Advertisements
Advertisements
Question
Write the truth value of the negation of the following statement :
cos2 θ + sin2 θ = 1, for all θ ∈ R
Advertisements
Solution
p : cos2 θ + sin2 θ = 1, for all θ ∈ R
The truth value of p is T.
∴ The truth value of ~p is F
APPEARS IN
RELATED QUESTIONS
The negation of p ∧ (q → r) is ______________.
Rewrite the following statement without using if ...... then.
It 2 is a rational number then `sqrt2` is irrational number.
Without using truth table prove that:
(p ∨ q) ∧ (p ∨ ∼ q) ≡ p
Using rules in logic, prove the following:
p ↔ q ≡ ∼(p ∧ ∼q) ∧ ∼(q ∧ ∼p)
Using rules in logic, prove the following:
∼ (p ∨ q) ∨ (∼p ∧ q) ≡ ∼p
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.
Without using truth table, show that
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Without using truth table, show that
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
Using the algebra of statement, prove that
(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)
For any two statements p and q, the negation of the expression (p ∧ ∼q) ∧ ∼p is ______
The negation of the Boolean expression (r ∧ ∼s) ∨ s is equivalent to: ______
∼ ((∼ p) ∧ q) is equal to ______.
Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.
The simplified form of [(~ p v q) ∧ r] v [(p ∧ ~ q) ∧ r] is ______.
The statement p → (q → p) is equivalent to ______.
Show that the simplified form of (p ∧ q ∧ ∼ r) ∨ (r ∧ p ∧ q) ∨ (∼ p ∨ q) is q ∨ ∼ p.
The logically equivalent statement of \[\left(\sim p\wedge q\right)\vee\left(\sim p\wedge\sim q\right)\] \[\vee\left(\ p\wedge\sim q\right)\] is
