Tag: different forms of theoretical statements

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

Solve it:-
$\left( {p \to q} \right) \to [\left( { \sim p \to q} \right) \to q]$

  1. Tautology

  2. Contradiction

  3. Contingent

  4. Not statement
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$Y=\left( p\longrightarrow q \right) \longrightarrow \left[ \left( \sim p\longrightarrow q \right) \longrightarrow q \right]$
Method : TRUTH TABLE [  ALWAYS PREFERABLE]

 $p$  $q$  $p\longrightarrow q$  $\left( \sim p\longrightarrow q \right) $  $\left[ \left( \sim p\longrightarrow q \right) \longrightarrow q \right]$  $Y$
 $1$  $0$  $0$  $1$  $1$ $1$ 
 $1$  $1$  $1$  $1$  $1$  $1$
 $0$  $0$  $1$  $0$  $1$  $1$
 $0$  $1$  $1$  $1$  $1$  $1$
As the result is always TRUE $\left(i.e. 1\right)$;
$\left( p\longrightarrow q \right)\longrightarrow \left[ \left( \sim p\longrightarrow q \right) \longrightarrow q \right]$ is Tautology.

A. Tautology



















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

Let $p$ and $q$ be two propositions given by
$p$ : The sky is blue.
$q$ : The milk is white.
Then $p\wedge q$ will be

  1. The sky if blue or milk is white

  2. The sky is blue and milk is white

  3. The sky is white and milk is blue

  4. If the sky is blue then milk is white
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$p \wedge q$ means statement p and q
=> The sky is blue and  milk is white.

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

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