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

Let  $P , Q , R$  and  $S$  be statements and suppose that  $P \rightarrow Q \rightarrow R \rightarrow P.$  If  $\sim S \rightarrow R,$  then

  1. $S \rightarrow \sim Q$
  2. $\sim Q \rightarrow S$
  3. $\sim S \rightarrow \sim Q$
  4. $Q \rightarrow \sim S$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given P -> Q, Q -> R, R -> P, we have P <-> Q <-> R. If ~S -> R, then by contrapositive ~R -> S. Since R is equivalent to Q, ~Q -> S.

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

$(p\rightarrow q)\leftrightarrow (q\vee \sim p)$ is - 

  1. Equivalent to $p\wedge q$

  2. Tautology

  3. Fallacy

  4. Neither tautology nor fallacy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The implication p implies q is logically equivalent to not p or q. The given statement equates this with (q or not p), which is identical by the commutative law. Thus, both sides are always equal, making the bi-implication a tautology.

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

Let S be a set of n persons such that:(i)any person is acquainted to exactly k other persons in s;(ii)any two persons that are acquainted have exactly $\displaystyle l $ common acquaintances in s;(iii)any two persons that are not acquainted have exactly m common acquaintances in S.Prove that $\displaystyle m\left ( n-k \right )-k\left ( k-1 \right )+k-m= 0.$

  1. $\displaystyle k\left ( k-1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.
  2. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.
  3. $\displaystyle k\left ( k-1+l \right )= m\left ( n-k-1 \right )$ is equivalent to the desired zero.
  4. $\displaystyle k\left ( k+1-l \right )= m\left ( n-k+1 \right )$ is equivalent to the desired one.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let a be a fixed element os S. Let us count the triples (a, x, y) such that a, x are acquainted, x, y are acquainted and a, y are not acquainted. Because a isacquainted to exactly k other persons in S, x can be chosen in k ways and for fixeda and x, y can be chosen in $\displaystyle k-1-l $ ways. Thus the number of such triples is $\displaystyle k\left ( k-1-l  \right ).$ Let us count again, choosing y first. The number of persons not acquainted to a equals n-k-1, hence y can be chosen in n-k-1 ways. Because x is a commonacquaintance of a and y, it can be chosen in m ways, yielding a total of $\displaystyle m\left ( n-k-1 \right )$ triples. It is not difficult to see that the equality $\displaystyle k\left ( k-1-l  \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

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

The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim  r]$ is:

  1. $\sim p \vee [\sim q \vee (p \vee q) \vee \sim r]$
  2. $ p \vee [q \vee (\sim p \wedge \sim q) \vee r]$
  3. $ \sim p \vee [\sim q \vee (p\wedge q) \vee \sim r]$
  4. $ \sim p \vee [\sim q \wedge (p\wedge q) \wedge \sim r]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

$\equiv  \sim p \vee [\sim q \vee (p\wedge q) \vee  \sim r]$

Note: For dual of a statement just replace $\vee$ by $\wedge$ and vice versa.

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

The equivalent statement of (p $\leftrightarrow$ q) is

  1. $(p \wedge q) \vee (p \vee q)$
  2. $(p \rightarrow q) \vee (q \rightarrow p)$
  3. $(\sim p \vee q) \vee (p \vee \sim q)$
  4. $(\sim p \vee q) \wedge (p \vee \sim q)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$p\rightarrow q \equiv (\sim p\vee q)$

$q\rightarrow p \equiv (\sim q \vee p)$
$\therefore$
$p\leftrightarrow q \equiv (p\rightarrow q)\wedge(q\rightarrow p)$
$\Rightarrow p\leftrightarrow q\equiv (\sim p\vee q)\wedge(p\vee\sim q)$

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

Which of the following is correct?

  1. $(~p \vee ~q) \equiv (p \wedge q)$
  2. $(p \rightarrow q) \equiv (~q \rightarrow ~p)$
  3. $~(p \rightarrow ~q) \equiv (p \wedge ~q)$

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

Clearly, the statements $p \vee q$ and $p\wedge q$ cannot be equivalent as they one operator means "OR" and the other operator means "AND".

$p$ $q$ $p\rightarrow q$ $q\rightarrow p$
T T T T
T F F T
F T T F
F F T T

Option B is also incorrect.

$p$ $q$ $p\rightarrow q$ $p\wedge q$
T T T T
T F F F
F T T F
F F T F


Hence, option C is also incorrect.

Option D is also incorrect as
$p \leftrightarrow q=(p\rightarrow q)\wedge (q\rightarrow p)$


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

Which of the following statement are NOT logically equivalent?

  1. $ \sim (p \vee \sim q)$ and $ (\sim p \wedge q )$
  2. $\sim (p \rightarrow q )$ and $(p \wedge \sim q )$
  3. $(p \rightarrow q) $ and $(\sim q \rightarrow \sim p) $
  4. $(p \rightarrow q )$ and $(\sim p \wedge q)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We make an option wise check for this.

Option A: $\sim \left( p\vee \sim q \right) \quad and\quad \left( \sim p\wedge q \right) $
By application of Demorgan's Law on $\sim \left( p\vee \sim q \right) $ we get, $\sim \left( p\wedge q \right) $ 
So this option is logically equivalent.

Option B: $\sim \left( p\longrightarrow q \right) \quad and\quad \left( p\wedge \sim q \right) $
Again by application Conditional Disjunction rule, we see that this option is also logically equivalent.

Option C: $\left( p\longrightarrow q \right) \quad and\quad \left( \sim q\longrightarrow \sim p \right) $
This is again true by Contrapositive tautology.

Option D:$\left( p\longrightarrow q \right) \quad and\quad \left( \sim p\wedge q \right) $
This is not logically equivalent.