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Published on: 13/05/2022
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Take MCQ Maths Test1.
Let \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) ,C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)be any three boolean matrices of the same type.
Find (A∧B)∨C
2.
Let \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) ,C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)be any three boolean matrices of the same type.
Find AΛB
3.
Establish the equivalence property p ➝ q ≡ ㄱp ν q
4.
Show that ¬(p↔️q) ≡ p↔️¬q
5.
Show that p ➝ q and q ➝ p are not equivalent
6.
Verify whether the following compound propositions are tautologies or contradictions or contingency
( p ⟶ q) ↔️ (~ p ⟶ q)
7.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(( p V q)∧ ¬ p) ➝ q
8.
Verify whether the following compound propositions are tautologies or contradictions or contingency
(p ∧ q) ∧ ¬ (p ∨ q)
9.
Prove that q ➝ p ≡ ¬p ➝ ¬q
10.
Show that ¬( p ∧ q) ≡ ¬p V ¬q
1.
Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\((A\wedge B)\vee C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
2.
Given \(A=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) ,B=\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) and\quad C=\left( \begin{matrix} 1 & 1 \\ 0 & 1 \\ 1 & 1 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 0 \end{matrix}\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{matrix} \right) \vee \left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 1\wedge 1 & 0\wedge 0 \end{matrix}\begin{matrix} 1\wedge 0 & 0\wedge 1 \\ 0\wedge 1 & 1\wedge 0 \\ 0\wedge 0 & 1\wedge 1 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}\begin{matrix} 0 & 0 \\ 0 & 0 \\ 0 & 1 \end{matrix} \right) \) [∵ a៱b=max(a,b)]
3.
| p | q | ㄱp | p ➝ q | ㄱp ν q |
| T | T | F | T | T |
| T | F | F | F | F |
| F | T | T | T | T |
| F | F | T | T | T |
The entries in the columns corresponding to p → q and ㄱp ν q are identical and hence they are equivalent.
4.
| p | q | p↔️q | ~(p↔️q) | ~q | p↔️~q |
| T | T | T | F | F | F |
| T | F | F | T | T | T |
| F | T | F | T | F | T |
| F | F | T | F | T | F |
The entries in column (4) and (6) are identical ~(p↔️q) ≡ p↔️~q
5.
| p | q | p ➝ q | q ➝ p |
| T | T | T | T |
| T | F | F | T |
| F | T | T | F |
| F | F | T | T |
The entries in column (3) and column (4) are not identical.
6.
| p | q | p ⟶ q | ~p | ~p ⟶ q | ( p ⟶ q) ↔️ (~p ⟶ q) |
| T | T | T | F | T | T |
| T | F | F | F | T | F |
| F | T | T | T | T | T |
| F | F | T | T | F | F |
Since this is neither a tautology not a contradiction
( p ⟶ q) ↔️ (~p ⟶ q) is a contingency.
7.
(( p V q)∧ ~p)) ➝ q
| p | q | p V q | ~p | ( p V q) ∧ ~q | ( p V q) ∧ ~q |
| T | T | T | F | F | T |
| T | F | T | F | F | T |
| F | T | T | T | T | T |
| F | F | F | T | F | T |
The statement (( p V q)∧ ~p) ➝ q is a tautology.
8.
| p | q | p ∧ q | p ∨ q | ~p ∨ q | (p ∧ q) ∧ ~(p ∨ q) |
| T | T | T | T | F | F |
| T | F | F | T | F | F |
| F | T | F | T | F | F |
| F | F | F | F | T | F |
The statement (p ∧ q) ∧ ~(p ∨ q)
9.
| p | q | q ➝ p | ~p | ~q | ~q ➝ ~p |
| T | T | T | F | F | T |
| T | F | T | F | T | T |
| F | T | F | T | F | F |
| F | F | T | T | T | T |
The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved
10.
~( p ∧ q) ≡ ~p V ~q
| p | q | p ∧ q | ~(p ∧ q) | ~p | ~p | ~p V ~q |
| T | T | T | F | F | F | F |
| T | F | F | T | F | T | T |
| F | T | F | T | T | F | T |
| F | F | F | T | T | T | T |
From column (4) and column (7), the entries are Identical.
ஃ ~( p ∧ q) ≡ ~p V ~q are identical and hence they are equivalent.
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Tamilnadu Stateboard 12th Standard Subjects
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