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Animated Solution for Mathematics - Basic Mathematics: If $p \to (p \land \sim q)$ false, then the truth values of $p$ and $q$ are respectively :

Q: I got $q=F$ because $\sim q = F$. Did I miss something?

It is a very common trap! You correctly identified that $\sim q$ must be False. However, $\sim q$ is the negation of $q$. If the negation of a statement is False, the original statement must be True. Think of it as a double negative: if 'not $q$' is False, then $q$ itself must be True. Keep practicing these negations; they are the most common place to lose marks!

Select Answer:

Visualized Solution

The Given Logical Expression

We are given the logical statement: $p \to (p \land \sim q)$
The truth value of this entire expression is False, meaning it equals $F$ .

The Implication Rule

Recall the truth table for implication: $A \to B$
$A \to B$ is False in only one specific case:
When the antecedent $A$ is True ( $A = T$ ) and the consequent $B$ is False ( $B = F$ ).

Deducing the Value of $p$

Applying the rule to $p \to (p \land \sim q)$ :
The antecedent is $p$ .
Therefore, $p$ must be True, so $p = T$ .

Deducing the Consequent

The consequent is the expression inside the brackets: $(p \land \sim q)$ .
According to our rule, the consequent must be False.
So, $(p \land \sim q) = F$ .

Substituting the Value of $p$

We already know that $p = T$ .
Let's substitute this value into our consequent equation:
$T \land \sim q = F$ .

The Conjunction Rule

We have a conjunction (AND operation): $T \land X = F$ .
The truth value of $T \land X$ depends entirely on $X$ .
For the result to be False, $X$ must be False.

Solving for $\sim q$

In our equation, the unknown part is $\sim q$ .
Therefore, $\sim q$ must be False, meaning $\sim q = F$ .

Finding the Value of $q$

We found that $\sim q = F$ (Not $q$ is False).
The negation of False is True.
Therefore, $q$ must be True, so $q = T$ .

Final Conclusion

We have successfully deduced the truth values:
Value of p: $p = T$
Value of q: $q = T$
The correct option is T, T.

00:00 / 00:00

The Sigma Insight: Linear Inequalities

Solution Diagram

Analyzing the Setup

In the realm of formal logic, the implication operator

p \to q

is a strict gatekeeper. It has one, and only one, way to be False: the antecedent must be True, and the consequent must be False.

We are given the statement

p \to (p \land \sim q)

and told that it is False. Applying the rule of implication, we conclude:

p = T

(p \land \sim q) = F

The Consequent's Dilemma

We now focus on the consequent

(p \land \sim q) = F

. Since we have already established that

p = T

, we substitute this value into the expression:

T \land \sim q = F

The conjunction operator

\land

(AND) only returns True if both sides are True. Since the left side is already True, the result depends entirely on the right side. For the entire conjunction to be False, the term

\sim q

must be False:

\sim q = F

The Final Negation

We have deduced that

\sim q = F

. The symbol

\sim

represents the negation operator, which flips the truth value of the variable.

If the negation of

q

is False, then

q

itself must be the opposite. Therefore:

q = T

Final Conclusion

By systematically breaking down the logical structure, we have determined the truth values for the variables:

$p = T$

$q = T$

The final result is the pair $(T, T)$ .

Common Traps & FAQs

Wait, why do we reject the case where the antecedent

p

is False?

That is a fantastic question! Remember the truth table for implication:

A \to B

is only False when

A

is True and

B

is False. If the antecedent

A

(in this case,

p

) were False, the implication

p \to (p \land \sim q)

would be True regardless of the consequent. Since the problem explicitly states the expression is False, we are forced into the only scenario where that is possible:

p

must be True.

Why can't we just use a truth table for all options?

You absolutely can! Truth tables are a robust, foolproof method. However, in competitive exams like JEE, time is your most precious resource. Using the logical rules of implication and conjunction allows you to 'short-circuit' the process and arrive at the answer in seconds rather than minutes. Both methods are valid, but mastering the rules is a superpower for speed.

I got

q = F

because

\sim q = F

. Did I miss something?

It is a very common trap! You correctly identified that

\sim q

must be False. However,

\sim q

is the negation of

q

. If the negation of a statement is False, the original statement must be True. Think of it as a double negative: if 'not

q

' is False, then

q

itself must be True. Keep practicing these negations; they are the most common place to lose marks!

Animated Solution for Mathematics - Basic Mathematics: If $p \to (p \land \sim q)$ false, then the truth values of $p$ and $q$ are respectively :

Select Answer:

Visualized Solution

The Given Logical Expression

The Implication Rule

Deducing the Value of $p$

Deducing the Consequent

Substituting the Value of $p$

The Conjunction Rule

Solving for $\sim q$

Finding the Value of $q$

Final Conclusion

The Sigma Insight: Linear Inequalities

Analyzing the Setup

The Consequent's Dilemma

The Final Negation

Final Conclusion

Similar Questions

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Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If p→(p∧∼q) false, then the truth values of p and q are respectively :

Select Answer:

Visualized Solution

The Given Logical Expression

The Implication Rule

Deducing the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Deducing the Consequent

Substituting the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

The Conjunction Rule

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∼q

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Final Conclusion

The Sigma Insight: Linear Inequalities

Analyzing the Setup

The Consequent's Dilemma

The Final Negation

Final Conclusion

Similar Questions

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Animated Solution for Mathematics - Basic Mathematics: If $p \to (p \land \sim q)$ false, then the truth values of $p$ and $q$ are respectively :

Deducing the Value of $p$

Substituting the Value of $p$

Solving for $\sim q$

Finding the Value of $q$

If $p \to (p \land \sim q)$ is false. Then the truth values of $p$ and $q$ are respectively

The Boolean expression $(p \land \sim q) \Rightarrow (q \lor \sim p)$ is equivalent to:

Which of the following is the negation of the statement "for all $M > 0$ , there exists $x \in S$ such that $x \geq M$ "?

If $a, b, c$ are distinct +ve real numbers and $a^{2} + b^{2} + c^{2} = 1$ then $ab + b c + c a$ is

If $S$ is the set of all real $x$ such that $\frac{2 x - 1}{2 x ^{3} + 3 x ^{2} + x}$ is positive, then $S$ contains

Prove by mathematical induction that $\frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \leq \frac{1}{( 3 n + 1 ) ^{1/2}}$ for all positive integers $n$ .

Let $y = \frac{( x + 1 ) ( x - 3 )}{( x - 2 )}$ . Find all the real values of $x$ for which $y$ takes real values.