What is the binary number for decimal number 9?

Concept — In a base-2 (binary) positional system each digit position carries a weight that is a power of 2. Reading a numeral from right to left the weights are 20, 21, 22, 23, … , and the value of the numeral is the sum of the weights whose digit is 1. Converting a decimal number to binary therefore means writing it as a sum of distinct powers of 2; the mechanical way to find those powers is to divide repeatedly by 2 and read the remainders from the last division backwards.

Application — Divide 9 by 2 repeatedly and record every remainder:

1. 9 ÷ 2 = 4, remainder 1 — this remainder is the least-significant bit.

2. 4 ÷ 2 = 2, remainder 0.

3. 2 ÷ 2 = 1, remainder 0.

4. 1 ÷ 2 = 0, remainder 1 — this remainder is the most-significant bit.

Reading the remainders from the last division back to the first gives 10012. The sum-of-powers view gives the same thing: 9 = 8 + 1 = 23 + 20, so the bits at the weights 8 and 1 are set and the bits at the weights 4 and 2 are clear.

Cross-check — Expand the result back: 1×8 + 0×4 + 0×2 + 1×1 = 9, the number we started from. Comparing the neighbouring 4-bit patterns shows how each differs:

• 11002 sets the weights 8 and 4, giving 1210.

• 01112 sets the weights 4, 2 and 1, giving 710.

• 10112 sets the weights 8, 2 and 1, giving 1110.