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Rewrite each of the following statements in the form of conditional statements:

i. The square of an odd number is odd.

ii. You will get a sweet dish after the dinner.

iii. You will fail if you do not study.

iv. The unit digit of an integer is 0 or 5 if it is divisible by 5.

v. The square of a prime number is not prime.

vi. 2b = a + c if a, b, and c are in A.P.

0 2 Views | Posted 3 months ago
Asked by Shiksha User

  • 1 Answer

  • P

    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

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

    ( i ) I f     p ,     t h e n     q ⇒ I f     t h e     n u m b e r     i s     o d d ,     t h e n     i t s     s q u a r e     i s     o d d     n u m b e r . ( i i ) q     i f     p ⇒ I f     t a k e     t h e     d i n n e r ,     t h e n     y o u     w i l l     g e t     s w e e t     d i s h . ( i i i ) p     o n l y     i f     q ⇒ I f     y o u     d o     n o t     s t u d y ,     t h e n     y o u     w i l l     f a i l . (iv)p  is  sufficient  for  q⇒If  an  integer  is  divisible  by  5,  then  its  unit  digits  are  0  and  5. ( v ) q     i s     n e c e s s a r y     f o r     p ⇒ I f     a n y     n u m b e r     i s     p r i m e ,     t h e n     i t s     s q u a r e     i s     n o t     p r i m e . ( v i ) q     i m p l i e s     p ⇒ I f     a , b , c     a r e     i n     A . P     t h e n     2 b = a + c .

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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