Electronics Class Notes — Boolean Algebra

Teacher: Prof P. M. Sarun • NPHC206 • WINTER - 2025-2026 • Last updated:

Boolean Algebra

Digital circuits perform the binary arithmetic operations with binary digits \(1\) and \(0\). These opera tions are called logic functions or logical operations. The algebra used to symbolically describe logic functions is called Boolean algebra. Boolean algebra is a set of rules and theorems by which logical operations can be expressed symbolically in equation form and be manipulated mathemati cally. As with the ordinary algebra, the letters of alphabet (e.g. \(A\), \(B\), \(C\) etc.) can be used to represent the variables. Boolean algebra differs from ordinary algebra in that Boolean constants and variables can have only two values ; \(0\) and \(1\). There are four connecting symbols used in Boolean algebra viz.

  • (a) equals sign (\(=\))
  • (b) plus sign (\(+\))
  • (c) multiply sign (\(·\))
  • (d) bar (\(–\))

Equals sign (=)

The equals sign in Boolean algebra refers to the standard mathematical equality. In other words, the logical value on one side of the sign is identical to the logical value on the other side of the sign. Suppose we are given two logical variables such that \(A = B\). Then if \(A = 1\), then \(B = 1\) and if \(A = 0\), then \(B = 0\).

Plus sign (+)

The plus sign in Boolean algebra refers to the logical OR operation. Thus, when the statement \(A + B = 1\) appears in Boolean algebra, it means \(A\) ORed with \(B\) equals \(1\). Consequently, either \(A = 1\) or \(B = 1\) or both equal \(1\).

Multiply sign (·)

The multiply sign in Boolean algebra refers to AND operation. Thus, when the statement \(A · B = 1\) appears in Boolean algebra, it means \(A\) ANDed with \(B\) equals \(1\). Consequently, \(A = 1\) and \(B = 1\). The function \(A · B\) is often written as \(AB\), omitting the dot for convenience.

Bar sign (–)

The bar sign in Boolean algebra refers to NOT operation. The NOT has the effect of inverting (complementing) the logical value. Thus, if A = 1, then \(\overline{A}\) = 0.