Tag: discrete mathematics

Quizzes (6) Questions (61)

Questions Related to discrete mathematics

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

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

  1. tautology

  2. contradiction

  3. dualoty

  4. double implication
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Evaluating the expression using truth tables or logical equivalences shows that it evaluates to false for all truth values of the variables. The components simplify such that a proposition and its negation are strictly contradicted, yielding a contradiction.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following is not correct ?

  1. $p \vee \sim p $ is a tautology.
  2. $\sim (\sim p) \leftrightarrow p$ is a tautology.
  3. $p \wedge \sim p $ is a contradiction.
  4. $([(p \wedge p ) \rightarrow q] \rightarrow p)$ is a tautology.
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

$\because [(p \wedge p ) \rightarrow q ] \rightarrow p \equiv ( p

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

when $p$ is false and $q$ is true (or false) then

$(p \rightarrow q) $ is true i.e. $(p \rightarrow q ) ]\rightarrow p$ is false

Hence $[(p \wedge p ) \rightarrow q] \rightarrow p$ is not a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following statement is a tautology?

  1. $(\sim p \vee \sim q ) \vee ( p \vee \sim q )$
  2. $(\sim p \vee \sim q ) \wedge (p \vee \sim q )$
  3. $\sim p \wedge (\sim p \vee \sim q )$
  4. $\sim q \wedge (\sim p \vee \sim q )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\because (\sim p  \vee \sim q) \vee (p  \vee \sim q) $

$\equiv (\sim p \vee p ) \vee (p \vee \sim q )$    (by distributive law)

$ \equiv t \vee \sim q \equiv t $                      t is a tautology

Hence $(\sim p  \vee \sim, q ) \vee (p  \vee \sim q )$ is a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$ p\Rightarrow p \vee q$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$p$       $q$      $p\vee q$       $p\rightarrow (p\vee q)$
T T     T            T
T F    T            T
F T    T              T
F F     F            T

Since, all the entries in the last column has true value. So, the given statement is a tautology

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$p\Rightarrow \sim p$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
 $p$  $\sim p$  $p\Rightarrow \sim p$
 T  F  F
 F  T  T

So the result of the Truth table show that $p\Rightarrow \sim p$ is neither a tautology not a contradiction

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

$ p\wedge  (\sim p)$ is

  1. a tautology.

  2. a contradiction.

  3. a tautology and a contradiction.

  4. neither a tautology nor a contradiction.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Truth table is,

$p$         $\sim p$ $p\wedge (\sim p)$
T                      F                      F
F T F

Hence given statement is contradiction.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following is a contradiction?

  1. $p\vee q$
  2. $p\wedge q$
  3. $p\vee (\sim p)$
  4. $p\wedge (\sim p)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$p$  $q$  $p\vee q$  $p\wedge q$
$ T$ $ T$ $ T$ $ T$
$ T$ $ F$ $ T$ $ F$
$ F$ $T$ $T$ $F$
$ F$ $F$ $ F$ $ F$
So $p\vee q$ and $p\wedge q$ are not contradiction 

Now check other options:

| $p$ |  $\sim p$ |  $p\vee (\sim p)$ |  $p\wedge (\sim p)$ | | --- | --- | --- | --- | |  $T$ | $ F$ | $ T$ | $ F$ | | $ F$ | $T$ | $T$ | $F$      |
Hence $p\wedge (\sim p)$ is contradiction 

Note: If a compound statement is always False , then it is called contradiction 
Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The proposition $(p\rightarrow \sim p)\wedge (\sim p\rightarrow q)$ is

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. a tautology and a contradiction
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$p$      $q$      $\sim p$      $p\rightarrow \sim p$     $\sim p \rightarrow q$     $(p\rightarrow \sim p)\wedge(\sim p\rightarrow q)$
T T    F      F       T         F
T F    F      F       T         F
F T    T      T       T         T
F F    T      T       F         F
Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The statement $(p-q)\rightarrow [(\sim p \rightarrow q)\rightarrow q]$ is 

  1. a tautology
  2. equivalent to $\sim p \rightarrow q$
  3. equivalent to $p\rightarrow \sim q$

  4. a fallacy
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The statement (p -> q) -> ((~p -> q) -> q) is a tautology. It can be verified using a truth table or logical equivalences.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Which of the following proposition is a contradiction?

  1. $(\sim p\vee \sim q)\vee (p\vee \sim q)$
  2. $(p\rightarrow q)\vee (p\wedge \sim q)$
  3. $(\sim p\wedge q)\wedge (\sim q)$
  4. $(\sim p\wedge q)\vee (\sim q)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Contradiction is a preposition which is always false(F).
Here $\sim \equiv negation$ and $\wedge \equiv AND$
Let $ x=(\sim p \wedge q) \wedge (\sim q)$
If $p=F$ and $q=F$ then $x=F$
If $p=F$ and $q=T$ then $x=F$
If $p=T$ and $q=F$ then $x=F$
If $p=T$ and $q=T$ then $x=F$
Hence option (c) is correct