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Using the contrapositive method, prove that if n 2  is an even integer, then n  is also an even integer.

0 3 Views | Posted 3 months ago
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  • 1 Answer

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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

    Let  p:n2  is  an  eveninteger  .          q:n  is  also  an  even  integer.Also  let  ∼q  is  true  i.e.,  n  is  not  an  even  integer.⇒n2  is  not  an  even  integer.⇒∼p  is  true.                                          [?Square  of  an  odd  integer  is  odd]Hence,  ∼q  is  true⇒∼p  is  true.

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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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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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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 )  

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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 )  

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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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