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

The statement $p \to (q \to p)$ is equivalent to 

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

The statement p -> (q -> p) is equivalent to p -> (~q OR p), which is ~p OR (~q OR p). This simplifies to (~p OR p) OR ~q, which is True OR ~q = True. Option B is also a tautology, but the equivalence is not standard.

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. $~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation


$~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$ is true, we show it by truth table using boolean expression.

1.$p\rightarrow q$=min(1,1+q-p)
2.$p\wedge q$=min(p,q)
3.$p\leftrightarrow q$=1-|p-q|

Now we draw or make truth table using these operations
L.H.S  

 p  q $p\leftrightarrow q$ 
 1 1  1 
1  0  0 


R.H.S 

p q  $p\rightarrow q$  $q\rightarrow p$   $(p \rightarrow q) \wedge (q \rightarrow p)$
1  1  1  1  1
1  1  0 1  0 

L.H.S =R.H.S

$~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$

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

$(p \wedge q) \vee  \sim p$ is equivalent to 

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

By distributive law, (p and q) or not p is equivalent to (p or not p) and (q or not p). Since (p or not p) is a tautology (True), the expression simplifies to (q or not p), which is not p or q.

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

$\sim (p \wedge q)\Rightarrow (\sim p)\vee (\sim p \vee q)$ is equal to

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

The expression ~(p AND q) -> (~p OR (~p OR q)) simplifies to ~(p AND q) -> (~p OR q). This is (~p OR ~q) -> (~p OR q). This is equivalent to (~(~p OR ~q) OR (~p OR q)) = (p AND q) OR (~p OR q). This simplifies to (~p OR q).

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

The equivalent of $(p \rightarrow \sim p) \vee (\sim p \rightarrow p)$ is 

  1. $p \vee \sim p$
  2. $T \rightarrow F$
  3. $T \leftrightarrow F$
  4. $p \wedge \sim p$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (p -> ~p) is equivalent to ~p, and (~p -> p) is equivalent to p. Therefore, the disjunction (~p) v p is a tautology, which is equivalent to p v ~p.

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

Identify which of the following statement is not equivalent to the others

  1. If $x$ is bass then $x$ is bad.

  2. Boss implies bad,

  3. Bad is necessary condition for bass.
  4. $x$ is boss iff $x$ is bad.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Statements A, B, and C all translate to the conditional statement that boss implies bad, or bad is a necessary condition for boss. Option D states an iff (biconditional) condition, which is a stronger claim and thus not equivalent to the others.