Reasoning

Logic and Fallacies

1,803 Questions

Understand the fundamentals of propositional logic, logical inference rules, and paradoxes. This set includes identifying logical fallacies, including those found in classical Nyaya logic. Strong grasp of these concepts is crucial for scoring well in the reasoning sections of competitive tests.

Propositional logic modelsLogical inference rulesTypes of logical fallaciesParadoxes and contingent truthsNyaya logic concepts

Logic and Fallacies Questions

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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. $p \rightarrow \sim p$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A compound proposition is always false if it is a contradiction. Option C equates (not p or q) with (p and not q), which are exact opposites, meaning their biconditional is always false.

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A conditional statement p implies q is false only in the single case where the hypothesis p is true and the conclusion q is false. In all other cases, the implication evaluates to true.

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$'\Lambda '$ stands for logical and 

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

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

The dual of the statement $\left[ p\wedge \left( \sim q \right)  \right] \wedge \left( \sim p \right)] $ is

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

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The original statement has mismatched brackets: [p ∧ (¬q)] ∧ (¬p)] - this is grammatically incorrect. Assuming the intended statement is p ∧ (¬q) ∧ (¬p), its dual is formed by replacing ∧ with ∨: p ∨ ¬(q) ∨ ¬(p). Using De Morgan's law, ¬(q) ∨ ¬(p) ≡ ¬(q ∧ p), so the dual can also be written as (p ∨ ¬q) ∨ ¬p, which matches Option B.

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

The contrapositive of the sentence $\sim p \rightarrow q$ is equivalent to

  1. $p \rightarrow \sim q$
  2. $q \rightarrow \sim p$
  3. $q \rightarrow p$
  4. $\sim p \rightarrow \sim q$
  5. $\sim q \rightarrow \sim p$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

For a conditional statement p → q, Its converse statement (q → p) and inverse statement (∼p → ∼q) are equivalent to each other. p → q and its contrapositive statement (∼q → ∼p) are equivalent to each other.

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of following is the negation of $(P \ \vee\sim Q).$

  1. $\sim P\vee Q$
  2. $\sim P\wedge Q$
  3. $\sim Q\wedge P$
  4. $\sim Q\vee P$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 P  Q  $\sim P$  $\sim Q$  $P\vee \sim Q$ $\sim \left( P\vee \sim Q \right) $  $\sim P\wedge Q$ 
T   T  F  F  T  F  F
T F   F  T  T  F  F
F T   T  F  F  T  T
F F  T  T  F  T  F
Therefore, $\sim \left( P\vee \sim Q \right) $ is $\sim P\wedge Q$ 
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Is $(p\rightarrow q)\vee (q\rightarrow p)$  a tautology ?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$p$ $q$ $(p\rightarrow q)$ $(q\rightarrow p)$ $(p\rightarrow q)\vee(q\rightarrow p)$
T T             T              T                               T
T F             F              T                               T
F T             T              F                               T
F F             T              T                               T               

The given statement is a tautology as the truth table has all the values as true in the output which is the property of tautology
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Identify the Law of Logic
$(p \vee q) \vee r \equiv p \vee (q \vee r) \equiv p \vee q \vee r$

  1. Associative law

  2. Commutative Law

  3. Involution Law

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

Associative Law

This law allows the removal of brackets from an expression and regrouping of the variables.
$(p\vee q)\vee r \equiv p \vee (q \vee r)\equiv p\vee q\vee r$

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Identify the Law of Logic
$\sim(p \wedge q) \equiv \sim p \vee \sim q$

  1. Commutative Law

  2. DeMorgan's Law

  3. Complement Law

  4. Conditional Law
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given 
$\sim (p\wedge q)=\sim p \vee \sim q$

It is Demorgan's law 
according to the if we take transpose or negation of any quatity then all the relation get opposite

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Identify the Law of Logic
$\sim(p \vee q) \equiv \sim p \wedge \sim q$

  1. Conditional Law

  2. Demorgan's Law

  3. Absorption Law

  4. Identity Law
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given 
$\sim (p\wedge q)=\sim p \vee \sim q$

It is Demorgan's law 
according to the if we take transpose or negation of any quatity then all the relation get opposite

Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Identify the Law of Logic
$p \rightarrow q \equiv \sim p \vee q$

  1. Idempotent Law

  2. Conditional Law

  3. Involution Law

  4. Commutative Law
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
|  $p$ |  $q$ |  $p\rightarrow q$ |  $\sim p$ |  $(\sim p)\vee q$ | | --- | --- | --- | --- | --- | |  $T$ |   $T$ |   $T$ |   $F$ |   $T$ | |   $T$ |   $F$ |   $F$ |   $F$ |   $F$ | |  $F$ |   $T$ |   $T$ |   $T$ |   $T$ | |   $F$ |   $F$ |   $T$ |   $T$ |   $T$ |
We can say that if $p$ ,then $q$ or $p$ implies $q$ .
'$\rightarrow$' is called a conditional operator.
So, the giving logical equivalence is the conditional law.
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Let p and q be any two logical statements and $r : p \rightarrow (\sim p \vee q)$. If r has a truth value F, then the truth values of p and q are respectively

  1. F, F

  2. T, T

  3. F, T

  4. T, F
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
p q $\sim$p $\sim$ p $\vee$ q r
T T F T T
F F T T T
T F F F F
F T T T T

$\therefore$ Clearly from above able, If r has a truth value F, then the truth values of p and core T and F respectively.