To prove the statement by contraposition, we first need to recall what contraposition means. The contrapositive of an implication of the form:

$P \implies Q$

is:

$\neg Q \implies \neg P$

where:

• $P$ is the hypothesis (the condition),
• $Q$ is the conclusion.

So, the original statement is:

If the product of two positive real numbers is greater than 100, then at least one of the numbers is greater than 10.

In symbolic form, let $a$ and $b$ be the two positive real numbers. The original statement is:

$(a \cdot b > 100) \implies (a > 10 \text{ or } b > 10)$

Step 1: Contrapositive of the Statement

To prove the statement by contraposition, we need to prove the contrapositive, which is:

$\neg (a > 10 \text{ or } b > 10) \implies \neg (a \cdot b > 100)$

Step 1.1: Simplifying the Contrapositive

The contrapositive involves negating the statement on both sides:

1. Negating $(a > 10 \text{ or } b > 10)$ gives us: $\neg (a > 10 \text{ or } b > 10) \equiv (a \leq 10 \text{ and } b \leq 10)$ This is because the negation of "or" becomes "and," and we must have both $a \leq 10$ and $b \leq 10$.

2. Negating $(a \cdot b > 100)$ gives us: $\neg (a \cdot b > 100) \equiv (a \cdot b \leq 100)$

Thus, the contrapositive is:

$(a \leq 10 \text{ and } b \leq 10) \implies (a \cdot b \leq 100)$

Step 2: Proving the Contrapositive

Now, we need to prove the contrapositive statement: If both $a \leq 10$ and $b \leq 10$, then $a \cdot b \leq 100$.

If $a \leq 10$ and $b \leq 10$, the product $a \cdot b$ is maximized when both $a$ and $b$ are equal to 10. So:

$a \cdot b \leq 10 \cdot 10 = 100$

Thus, if both $a \leq 10$ and $b \leq 10$, it follows that $a \cdot b \leq 100$, which proves the contrapositive.

Conclusion

Since we've proved the contrapositive of the original statement, we have thereby proven the original statement as well:

If the product of two positive real numbers is greater than 100, then at least one of the numbers is greater than 10.

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Would you like further clarification or details on any of the steps? Here are some related questions to deepen your understanding:

1. How can the contrapositive approach be useful in other types of proofs?
2. What would happen if we used a different method to prove the statement (like direct proof)?
3. Can the contrapositive method be applied to statements involving inequalities like $a^2 > 25 \implies a > 5$?
4. How do logical equivalences like De Morgan’s laws play a role in contraposition proofs?
5. What happens if one of the numbers $a$ or $b$ is negative or zero in this case?

Tip: When proving by contraposition, always ensure you correctly negate the "or" and "and" statements, as De Morgan's laws will guide you through this transformation.