1. **Problem Statement:** We are asked to write true biconditionals and show that both implied conditionals are true for each case. 2. **Biconditional Definition:** A biconditional statement "P if and only if Q" means both "If P then Q" and "If Q then P" are true. --- ### A. Biconditional: $\sqrt{x} \geq x$ if and only if $0 < x < 1$ 1. First implied conditional: If $\sqrt{x} \geq x$, then $0 < x < 1$. 2. Second implied conditional: If $0 < x < 1$, then $\sqrt{x} \geq x$. **Explanation:** - For $x \geq 0$, $\sqrt{x}$ is defined. - For $0 < x < 1$, $\sqrt{x} > x$ because the square root function grows slower than the linear function in this interval. - For $x > 1$, $\sqrt{x} < x$. Thus, both conditionals hold true. --- ### B. Biconditional: $x^2 < x^3$ if and only if $x > 0$ 1. First implied conditional: If $x^2 < x^3$, then $x > 0$. 2. Second implied conditional: If $x > 0$, then $x^2 < x^3$. **Explanation:** - For $x > 0$, dividing both sides of $x^2 < x^3$ by $x^2$ (which is positive) gives: $$\cancel{x^2} < \cancel{x^2} x \Rightarrow 1 < x$$ - So the inequality holds for $x > 1$. - For $0 < x < 1$, $x^3 < x^2$, so $x^2 < x^3$ is false. - For $x \leq 0$, the inequality does not hold. Hence, the biconditional is true for $x > 1$, but the problem states $x > 0$, so the biconditional is true only for $x > 1$. --- ### C. Biconditional: A whole number is divisible by 3 if and only if it is divisible by 6 1. First implied conditional: If a whole number is divisible by 3, then it is divisible by 6. 2. Second implied conditional: If a whole number is divisible by 6, then it is divisible by 3. **Explanation:** - Divisible by 6 implies divisible by 3 because 6 is a multiple of 3. - However, divisible by 3 does not imply divisible by 6 (e.g., 3 is divisible by 3 but not by 6). Therefore, this biconditional is false. --- ### D. Biconditional: A whole number is even if and only if it is divisible by 2 1. First implied conditional: If a whole number is even, then it is divisible by 2. 2. Second implied conditional: If a whole number is divisible by 2, then it is even. **Explanation:** - By definition, even numbers are integers divisible by 2. - Both conditionals are true. --- **Final answers:** - A and D are true biconditionals. - B is true only for $x > 1$, not all $x > 0$. - C is false.