Math Problem Statement

A'(a+b)+(b+aa)(a+b') computer organisation Boolean algebra

Solution

Let's break down the given Boolean expression step by step, applying Boolean algebra rules.

Boolean Expression:

A(a+b)+(b+aa)(a+b)A'(a + b) + (b + aa)(a + b')

  1. First part: A(a+b)A'(a + b)

    • Distribute AA' over the sum (a+b)(a + b): Aa+AbA'a + A'b
  2. Second part: (b+aa)(a+b)(b + aa)(a + b')

    • Simplify aaaa. In Boolean algebra, aa=aaa = a, so: (b+a)(a+b)(b + a)(a + b')
    • Now, distribute over the two terms: ba+bb+aa+abba + bb' + aa + ab'
    • Using the complement rule bb=0bb' = 0: ba+0+a+abba + 0 + a + ab'
    • Simplifying gives: ba+a+abba + a + ab'
  3. Final step: Combine both parts: Aa+Ab+(ba+a+ab)A'a + A'b + (ba + a + ab')

    • Since a+a=aa + a = a (idempotent law), we simplify further: Aa+Ab+ba+abA'a + A'b + ba + ab'

This is the simplified Boolean expression: Aa+Ab+ba+abA'a + A'b + ba + ab'

Let me know if you'd like more details on the steps or have any specific questions!


Here are 5 related questions to expand your understanding:

  1. What is the idempotent law in Boolean algebra?
  2. How does the complement rule work in Boolean algebra?
  3. Can you explain the distribution property in Boolean expressions?
  4. What is the significance of the absorption law in simplifying expressions?
  5. How would the expression change if A=1A = 1?

Tip: Remember that simplifying Boolean expressions often involves applying rules multiple times, so recognizing patterns is key!

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Math Problem Analysis

Mathematical Concepts

Boolean Algebra
Simplification of Boolean Expressions

Formulas

Distributive Law: A(B + C) = AB + AC
Complement Law: A + A' = 1, A * A' = 0
Idempotent Law: A + A = A, A * A = A

Theorems

De Morgan's Theorem
Absorption Law

Suitable Grade Level

Grades 10-12