. Let’s see how it would look:

Convert $1001001_2$ to base-4

• First Digit: $01_2 \Rightarrow 1_4$
• Second Digit: $10_2 \Rightarrow 2_4$
• Third Digit: $00_2 \Rightarrow 0_4$
• Fourth Digit: $1_2 \Rightarrow 1_4$
• Answer: $1021_4$

Let’s look at an example where the converting base is that of the original base cubed (so you would take it in groups of 3):

Convert $110001011_2$ to base-8

• First Digit: $011_2 \Rightarrow 3_8$
• Second Digit: $001_2 \Rightarrow 1_8$
• Third Digit: $110_2 \Rightarrow 6_8$
• Answer: $613_8$

Similarly, you can perform the method in reverse. So when converting from base-9 to base-3 you would take each digit in base-9 and convert it to two-digit base-3 representation. For example:

Convert $643_9$ to base-3

• First/Second Digits: $3_9 \Rightarrow 10_3$
• Third/Fourth Digits: $4_9 \Rightarrow 11_3$
• Fifth/Sixth Digits: $6_9 \Rightarrow 20_3$
• Answer: $201110_3$

Problem Set 3.2.4