Advertisements
Advertisements
प्रश्न
Write the truth value of the negation of the following statement :
cos2 θ + sin2 θ = 1, for all θ ∈ R
Advertisements
उत्तर
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
संबंधित प्रश्न
The negation of p ∧ (q → r) is ______________.
Without using truth tabic show that ~(p v q)v(~p ∧ q) = ~p
Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)
Using the rules of negation, write the negatlon of the following:
(a) p ∧ (q → r)
(b) ~P ∨ ~q
Rewrite the following statement without using if ...... then.
It 2 is a rational number then `sqrt2` is irrational number.
Rewrite the following statement without using if ...... then.
It f(2) = 0 then f(x) is divisible by (x – 2).
Using rules in logic, prove the following:
∼p ∧ q ≡ (p ∨ q) ∧ ∼p
Using the rules in logic, write the negation of the following:
p ∧ (q ∨ r)
Let p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r). Then, this law is known as ______.
Without using truth table, show that
~ [(p ∧ q) → ~ q] ≡ p ∧ q
Without using truth table, show that
(p ∨ q) → r ≡ (p → r) ∧ (q → r)
Using the algebra of statement, prove that
(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)
Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q).
The logically equivalent statement of (p ∨ q) ∧ (p ∨ r) is ______
Without using truth table, prove that : [(p ∨ q) ∧ ∼p] →q is a tautology.
Without using truth table prove that
[(p ∧ q ∧ ∼ p) ∨ (∼ p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ∼ q ∧ r) ≡ (p ∨ q) ∧ r
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.
