Tag: proofs in mathematics

Questions Related to proofs in mathematics

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

  1. tautology.

  2. contradiction.

  3. neither a tautology nor a contradiction.

  4. tautology and contradiction.


Correct Option: B
Explanation:

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

A contradiction.

Negation of the statement $p:\dfrac {1}{2}$ is rational and $\sqrt {3}$ is irrational is

  1. $\dfrac {1}{2}$ is rational or $\sqrt {3}$ is irrational

  2. $\dfrac {1}{2}$ is not rational or $\sqrt {3}$ is not irrational

  3. $\dfrac {1}{2}$ is not rational or $\sqrt {3}$ is irrational

  4. $\dfrac {1}{2}$ is rational and $\sqrt {3}$ is irrational


Correct Option: A

P: he studies hard, q: he will get good marks. The symbolic form of " If he studies hard then he will get good marks "is_____

  1. $\sim q\Rightarrow p$

  2. $p\Rightarrow q$

  3. $\sim p\vee q$

  4. $p\Leftrightarrow q$


Correct Option: B

Disjunction of two statements p and q is denoted by

  1. $p \leftrightarrow q$

  2. $p \rightarrow q$

  3. $p \leftarrow q$

  4. $p \vee q$


Correct Option: D

An implication or conditional "if p then q "is denoted by

  1. $p \vee q$

  2. $p \rightarrow q$

  3. $p \leftarrow q$

  4. None of these


Correct Option: B

The truth values of p, q and r for which $(pq)(∼r)$ has truth value F are respectively

  1. F, T, F

  2. F, F, F

  3. T, T, T

  4. T, F, F


Correct Option: C

 The negation of the compound proposition $p \vee (p \vee q)$ is

  1. $(p\wedge ∼q)\wedge ∼p$

  2. $(p\wedge ∼q)\vee  ∼p$

  3. $(p\wedge ∼q)\vee ∼p$

  4. none of these


Correct Option: A

Given, "If I have a Siberian Husky, then I have a dog." Identify the converse

  1. If I do not have a Siberian Husky, then I do not have a dog.

  2. If I have a dog, then I have a Siberian Husky.

  3. If I do not have a dog, then I do not have a Siberian Husky.

  4. If I do not have a Siberian Husky, then I have a dog.


Correct Option: B

$[(p)\wedge q]$ is logically equivalent to

  1. $(p\vee q)$

  2. $[p\wedge(q)]$

  3. $p\wedge(q)$

  4. $p\vee(q)$


Correct Option: D

$∼(p⇒q)⟺∼p\vee ∼q  \, is$

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. cannot come to any conclusion


Correct Option: C