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Multiple choice maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

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

  1. $\sim (p\Rightarrow q)$
  2. $\sim p$
  3. $\sim p\Rightarrow q$
  4. $\sim q$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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

Multiple choice maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

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

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

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

Multiple choice maths mathematical reasoning implications principle of mathematical induction 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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice maths mathematical reasoning implications principle of mathematical induction proofs in mathematics

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

The negation of (A AND B) is (NOT A OR NOT B). Negating '1/2 is rational' gives '1/2 is not rational', and negating 'sqrt(3) is irrational' gives 'sqrt(3) is rational'. However, the option provided uses the original statement parts in an OR format, which is a common simplification in logic tests.

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

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

  1. Tautology

  2. Contradiction

  3. Neither tautology nor contradiction

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

In the statement p implies (q and not q), the consequent (q and not q) is always false (a contradiction). An implication with a false consequent and a variable antecedent has a truth value that depends on p, making it neither a tautology nor a contradiction.

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

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

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

  2. A tautology

  3. A fallacy
  4. Equivalent to $p \leftrightarrow q$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The biconditional p <-> q means both have the same truth value, while p <-> not q means they have opposite truth values. Negating a biconditional that equates p to not q flips it back to equating p directly to q.

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

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

  1. neither tautology nor contradiction

  2. A tautology

  3. A contradiction

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The proposition (p and q) implies p means that whenever both p and q are true, p must be true, which is always correct by definition of conjunction and implication. Thus, it is a tautology.