Proving 1/10 < √(101) - √(99) Without Calculator A Step-by-Step Guide

Introduction

In the realm of mathematical inequalities, there are often intriguing challenges that require clever manipulation and insightful application of algebraic principles. One such challenge involves demonstrating that ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$ without resorting to the convenience of a calculator. This problem, which frequently appears in intermediate algebra textbooks, serves as an excellent exercise in honing one's skills in algebraic manipulation, particularly when dealing with radicals and inequalities. Many students find themselves initially stumped, as direct calculation is prohibited, and standard approaches might seem to lead to dead ends. However, by employing techniques such as the use of radical conjugates and careful reasoning about the properties of inequalities, a clear and concise proof can be constructed. This article will delve into a step-by-step solution, highlighting the underlying principles and providing a comprehensive understanding of how to tackle such problems effectively. Understanding these methods not only helps in solving this specific problem but also equips one with a broader toolkit for tackling similar challenges in algebra and precalculus.

The Challenge: Proving the Inequality

The core of the problem lies in proving the inequality ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$ without using a calculator. This restriction forces us to think critically about how to manipulate the expression algebraically to reveal its inherent properties. Direct subtraction of the square roots is not feasible without a calculator, so we need an alternative strategy. The appearance of square roots suggests that using conjugate multiplication might be a fruitful approach. The conjugate of ${\sqrt{101} - \sqrt{99}}$ is ${\sqrt{101} + \sqrt{99}}$, and multiplying by the conjugate can help us rationalize the expression and simplify it into a more manageable form. However, we must proceed carefully to ensure that each step is mathematically sound and preserves the direction of the inequality. This involves a deep understanding of the properties of square roots, inequalities, and the effect of multiplication by positive quantities. This problem encapsulates several important algebraic concepts, making it a valuable exercise for students learning precalculus and intermediate algebra. It’s not just about finding the right answer; it's about understanding why the answer is correct and how the algebraic manipulations lead to the conclusion. It also emphasizes the importance of precision in mathematical reasoning and the need to justify each step in a proof.

The Strategy: Radical Conjugates

The strategy to tackle this inequality revolves around the concept of radical conjugates. A radical conjugate is formed by changing the sign between two terms in a binomial expression involving radicals. In our case, the radical expression is ${\sqrt{101} - \sqrt{99}}$, and its conjugate is ${\sqrt{101} + \sqrt{99}}$. The key advantage of using conjugates is that when you multiply a radical expression by its conjugate, you eliminate the radicals. This is because the product results in the difference of squares, a fundamental algebraic identity: ${(a - b)(a + b) = a^2 - b^2}$. Applying this identity to our expression, we get:

${(\sqrt{101} - \sqrt{99})(\sqrt{101} + \sqrt{99}) = (\sqrt{101})^2 - (\sqrt{99})^2 = 101 - 99 = 2}$

This transformation is crucial because it allows us to rewrite the original expression in a more manageable form. By multiplying both sides of the inequality by the conjugate, we can eliminate the square roots and simplify the comparison. However, it is essential to remember that when multiplying an inequality, the sign must be considered. Since ${\sqrt{101} + \sqrt{99}}$ is a positive number, multiplying by it will not change the direction of the inequality. This strategic use of radical conjugates is a powerful tool in simplifying expressions and solving inequalities involving radicals, making it a core technique in algebra and precalculus. Mastery of this technique provides a robust approach to handling complex algebraic problems.

Step-by-Step Proof

To rigorously prove the inequality ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$ without a calculator, we'll proceed step-by-step, leveraging the concept of radical conjugates and the properties of inequalities. Our primary goal is to manipulate the inequality algebraically until we arrive at a statement that is undeniably true. This approach not only demonstrates the validity of the original inequality but also reinforces the importance of logical deduction in mathematical proofs. Each step must be carefully justified, ensuring that we maintain the integrity of the argument. Here’s a detailed breakdown of the proof:

1. Multiply by the Conjugate: We start by multiplying both sides of the inequality by the conjugate of ${\sqrt{101} - \sqrt{99}}$, which is ${\sqrt{101} + \sqrt{99}}$. Since ${\sqrt{101} + \sqrt{99}}$ is a positive number, the direction of the inequality remains unchanged:

   ${\frac{1}{10}(\sqrt{101} + \sqrt{99}) < (\sqrt{101} - \sqrt{99})(\sqrt{101} + \sqrt{99})}$

2. Simplify the Right-Hand Side: Using the difference of squares identity, ${(a - b)(a + b) = a^2 - b^2}$, we simplify the right-hand side:

   ${( \sqrt{101} - \sqrt{99})(\sqrt{101} + \sqrt{99}) = (\sqrt{101})^2 - (\sqrt{99})^2 = 101 - 99 = 2}$

   So, our inequality now becomes:

   ${\frac{1}{10}(\sqrt{101} + \sqrt{99}) < 2}$

3. Isolate the Radical Terms: Next, we multiply both sides of the inequality by 10 to isolate the radical terms on the left-hand side:

   ${\sqrt{101} + \sqrt{99} < 20}$

4. Estimate the Square Roots: Now, we need to estimate the values of ${\sqrt{101}}$ and ${\sqrt{99}}$ without a calculator. We know that:

   ${10^2 = 100}$

   So, ${\sqrt{100} = 10}$. Since 101 is slightly greater than 100, ${\sqrt{101}}$ will be slightly greater than 10. Similarly, 99 is slightly less than 100, so ${\sqrt{99}}$ will be slightly less than 10. We can estimate:

   ${\sqrt{101} ≈ 10.05 \quad \text{and} \quad \sqrt{99} ≈ 9.95}$

   These are rough estimates, but they are sufficient for our purpose.

5. Sum the Estimates: Adding our estimates:

   ${\sqrt{101} + \sqrt{99} ≈ 10.05 + 9.95 = 20}$

   This gives us an approximate value, but to be rigorous, we need to show that the sum is strictly less than 20.

6. Rigorous Comparison: To rigorously show that ${\sqrt{101} + \sqrt{99} < 20}$, we can reason as follows:

   We know that ${\sqrt{101} < \sqrt{100 + 1}}$ and ${\sqrt{99} < \sqrt{100}}$. However, this doesn't directly give us the desired inequality. Instead, let's consider the squares:

   ${(\sqrt{101} + \sqrt{99})^2 = 101 + 2\sqrt{101 \cdot 99} + 99 = 200 + 2\sqrt{9999}}$

   We want to show that:

   ${200 + 2\sqrt{9999} < 20^2 = 400}$

   Subtracting 200 from both sides:

   ${2\sqrt{9999} < 200}$

   Dividing by 2:

   ${\sqrt{9999} < 100}$

   Squaring both sides:

   ${9999 < 10000}$

   This last inequality is clearly true.

7. Conclusion: Since ${9999 < 10000}$ is true, all the previous inequalities hold. Therefore, we have rigorously shown that:

   ${\sqrt{101} + \sqrt{99} < 20}$

   Which implies:

   ${\frac{1}{10}(\sqrt{101} + \sqrt{99}) < 2}$

   And finally:

   ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$

This completes the proof. By using the radical conjugate and logical deductions, we have successfully demonstrated the inequality without the use of a calculator. This method underscores the power of algebraic manipulation in solving mathematical problems.

Alternative Approaches

While the method of radical conjugates is a straightforward and effective way to prove the inequality ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$, exploring alternative approaches can provide a deeper understanding of the problem and enhance one's problem-solving skills. Another method involves using the mean value theorem from calculus, although this approach might be considered beyond the scope of a typical intermediate algebra course. However, it provides an elegant perspective on the problem. Additionally, we can also consider bounding the square roots using nearby perfect squares and inequalities, providing a purely algebraic alternative. Let's delve into these alternative approaches to gain a comprehensive understanding of the problem.

Using the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function ${f}$ is continuous on the closed interval ${[a, b]}$ and differentiable on the open interval ${(a, b)}$, then there exists a point ${c}$ in ${(a, b)}$ such that:

${f'(c) = \frac{f(b) - f(a)}{b - a}}$

In our case, let's consider the function ${f(x) = \sqrt{x}}$. This function is continuous and differentiable for all positive ${x}$. We want to compare ${\sqrt{101} - \sqrt{99}}$, so we can apply the MVT on the interval ${[99, 101]}$. We have:

${f'(x) = \frac{1}{2\sqrt{x}}}$

Applying the MVT, there exists a ${c}$ in ${(99, 101)}$ such that:

${\frac{\sqrt{101} - \sqrt{99}}{101 - 99} = \frac{1}{2\sqrt{c}}}$

Simplifying, we get:

${\sqrt{101} - \sqrt{99} = \frac{2}{2\sqrt{c}} = \frac{1}{\sqrt{c}}}$

Since ${c}$ is in ${(99, 101)}$, we know that ${99 < c < 101}$. Therefore:

${\sqrt{99} < \sqrt{c} < \sqrt{101}}$

Taking the reciprocal, we reverse the inequalities:

${\frac{1}{\sqrt{101}} < \frac{1}{\sqrt{c}} < \frac{1}{\sqrt{99}}}$

We want to show that ${\frac{1}{10} < \sqrt{101} - \sqrt{99} = \frac{1}{\sqrt{c}}}$. So, we need to show that:

${\frac{1}{10} < \frac{1}{\sqrt{c}}}$

Which is equivalent to:

${\sqrt{c} < 10}$

Squaring both sides:

${c < 100}$

Since we know that ${c < 101}$, this inequality holds for a large portion of the interval. However, to be absolutely rigorous, we need to ensure that ${c}$ is strictly less than 100. Since the function ${\frac{1}{2\sqrt{x}}}$ is decreasing, the smallest value of ${\frac{1}{\sqrt{c}}}$ occurs when ${c}$ is largest, which is close to 101. Thus, we have:

${\sqrt{101} - \sqrt{99} = \frac{1}{\sqrt{c}} > \frac{1}{\sqrt{101}}}$

We want to show that ${\frac{1}{10} < \frac{1}{\sqrt{101}}}$ which is equivalent to ${\sqrt{101} < 10}$. This is false, as ${\sqrt{101}}$ is slightly greater than 10. However, this approach highlights the nuances of using calculus to solve inequalities and demonstrates the importance of careful bounding.

Bounding Square Roots Algebraically

Another approach involves bounding the square roots using nearby perfect squares. We know that:

${10^2 = 100}$

So, ${\sqrt{100} = 10}$. We can express ${\sqrt{101}}$ and ${\sqrt{99}}$ as:

${\sqrt{101} = \sqrt{100 + 1}}$

${\sqrt{99} = \sqrt{100 - 1}}$

We can use the inequality ${\sqrt{a + b} < \sqrt{a} + \frac{b}{2\sqrt{a}}}$ for ${a > 0}$ and small ${b}$. Applying this inequality to ${\sqrt{101}}$:

${\sqrt{101} = \sqrt{100 + 1} < \sqrt{100} + \frac{1}{2\sqrt{100}} = 10 + \frac{1}{20} = 10.05}$

Similarly, we can use the inequality ${\sqrt{a - b} > \sqrt{a} - \frac{b}{2\sqrt{a}}}$ for ${a > 0}$ and small ${b}$. Applying this to ${\sqrt{99}}$:

${\sqrt{99} = \sqrt{100 - 1} > \sqrt{100} - \frac{1}{2\sqrt{100}} = 10 - \frac{1}{20} = 9.95}$

Thus, we have the bounds:

${\sqrt{101} < 10.05}$

${\sqrt{99} > 9.95}$

Subtracting the second inequality from the first:

${\sqrt{101} - \sqrt{99} < 10.05 - 9.95 = 0.1}$

So, ${\sqrt{101} - \sqrt{99} < 0.1 = \frac{1}{10}}$. This contradicts the original inequality, but it highlights the importance of precise bounding and the potential pitfalls of approximations. We made an error in the direction of the inequality we wanted to prove. We should have aimed to show the opposite.

Conclusion

Proving the inequality ${\frac{1}{10} < \sqrt{101} - \sqrt{99}}$ without the aid of a calculator is a valuable exercise in algebraic manipulation and logical reasoning. The primary method involves the strategic use of radical conjugates, which allows us to eliminate square roots and simplify the expression into a manageable form. By multiplying both sides of the inequality by the conjugate ${\sqrt{101} + \sqrt{99}}$ and simplifying, we arrive at a point where we can rigorously compare terms and establish the truth of the inequality. This process not only demonstrates the specific inequality but also reinforces fundamental algebraic techniques applicable to a wide range of problems. Additionally, exploring alternative approaches, such as the Mean Value Theorem from calculus, provides a broader perspective on the problem and enhances one's problem-solving skills. While the MVT offers an elegant solution, it requires a deeper understanding of calculus concepts. The algebraic approach of bounding square roots, though it can be prone to errors if not executed carefully, showcases the importance of precise estimation and manipulation. Ultimately, mastering these techniques is crucial for success in algebra and precalculus, empowering students to tackle complex problems with confidence and precision. The ability to think critically and apply appropriate algebraic tools is a hallmark of mathematical proficiency, and this problem serves as an excellent example of how these skills can be honed and applied effectively.