Tag: proofs in mathematics

Questions Related to proofs in mathematics

Which of the following is always true ? 

  1. $\left( {p \to q} \right) \cong \left( { \sim q \to \sim p} \right)$

  2. $ \sim \left( {p \vee q} \right) \cong \left( { \sim p \vee \sim q} \right)$

  3. $ \sim \left( {p \to q} \right) \cong \left( {p \vee \sim q} \right)$

  4. $ \sim \left( {p \wedge q} \right) \cong \left( { \sim p \wedge \sim q} \right)$


Correct Option: C
Explanation:

We know that 


$p\rightarrow q\equiv \sim p\wedge q$

$\sim (p\rightarrow q)\equiv \sim (\sim p\wedge q)$

$\sim (p\rightarrow q)\equiv p\vee \sim q$               (De morgan's law)

$C$ is correct

The Boolean expression $ \sim\ ( p \vee q ) \vee ( \sim\ p \wedge q ) $ is equivalent to:

  1. $p$

  2. $q$

  3. $ \sim q $

  4. $ \sim p $


Correct Option: D
Explanation:
$\sim (p\vee q) \vee  (\sim p\wedge q)$
$=(\sim p \wedge \sim q) \vee (\sim p\wedge q)$
$=\sim p\wedge (\sim q\vee q)$
$=\sim p\wedge T$
$=\sim p$

$p \leftrightarrow q \equiv  \sim \left( {p\Delta  \sim q} \right)\Delta  \sim \left( {q\Delta  \sim p} \right)$

  1. True

  2. False


Correct Option: A

Let $p$ and $q$ be two statements, then $ \sim ( \sim p \wedge q) \wedge (p \vee q)$ is logically equivalent to 

  1. $q$

  2. $p\vee q$

  3. $p$

  4. $p\vee \sim q$


Correct Option: D

$ \sim (p \wedge q) \to ( \sim p \vee ( \sim p \vee q))$  is equivalent to 

  1. $p \vee \sim q$

  2. $p \wedge \sim q$

  3. $ \sim p \vee q$

  4. $ \sim p \wedge q$


Correct Option: B

$\left( { \sim p\Delta q} \right)V\left( { \sim p\Delta  \sim q} \right)V\left( { \sim p\Delta  \sim q} \right) \equiv  \sim pV \sim q$

  1. True

  2. False


Correct Option: A

The compound proposition which is always false is:

  1. $\left(p \rightarrow q\right)\leftrightarrow \left( \sim q \rightarrow \sim p \right) $

  2. $\left[ \left( p\rightarrow q \right) \wedge \left( q\rightarrow r \right) \right]\rightarrow \left( p\rightarrow r \right) $

  3. $\left( \sim p\vee q \right) \leftrightarrow \left( p\wedge \sim q \right) $

  4. $p \rightarrow \sim p$


Correct Option: A

$p \wedge ( q \wedge r )$  is logically equivalent to

  1. $p \vee ( q \wedge r )$

  2. $( p \wedge q ) \wedge r$

  3. $( p \vee q ) \vee r$

  4. $p \rightarrow ( q \wedge r )$


Correct Option: B

If  $p$ and  $q$ are two simple proposition then  $p \rightarrow q$  is false when

  1. $p \text { is true and } q \text{ is true}$

  2. $p \text { is false and } q  \text{ is true}$

  3. $p \text { is true and } q \text{ is false}$

  4. both $p$ and $q$ are false


Correct Option: B

Let  $p :$  Mathematics is interesting and let  $q:$  Mathematics is difficult, then the symbol  $p\wedge q$  means

  1. Mathematics is interesting implies that Mathematics is difficult

  2. Mathematics is interesting implies and is implied by Mathematics is difficult

  3. Mathematics is interesting and Mathematics is difficult

  4. Mathematics is interesting or Mathematics is difficult


Correct Option: C
Explanation:

$'\Lambda '$ stands for logical and 

$\therefore$    $p\Lambda q$ means 
Mathematics is interesting and Mathematics is difficult.