Tag: maths

Questions Related to maths

If $p$ is false, $q$ is true, then which of the following is/are false?

maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

  1. $\sim (p\Rightarrow q)$

  2. $\sim p$

  3. $\sim p\Rightarrow q$

  4. $\sim q$

Show answer
Correct Option: A,D
Explanation:
 $p$  $q$  $p\Rightarrow q$ $\sim \left( p\Rightarrow q \right) $  $\sim p$  $\sim q$   $\sim p\Rightarrow q$
 F  T  T  F  T  F  T

Here we see that $\sim \left( p\Rightarrow q \right) $  and $\sim q$  are false

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

maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

  1. $p$

  2. $\sim p$

  3. $\sim q$

  4. $q$

Show answer
Correct Option: A
Explanation:

$(p\vee q)\wedge (p\vee \sim q)$
$=p\vee(q\wedge \sim q)$ (distributive law)
$=p\vee 0$ (complement law)
$=p$ ($0$ is indentity for v)

The negation of the statement "No slow learners attend this school," is:

maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

  1. All slow learners attend this school.

  2. All slow learners do not attend this school.

  3. Some slow learners attend this school.

  4. Some slow learners do not attend this school.

  5. No slow learners do not attend this school.

Show answer
Correct Option: C
Explanation:

The negation is : It is false that no slow learners attend this school. Therefore, some slow learners attend this school.

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

maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

  1. tautology.

  2. contradiction.

  3. neither a tautology nor a contradiction.

  4. tautology and contradiction.

Show answer
Correct Option: B
Explanation:

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

A contradiction.

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

maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

  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

Show answer
Correct Option: A

Which one of the statement gives the same meaning of statement
If you watch television, then your mind is free and if your mind is free then you watch television

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

  1. You watch television if and only if your mind is free.

  2. You watch television and your mind is free.

  3. You watch television or your mind is free.

  4. None of these

Show answer
Correct Option: B
Explanation:
"You watch television and your mind is free".
The above statement gives or suits for the same meaning of the structure given because it is logically correct.

Which of the following is NOT true for any two statements $p$ and $q$?

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

  1. $\sim[p\vee (\sim q)]=(\sim p)\wedge q$

  2. $\sim(p\vee q)=(\sim p)\vee (\sim q)$

  3. $q\wedge \sim q$ is a contradiction

  4. $\sim (p\wedge (\sim p))$ is a tautology

Show answer
Correct Option: B
Explanation:
$p$ and $q$ are two statements.
$A) LHS = \sim [pv (\sim q)]$
By De morgon's laws
$\sim(pr (\sim q))= \sim pnq$
$\therefore (A) $ is true .

$B) \sim(p v q) = (\sim p) \vee (\sim q)$
According to demorgon's laws, this is false.
$\because \sim (p \vee q) = (\sim p)\wedge (\sim q)$. 
$\therefore (B)$ is false.

$C) q \wedge \sim  q$ is a contradiction because $'q'$ and $\sim q$ are opposite statements i.e, cannot be there at the same time.

$D) \sim (p \wedge (\sim p))$
$p \wedge (\sim p)$ is a contradiction, which is evident from option $(C)$. $\therefore $ opposite of a contradiction is a tautology .
$\therefore [B]$ is wrong.

If p and q are two statements, then statement $p\Rightarrow q\wedge \sim q$.

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

  1. Tautology

  2. Contradiction

  3. Neither tautology nor contradiction

  4. None of these

Show answer
Correct Option: A

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

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

  1. Equivalent to $\sim p \leftrightarrow q$

  2. A tautology

  3. A fallacy

  4. Equivalent to $p \leftrightarrow q$

Show answer
Correct Option: C

The proposition $\left( {p \wedge q} \right) \Rightarrow p$ is 

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

  1. neither tautology nor contradiction

  2. A tautology

  3. A contradiction

  4. Cannot be determined

Show answer
Correct Option: C