Binary Number

Taught in ECE 124

Binary Numbers §

Digital Circuits typically use binary number system where:

• The allowed symbols in the number representation are 0 and 1 (bits)
• The radix is 2
• For an m-bit number $b_{m-1},b_{m-2}\dots ,b_0$ where $b_{\ast }$’s are either 0 or 1

$V=b_{m-1}\ast 2^{m-1}+b_{m-2}\ast 2^{m-2}+\cdots +b_0\ast 2^0$

Example: $125=1\ast 2^6+1\ast 2^5+1\ast 2^4+1\ast 2^3+1\ast 2^2+0\ast 2^1\ast 1\ast 2^0$

Value of 125 is represented as $(125{)}_{10}$ in decimal and $(1111101{)}_2$

Conversion §

Given a positive value V in decimal, find it’s binary representation

• If we divide V by 2 we get a quotient and a remainder. The remainder is 0 or 1. Denoting the Quotient as $Q_1$ and the remainder as $b_0$, we can write
  • $V=Q_1\ast 2+b_0$ (e.g., $27=13\ast 2+1$)
• If $Q_1>0$ we divide it by 2 and can write $Q_1=Q_2\ast 2+b_1$ where $Q_2$ and $b_1$ denote the resulting quotient and remainder, respectively. Now plugging this expression of $Q_1$ into that of aforementioned V, we have

$V=(Q_2\ast 2+b_1)\ast 2+b_0=Q_2\ast 2^2+b_1\ast 2+b_0$

E.g., $27=6\ast 2^2+1\ast 2+1$

We repeat the process until resulting quotient becomes 0

• Suppose this happens after m iterations, i.e., $Q_m$ is 0
• We denote the remainders resulted in the process as $b_0,b_1,b_2,\dots ,b_{m-1}$ which are 0 or 1
• Then
  • $V=b_{m-1}\ast 2^{m-1}+b_{m-2}\ast 2^{m-2}+\cdots +b_1\ast 2^1+b_0$
  • e.g., $27=1\ast 2^4+1\ast 2^3+0\ast 2^2+1\ast 2+1$