Tag: proofs in mathematics

Quizzes (2) Questions (121)

Questions Related to proofs in mathematics

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
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.