Tag: discrete mathematics

Quizzes (6) Questions (61)

Questions Related to discrete mathematics

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The only statement among the following taht is a tautology is -

  1. $A\wedge (A\vee B)$
  2. $A\vee (A\wedge B)$
  3. $[A\wedge (A\rightarrow B)]\rightarrow B$
  4. $B\rightarrow [A\wedge (A\rightarrow B)]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Option C represents the standard logical rule of detachment, or modus ponens, which is a tautology. Evaluating its truth table yields true in every row.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The contrapositive of the statement "if  $2 ^ { 2 } = 5 ,$  then  $1$  get first class" is

  1. If I do not get a first class, then $2 ^ { 2 } = 5$
  2. If I do not get a first class, then $2 ^ { 2 } \neq 5$
  3. If I get a first class, then $2 ^ { 2 } = 5$
  4. If I get a first class, then $2 ^ { 3 } = 5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P:{ 2 }^{ 2 }=5$

$q:I$ get first class
the contrapositive of $p\rightarrow q$ is $\sim q\rightarrow \sim p$. Hence the answer is if $I$ do not get a first class, then ${ 2 }^{ 2 }\neq 5$
Correct Answer : Option B.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The statement  $\sim ( p \wedge q ) \vee q$

  1. is a tautology
  2. is equivalent to $( p \wedge q ) \vee ( - q )$
  3. is equivalent to $p \vee q$

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

The expression ~(p ^ q) v q is equivalent to (~p v ~q) v q. By associativity, this is ~p v (~q v q), which is ~p v T, which is T. Thus, it is a tautology.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

Consider :
Statement - I :$(p\wedge \sim q)\wedge (\sim p\wedge q)$ is a fallacy.
Statement - II :$(p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)$ is a tautology.

  1. Statement - I is true: Statement - II is true: Statement - II is a correct explanation for Statement - I.

  2. Statement - I is true: Statement - II is true: Statement - II is not a correct explanation for Statement - I.

  3. Statement - I is true; Statement - II is false.

  4. Statement - I is false; Statement - II is true.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Statement I is true because the expression contains contradictory sub-parts making it a fallacy or contradiction. Statement II is true because a conditional is logically equivalent to its contrapositive, making it a tautology; however, Statement II does not explain Statement I.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The statement (p ^ q) ^ (-pv - q) is _______________.

  1. a tautology

  2. a contradiction

  3. a contingency

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

The expression contains a conjunction of a statement and its direct negation through De Morgan expansion of the second term, ensuring it is always false. Thus, the statement is a contradiction.

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The statement $\sim (p \rightarrow q) \leftrightarrow  (\sim p \vee \sim q)$ is 

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When $p$ and $q$ both are true then 

$(p \rightarrow  q) and (\sim p \vee \sim q)$ both are false

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is true when $p$ and $q$ both are false then  

$\sim (p \rightarrow  q)$ is false and $(\sim p \vee \sim q)$ is true

i.e. $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is false

Hence $\sim (p \rightarrow  q) \leftrightarrow  (\sim p \vee  \sim q)$ is neither tautology nor contradiction