Tag: principle of mathematical induction

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Questions Related to principle of mathematical induction

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. 

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

The statement $\sim (p\rightarrow \sim q)$ is equivalence to ___________.

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

$\sim\left({p} \rightarrow \sim{q} \right)$

We know that,
               $\sim\left({p} \rightarrow {q} \right)={p}\wedge\sim{q}$
          $\Rightarrow\sim\left({p}\rightarrow\sim{q}\right)=\sim{p}\wedge\sim\left(\sim{q}\right)$
                                    $=\sim{p}\wedge{q}$
Hence, $\left(\sim{p}\wedge{q}\right)$ is the correct answer.


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

Which of the following is always true?

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

$p \implies q \equiv \sim p \lor q  $
$\therefore \sim (p \implies q ) \equiv \sim (\sim p \lor q )$
$\therefore \sim (p \implies q ) \equiv (p \land \sim q )$

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

Which of the following is/are false?

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

We know that:
$p\rightarrow q \equiv \sim q\rightarrow \sim p$    {By logical equivalences }    
Hence $A$ is false


Now for option $B$
$\sim (p \ \rightarrow \ \sim q)$ $\equiv$ $\sim (\sim p\vee \sim q)=p\wedge q$   [By logical Equivalences ]
Hence $B$ is false

Now for option $C$
$\sim (\sim p\rightarrow \sim q)$ $\equiv \sim (p  \vee \sim q) $  $\equiv \sim p\wedge q$  [By Logical Equivalences]
Hence $C$ is true

Now for option $D$
$\sim (p\leftrightarrow q)$ $\equiv \sim ((p\rightarrow q)\wedge (q\rightarrow p))$ $\equiv \sim (p\rightarrow q)\vee \sim (q\rightarrow p)$
Hence $D$ is false                        [By logical Equivalences]