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Write the negation of the following simple statements:

i. The number 17 is prime.

ii. 2 + 7 = 6.

iii. Violets are blue.

iv. √ 5 is a rational number.

v. 2 is not a prime number.

vi. Every real number is an irrational number.

vii. Cow has four legs.

viii. A leap year has 366 days.

ix. All similar triangles are congruent.

x.         Area of a circle is the same as the perimeter of the circle.

0 4 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 )         T h e     n u m b e r     1 7     i s     n o t     p r i m e . ( i i )       2 + 7 ≠ 6 . ( i i i )     V i o l e t s     a r e     n o t     b l u e . ( i v )     5     i s     n o t     a     r a t i o n a l     n u m b e r . ( v )         2     i s     a     p r i m e     n u m b e r . ( v i )     E v e r y     r e a l     n u m b e r     i s     n o t     a n     i r r a t i o n a l     n u m b e r . ( v i i )     C o w     d o e s     n o t     h a v e     f o u r     l e g s . ( v i i i )   A     l e a p     y e a r     d o e s     n o t     h a v e     3 6 6     d a y s . ( i x )         T h e r e     e x i s t     s i m i l a r     t r i a n g l e s     w h i c h     a r e     n o t     c o n g r u e n t . ( x )           A r e a     o f     a     c i r c l e     i s     n o t     s a m e     a s     t h e     p e r i m e t e r     o f     t h e     c i r c l e .

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:
R
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

...more

Negation of the Boolean statement ( p ∨ q ) ⇒ ( ( ∼ r ) ∨ p )  is equivalent to
V
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:
R
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?
R
Raj Pandey

kindly consider the following Image 

 

The statement ( ∼ ( p ⇔ ∼ q ) ) ∧ q     i s :  
R
Raj Pandey

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

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

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