1. The problem asks to explain the relationship between Proof by Contrapositive and Proof by Contradiction, and how proof by contradiction contains the reasoning of contrapositive. 2. It also asks why the Base Case in Mathematical Induction cannot be skipped and what happens if the Inductive Step fails. --- 1. Proof by Contrapositive and Proof by Contradiction: 1.1. A statement "If p, then q" ($p \to q$) is logically equivalent to its contrapositive "If not q, then not p" ($\sim q \to \sim p$). 1.2. Proof by Contrapositive proves $p \to q$ by proving $\sim q \to \sim p$ directly. 1.3. Proof by Contradiction assumes $p$ and $\sim q$ (the negation of the conclusion) and shows this leads to a contradiction. 1.4. This assumption $p \wedge \sim q$ is exactly the negation of $p \to q$. 1.5. Therefore, showing $p \wedge \sim q$ leads to contradiction proves $p \to q$ must be true. 1.6. This means proof by contradiction contains the reasoning of contrapositive because it starts with the same assumption as the contrapositive's hypothesis and derives an impossibility. 1.7. Completing a contrapositive proof shows $\sim q \to \sim p$ is true, which means assuming $p \wedge \sim q$ is impossible, thus completing the contradiction. --- 2. Mathematical Induction: 2.1. The Base Case confirms the statement is true for the initial value $n = n_0$. 2.2. The Inductive Step proves that if $P(k)$ is true, then $P(k+1)$ is true. 2.3. We cannot skip the Base Case because the chain of implications starts there; without it, the chain has no starting point. 2.4. If the first link (Base Case) is missing, the statement is not established for any $n$, so the induction fails. 2.5. If the Inductive Step fails, it means the property does not hold for all $n > n_0$, so the statement cannot be proven for all $n \geq n_0$. 2.6. Induction works like a chain: the Base Case is the first link, and the Inductive Step connects each link to the next. 2.7. Without either, the chain breaks and the proof fails. --- Final answers: Proof by Contrapositive and Proof by Contradiction are related because both rely on the logical equivalence of $p \to q$ and $\sim q \to \sim p$. Proof by Contradiction assumes $p \wedge \sim q$ and shows it leads to contradiction, which is the negation of the contrapositive's hypothesis. Completing a contrapositive proof shows the assumption $p \wedge \sim q$ is impossible, thus completing the contradiction. In Mathematical Induction, the Base Case cannot be skipped because it establishes the starting point of the chain of implications. If the Base Case is missing, the chain has no first link and the statement is unproven. If the Inductive Step fails, the chain breaks and the statement cannot be proven for all $n \geq n_0$. Hence, both parts are essential for a valid induction proof.