Q−1: [2+1+1+2 marks] In the following statement, find the inverse, convers and contrapositive: (If 4x-2=10,then x=3) Evaluate the truth value of the following: ∀x(x<0)∨(x<2x) , Where the dom

Let’s break down your questions step by step. ### Step 1: Inverse, Converse, and Contrapositive **Given Statement:** If \(4x - 2 = 10\), then \(x = 3\). - **Inverse:** The inverse negates both parts of the statement: If \(4x - 2 \neq 10\), then \(x \neq 3\). - **Converse:** The converse swaps the hypothesis and the conclusion: If \(x = 3\), then \(4x - 2 = 10\). - **Contrapositive:** The contrapositive negates and swaps the hypothesis and conclusion: If \(x \neq 3\), then \(4x - 2 \neq 10\). ### Step 2: Truth Value Evaluation **Statement:** \(\forall x (x < 0) \lor (x < 2x)\), where the domain is the set of integers. To evaluate this statement, let’s rephrase it: For all integers \(x\), either \(x < 0\) or \(x < 2x\). 1. **Case 1:** If \(x < 0\), then the first condition is true. 2. **Case 2:** If \(x \geq 0\), we check \(x < 2x\): - If \(x = 0\), \(0 < 2(0)\) is false (0 is not less than 0). - If \(x = 1\), \(1 < 2(1)\), true. - If \(x = 2\), \(2 < 2(2)\), true. - In fact, for all integers \(x > 0\), \(x < 2x\) holds true. Since the only integer for which this statement does *not* hold true is \(x = 0\) and \(x < 0\) also holds for negative integers, the overall statement is: - **Truth Value:** False (due to \(x = 0\)). ### Step 3: Predicate in Symbols and Negation **Statement:** "Every Natural number is an Integer." **Domain:** Real numbers. - **Symbolic Representation:** \(\forall n (N(n) \rightarrow I(n))\), where \(N(n)\) means \(n\) is a natural number and \(I(n)\) means \(n\) is an integer. - **Negation:** \(\neg \forall n (N(n) \rightarrow I(n))\) is equivalent to: \(\exists n (N(n) \land \neg I(n))\), meaning: "There exists a natural number \(n\) that is not an integer." ### Step 4: Proof or Disprove \( (p \land (p \rightarrow q)) \rightarrow q \) To prove or disprove this statement, we can use a truth table. - **Variables**: \(p\), \(q\) \[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & p \rightarrow q & p \land (p \rightarrow q) & (p \land (p \rightarrow q)) \rightarrow q \\ \hline T & T & T & T & T \\ T & F & F & F & T \\ F & T & T & F & T \\ F & F & T & F & T \\ \hline \end{array} \] - **Analysis**: The last column which represents \((p \land (p \rightarrow q)) \rightarrow q\) is true in all cases. This means that the statement is a tautology, thus proving \( (p \land (p \rightarrow q)) \rightarrow q \) is **always true**. ### Summary Answers 1. Inverse: If \(4x - 2 \neq 10\), then \(x \neq 3\). Converse: If \(x = 3\), then \(4x - 2 = 10\). Contrapositive: If \(x \neq 3\), then \(4x - 2 \neq 10\). 2. Truth Value: False. 3. Symbol: \(\forall n (N(n) \rightarrow I(n))\); Negation: \(\exists n (N(n) \land \neg I(n))\). 4. Prove: \( (p \land (p \rightarrow q)) \rightarrow q \) is true.