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

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

The converse of 'If P then Q' is 'If Q then P'. Thus, the converse of 'If I have a Siberian Husky, then I have a dog' is 'If I have a dog, then I have a Siberian Husky'.

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

$∼(p⇒q)⟺∼p\vee ∼q  \, is$

  1. a tautology

  2. a contradiction

  3. neither a tautology nor a contradiction

  4. cannot come to any conclusion
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression ~(p -> q) is equivalent to (p AND ~q). The expression (~p OR ~q) is the negation of (p AND q). These are not equivalent, so the statement is neither a tautology nor a contradiction.

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

Consider the following statements 
$p$:you want to success
$q$:you will find way,
then the negation of $\sim (p\vee q)$ is

  1. you want of success and you find a way

  2. you want of success and you do not find a way

  3. if you do not want to succeed then you will find a way

  4. if you want of success then you cannot find a way
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice business maths proofs in mathematics implications principle of mathematical induction different forms of theoretical statements

Which of the following statements is a tautology

  1. $\left( { \sim p \vee q} \right) - \left( {p \vee \sim q} \right)$
  2. $\left( { \sim p \vee \sim q} \right) \to p \vee q$
  3. $\left( {p \vee \sim q} \right) \wedge \left( {p \vee q} \right)$
  4. $\left( { \sim p \vee \sim q} \right) \vee \left( {p \vee q} \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer