Exercises 5.2 Exercises
1.
Prove the identity property expressed by Equation (5.1.3) and Equation (5.1.4).
Solution
For Equation (5.1.3):
| \(x\) | \(x \cdot 1\) |
| \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{1}\) |
And for Equation (5.1.4):
| \(x\) | \(x + 0\) |
| \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{1}\) |
2.
Prove the commutative property expressed by Equation (5.1.5) and Equation (5.1.6).
Solution
For Equation (5.1.5):
| \(x\) | \(y\) | \(x \cdot y\) | \(y \cdot x\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
And for Equation (5.1.6):
| \(x\) | \(y\) | \(x + y\) | \(y + x\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
3.
Prove the annulment property expressed by Equation (5.1.7) and Equation (5.1.8).
Solution
For Equation (5.1.7):
| \(x\) | \(x \cdot 0\) |
| \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{0}\) |
And for Equation (5.1.8):
| \(x\) | \(x + 1\) |
| \(\binary{0}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) |
4.
Prove the complement property expressed by Equation (5.1.9) and Equation (5.1.10).
Solution
For Equation (5.1.9):
| \(x\) | \(x'\) | \(x \cdot x'\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |
And for Equation (5.1.10):
| \(x\) | \(x'\) | \(x + x'\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) |
5.
Prove the idempotent property expressed by Equation (5.1.11) and Equation (5.1.12).
Solution
For Equation (5.1.11):
| \(x\) | \(x\) | \(x \cdot x\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
And for Equation (5.1.12):
| \(x\) | \(x\) | \(x + x\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
6.
Prove the distributive property expressed by Equation (5.1.13) and Equation (5.1.14).
Solution
For Equation (5.1.13):
| \(x\) | \(y\) | \(z\) | \(y + z\) | \(x \cdot (y + z)\) | \(x \cdot y\) | \(x \cdot z\) | \(x \cdot y + x \cdot z\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
And for Equation (5.1.14):
| \(x\) | \(y\) | \(z\) | \(y \cdot z\) | \(x + y \cdot z\) | \(x + y\) | \(x + z\) | \((x + y) \cdot (x + z)\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) |
| \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{0}\) | \(\binary{0}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |
| \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) | \(\binary{1}\) |