Given that $a b$ and $a 4$, while $b

Question

Given that $a > b$ and $a > 4$, while $b < 5$, compare the following quantities:

**Quantity I:** $$\frac{1}{a} + \frac{1}{b} = 0$$

**Quantity II:** $$\left(\frac{1}{b} - \frac{1}{a}\right)^2 = 0$$

Determine the correct relationship between Quantity I and Quantity II by comparing their numerical values only.

Accepted Answer

**Solution:** 1. Given conditions: $a > b$, $a > 4$, and $b < 5$. 2. Consider **Quantity I:** $$\frac{1}{a} + \frac{1}{b} = 0$$ This implies: $$\frac{1}{a} = -\frac{1}{b} \implies a = -b$$ 3. Since $a > 4$ and $b < 5$, if $a = -b$, then $a$ and $b$ must be of opposite signs. 4. But from the given, $a > b$ and $a > 4$, so $a$ is positive and greater than 4. 5. If $a$ is positive and $a = -b$, then $b$ must be negative. 6. Now check **Quantity II:** $$\left(\frac{1}{b} - \frac{1}{a} ight)^2 = 0$$ This means: $$\frac{1}{b} - \frac{1}{a} = 0 \implies \frac{1}{b} = \frac{1}{a} \implies a = b$$ 7. But given $a > b$, so $a eq b$, thus Quantity II is not zero. 8. Therefore, Quantity I equals zero only if $a = -b$, which contradicts $a > b$ and $a > 4$ unless $b$ is negative. 9. Quantity II equals zero only if $a = b$, which is not possible here. 10. Since Quantity I can be zero under some conditions but Quantity II cannot be zero, Quantity I < Quantity II numerically (because Quantity II is a square and always non-negative, and zero only if $a = b$). **Answer:** Quantity I < Quantity II

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