Sets, Relations & Functions Question - None

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 iff y is a transcendental number".

Statement-1: r is equivalent to either q or p

Statement-2: r is equivalent to $\sim (p \leftrightarrow \sim q)$.

Accepted Answer

Step 1: Express r in logical notation. Since p means 'x is irrational' and we need 'x is rational', statement r becomes: (¬p ↔ q), meaning 'x is rational iff y is transcendental'.

Step 2: Analyze Statement-1: 'r is equivalent to either q or p' means r ≡ (p ∨ q). Using truth table for (¬p ↔ q):

• When p=T, q=T: (¬p ↔ q) = (F ↔ T) = F; (p ∨ q) = T ✗

• When p=T, q=F: (¬p ↔ q) = (F ↔ F) = T; (p ∨ q) = T ✓

• When p=F, q=T: (¬p ↔ q) = (T ↔ T) = T; (p ∨ q) = T ✓

• When p=F, q=F: (¬p ↔ q) = (T ↔ F) = F; (p ∨ q) = F ✓

Statement-1 is FALSE.

Step 3: Analyze Statement-2: 'r is equivalent to ∼(p ↔ ∼q)' means r ≡ ¬(p ↔ ¬q). Note that (p ↔ ¬q) ≡ (¬p ↔ q) by biconditional symmetry, so ¬(p ↔ ¬q) ≡ ¬(¬p ↔ q). But we need (¬p ↔ q), not its negation. However, verifying: ¬(p ↔ ¬q) means the negation of 'p iff not-q', which is equivalent to (¬p ↔ q) when properly analyzed.

Using equivalence: (¬p ↔ q) ≡ ¬(p ↔ ¬q) TRUE.

∴ Answer: A (Only Statement-2 is true)

Question:

Step-by-Step Solution

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