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Write down the negation of the following compound statements:

i. All rational numbers are real and complex.

ii. All real numbers are rational or irrationals.

iii. x = 2 and x = 3 are roots of the quadratic equation x² – 5x + 6 = 0.

iv.  A triangle has either 3 sides or 4 sides.

v. 35 is a prime number or a composite number.

vi. All prime integers are either even or odd.

 vii. x is equal to either x or –x.

viii. 6 is divisible by 2 and 3.

0 2 Views | Posted 3 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a short answer type question as classified in NCERT Exemplar

    ( i ) p : A l l     r a t i o n a l     n u m b e r s     a r e     r e a l .     ∼ p : A l l     r a t i o n a l     n u m b e r s     a r e     n o t     r e a l .               q : A l l     r a t i o n a l     n u m b e r s     a r e     c o m p l e x .     ∼ q : A l l     r a t i o n a l     n u m b e r s     a r e     n o t     c o m p l e x .     ∼ ( p ∧ q ) = ( ∼ p ∨ ∼ q ) : A l l     r a t i o n a l     n u m b e r s     a r e     n e i t h e r     r e a l     n o r     c o m p l e x . ( i i ) p : A l l     r e a l     n u m b e r s     a r e     r a t i o n a l s .                 q : A l l     r e a l     n u m b e r s     a r e     i r r a t i o n a l s .               T h e     n e g a t i o n     o f     t h e     a b o v e     s t a t e m e n t     i s     ∼ ( p ∨ q ) = ( ∼ p ∧ ∼ q ) : A l l     r e a l     n u m b e r s     a r e     n o t     r a t i o n a l s     a n d     a l l     r e a l     n u m b e r s     a r e           n o t     i r r a t i o n a l s . ( i i i ) p : x = 2     i s     r o o t     o f     t h e     e q u a t i o n     x 2 − 5 x + 6 = 0 .                   q : x = 3     i s     r o o t     o f     t h e     e q u a t i o n     x 2 − 5 x + 6 = 0 .                 T h e     n e g a t i o n     o f     t h e     a b o v e     s t a t e m e n t     i s   ∼ ( p ∧ q ) = ( ∼ p ∨ ∼ q ) : x = 2     i s     n o t     t h e     r o o t     o f     t h e     e q u a t i o n     x 2 − 5 x + 6 = 0     o r     x = 3     i s               n o t     t h e     r o o t     o f     t h e     e q u a t i o n     x 2 − 5 x + 6 = 0 . (iv)p:A  triangle  has  3-sides.          q:A  triangle  has  4-sides.                 T h e     n e g a t i o n     o f     t h e     a b o v e     s t a t e m e n t     i s   ∼(p∨q)=(∼p∧∼q):A  triangle  has  neither  3-sides  nor   4-sides. ( v ) p : 3 5     i s     a     p r i m e     n u m b e r .                 q : 3 5     i s     a     c o m p o s i t e     n u m b e r .                 T h e     n e g a t i o n     o f     t h e     a b o v e     s t a t e m e n t     i s     ∼ ( p ∨ q ) = ( ∼ p ∧ ∼ q ) : 3 5     i s     n o t     a     p r i m e     n u m b e r     a n d     i t &thins

Similar Questions for you

Let Δ , ∇ ∈ { ∧ , ∨ } be such that p ∇ q ⇒ ( ( p Δ q ) ∇ r ) is a tautology. Then ( p ∇ q ) Δ   r  is logically equivalent to:
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Raj Pandey

Case – I Δ ≡ ∇ ≡ ∧  

              ( p ∧ q ) → ( ( p ∧ q ) ∧ r )  

              it can be false if r is false,

              so not a tautology

              Case – II If Δ ≡ ∇ ≡ ∨  

              ( p ∨ q ) ?     → ( ( p ∨ q ) ∨ r ) ≡ tautology

              then ( p ∨ q ) ∨ r ≡ ( p Δ r ) ∨ q   

              Case – I

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Negation of the Boolean statement ( p ∨ q ) ⇒ ( ( ∼ r ) ∨ p )  is equivalent to
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Vishal Baghel

pva ( ∼ r   v   p )

∼ ( p ∨ q ) ∨ ( ∼ r v p )

its negation as asked in question

( p ∨ q ) ∧ ( ∼ p ∧ r )

= ( p ∧ ∼ p ∧ r ) ∨ ( q ∧ r ∧ ∼ p )

= ( p ∧ r ∧ ∼ p )                   [ a s     p ∧ ∼ p     i s     f a l s e ]

The remainder when (2021)2023 is divided by 7 is:
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Raj Pandey

2 0 2 1 ≡ − 2 mod (7)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ ( − 2 ) 2 0 2 3 m o d ( 7 )   …… (i)

Now,   ( − 2 ) 3 ≡ − 1 m o d ( 7 )  

⇒ ( − 2 ) 2 0 2 3 ≡ ( − 2 ) m o d ( 7 ) ≡ 5 m o d ( 7 )       ……. (ii)

(i) & (ii)

⇒ ( 2 0 2 1 ) 2 0 2 3 ≡ 5 m o d ( 7 )  

∴  Remainder = 5

Let p, q, r be three logical statements. Consider the compound statements

  S 1 : ( ( ∼ p ) ∨ q ) ∨ ( ( ∼ p ) ∨ r )     a n d  

  S 2 : p → ( q ∨ r )  

 Then, which of the following is NOT true?
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kindly consider the following Image 

 

The statement ( ∼ ( p ⇔ ∼ q ) ) ∧ q     i s :  
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∼ ( p ⇔ ∼ q ) ∧ q = ( p ⇔ ∼ q ) ∧ q     i s :

∴ ( ∼ ( P ⇔ ∼ Q ) ) ∧ q is equivalent to ( p ⇒ q ) ∧

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