Let p be the statement "x is an irrational number", q be the statement "y is a transcendental number

Q: Why is 'x is rational if y is transcendental' written as $q \to \sim p$ and not $\sim p \to q$?

This is the classic 'if' trap! In English, 'A if B' is logically equivalent to 'If B, then A'. The 'if' attaches to the condition. Since 'y is transcendental' ($q$) is the condition, it must be the antecedent. The result is 'x is rational' ($\sim p$). Always identify the condition first!

Q: How do I handle the 'either... or' phrasing in logic?

In formal logic, 'either... or' is treated as the inclusive OR ($p \lor q$), unless specified as 'either... or... but not both' (which would be XOR). In JEE problems, always assume the inclusive OR unless the context explicitly demands exclusivity.

Q: What do I do if I find a contradiction in the question text during the exam?

Stay calm! If you have rigorously checked your truth table and it doesn't match the options, you might have encountered a typo. In such cases, look for the 'intended' logic (often involving De Morgan's laws). If the question is flawed, the exam board usually awards marks to everyone, but your best bet is to identify the most likely intended logical form.

Question

Let p be the statement "x is an irrational number", q be the statement "y is a transcendental number", and r be the statement "x is a rational number if y is a transcendental number". 
Statement-1 : r is equivalent to either q or p 
Statement-2 : r is equivalent to ~(p <-> q).

Options:

(A) Statement -1 is false, Statement-2 is true

(B) Statement -1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1

(C) Statement -1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1

(D) Statement -1 is true, Statement-2 is false

Sigma Percentile · Accepted Answer

### The Architecture of Truth: Mastering Logical Implication

Welcome, future engineers. Today, we are not just solving a problem; we are peeling back the layers of the language of the universe—mathematical logic.

Logic is the bedrock upon which all of calculus, algebra, and physics is built. If you can master the structure of a statement, you can master any problem in the JEE Advanced paper.

### Phase 1: Decoding the Atomic Statements

Every complex logical argument is built from simple, atomic statements. Think of these as the 'atoms' of your reasoning. We are given two:

1. $p$: '$x$ is an irrational number'.
2. $q$: '$y$ is a transcendental number'.

These are our building blocks. Our goal is to analyze the compound statement $r$: '$x$ is a rational number if $y$ is a transcendental number'.

### Phase 2: The Implication Trap

Here is where most students stumble. The phrasing '$A$ if $B$' is a linguistic trick. It does not mean $A 	o B$; it means $B 	o A$. The 'if' clause is the condition, the antecedent.

In our case, '$y$ is a transcendental number' ($q$) is the condition. '$x$ is a rational number' is the result. Since $p$ is '$x$ is irrational', then '$x$ is rational' is simply $\sim p$.

Therefore, our statement $r$ is symbolically represented as:

$$r \equiv q 	o \sim p$$

This is the heart of the problem. If you get this translation wrong, the rest of the truth table will collapse like a house of cards. Always pause, identify the condition, and write the implication carefully.

### Phase 3: The Truth Table Construction

Now, let us build the truth table. We need to evaluate $r \equiv q 	o \sim p$ for all combinations of $p$ and $q$.

Remember the golden rule of implication: $A 	o B$ is only **False** when $A$ is **True** and $B$ is **False**. In all other cases, it is **True**.

- When $p$ is **True**, $\sim p$ is **False**. If $q$ is **True**, we get $T 	o F$, which is **False**. If $q$ is **False**, we get $F 	o F$, which is **True**.
- When $p$ is **False**, $\sim p$ is **True**. Regardless of $q$, we get $T 	o T$ or $F 	o T$, both of which are **True**.

This gives us our target column for $r$: $[F, T, T, T]$.

### Phase 4: Testing the Statements

Statement-1 claims $r \equiv p \lor q$. Let us test this. For $p=T, q=T$, $p \lor q$ is **True**, but $r$ is **False**. Mismatch! Statement-1 is **false**.

Statement-2 claims $r \equiv \sim(p \leftrightarrow q)$. The bi-conditional $p \leftrightarrow q$ is **True** when $p$ and $q$ match. Its negation is **True** when they differ.

Let us check the row where $p=F, q=F$. Here, $p \leftrightarrow q$ is **True**, so $\sim(p \leftrightarrow q)$ is **False**. But our $r$ is **True**. Mismatch! Statement-2 is also **false**.

### Phase 5: The JEE Reality Check

Now, I know what you are thinking: 'If both are false, why is this a question?' This is a vital lesson in exam strategy. Sometimes, questions in high-stakes exams contain typos.

The intended logic was likely $r \equiv \sim p \lor \sim q$. By De Morgan's Law, $\sim p \lor \sim q \equiv \sim(p \land q)$. If the question had been phrased this way, both statements would be true, and Statement-2 would be the perfect explanation for Statement-1.

In the exam hall, if you encounter this, trust your rigorous derivation. If your math is solid, you have done your job. The system will account for the error.

Keep your cool, trust your logic, and move forward with confidence. You are building the analytical mind that will define the future of engineering.

Visualized Solution

Decoding the Atomic Statements $p$ and $q$

Translating Statement $r$ into Symbolic Logic

Evaluating $r$ for True $p$

Evaluating $r$ for False $p$

Testing Statement-1: $r \equiv p \lor q$

Testing Statement-2: $r \equiv\sim (p \leftrightarrow q)$

Resolving the JEE Typo & Intended Logic

Final Verdict and Exam Strategy

The Sigma Insight: Types of Sets and Set Operations

The Architecture of Truth

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Among the two statements $(S 1) : (p \Rightarrow q) \land (p \land (\sim q))$ is a contradiction and $(S 2) : (p \land q) \lor ((\sim p) \land q) \lor (p \land (\sim q)) \lor ((\sim p) \land (\sim q))$ is a tautology

Among the statements: (S1) $(p \Rightarrow q) \lor (p \land (\sim q))$ is a tautology, (S2) $((\sim p) \Rightarrow (\sim q)) \land ((\sim p) \lor q)$ is a Contradiction. Then

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If the truth value of the statement $P \to (\sim p \lor r)$ is false(F), then the truth values of the statements p, q, r are respectively :

Consider Statement-1 : $(p \land \sim q) \land (\sim p \land q)$ is a fallacy. Statement-2 : $(p \to q) \leftrightarrow (\sim q \to\sim p)$ is a tautology.

Among the statements: (S1) $((p \lor q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$ , (S2) $((p \lor q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \lor (q \Rightarrow r))$

Statement $(P \Rightarrow Q) \land (R \Rightarrow Q)$ is logically equivalent to

If $q$ is false and $p \land q \leftrightarrow r$ is true, then which one of the following statements is a tautology?

Consider the two statements: (S1) : $(p \to q) \lor (\sim q \to p)$ is a tautology. (S2) : $(p \land \sim q) \land (\sim p \lor q)$ is a fallacy. Then :

Visualized Solution

Decoding the Atomic Statements .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Translating Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r into Symbolic Logic

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r for True .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r for False .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Testing Statement-1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r≡p∨q

Testing Statement-2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r≡∼(p↔q)