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Problem 45 Which of the numbers $$\frac{100001}{100002} \text { and } \frac{10000001}{10000002}$$ is bigger?

Short Answer

Expert verified
\( \frac{10000001}{10000002} \) is bigger.

Step by step solution

01

Express the Fractions

We need to compare the fractions \( \frac{100001}{100002} \) and \( \frac{10000001}{10000002} \). To do this effectively, we will rewrite each fraction in a slightly different form that makes comparison easier.
02

Simplify the Fractions

For both fractions, we will use the approximation \( \frac{a}{a+1} \approx 1 - \frac{1}{a} \) when \( a \) is a large number. This means:- For \( \frac{100001}{100002} \), it can be approximated as \( 1 - \frac{1}{100002} \).- For \( \frac{10000001}{10000002} \), it can be approximated as \( 1 - \frac{1}{10000002} \).
03

Compare the Approximations

Now we compare the approximations:- \( 1 - \frac{1}{100002} \) is slightly less than 1, with a correction term of \( \frac{1}{100002} \).- \( 1 - \frac{1}{10000002} \) is slightly less than 1, with a smaller correction term of \( \frac{1}{10000002} \) because it is a larger denominator.Since \( \frac{1}{100002} \) is greater than \( \frac{1}{10000002} \), the fraction with the smaller correction term will be closer to 1.
04

Conclusion

Given that \( 1 - \frac{1}{100002} \) and \( 1 - \frac{1}{10000002} \) are both close to 1, \( 1 - \frac{1}{10000002} \) is the larger fraction since \( \frac{1}{10000002} \) is smaller than \( \frac{1}{100002} \). Thus, \( \frac{10000001}{10000002} \) is greater than \( \frac{100001}{100002} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Approximation Method
The approximation method is a useful technique to simplify the comparison of fractions, especially when dealing with large numbers. It is based on simplifying the difference between the fraction and the number one. If you have a fraction of the form \( \frac{a}{a+1} \), and \( a \) is a large number, you can approximate this fraction as \( 1 - \frac{1}{a} \). This is because as \( a \) becomes larger, the fraction \( \frac{1}{a} \) becomes smaller, making the entire expression very close to 1.

This technique helps to quickly estimate and compare fractions without the need for complex calculations. The simplicity of this method makes it a powerful tool in mathematical problem solving, particularly in exam settings where time is limited.

In the given example, we applied this method to approximate both fractions:
  • \( \frac{100001}{100002} \approx 1 - \frac{1}{100002} \)
  • \( \frac{10000001}{10000002} \approx 1 - \frac{1}{10000002} \)
By comparing these approximations, it's easier to decide which fraction is closer to 1 and consequently larger.
Denominator Analysis
When comparing fractions closer to each other in size, denominator analysis becomes very insightful. This approach hinges on understanding how the size of the denominator affects the whole fraction, especially when it is slightly larger than the numerator.

The rule of thumb is: the larger the denominator in the fraction \( \frac{1}{b} \), the smaller the fraction becomes. Thus, a fraction such as \( 1 - \frac{1}{b} \) becomes larger as \( b \) increases. This is crucial when you have fractions where the numerators are nearly equal to their denominators plus a small increment.

In our problem, the key takeaway is:
  • If \( b_1 < b_2 \), then \( \frac{1}{b_1} > \frac{1}{b_2} \), making \( 1 - \frac{1}{b_1} < 1 - \frac{1}{b_2} \).
This principle helped us conclude which fraction is larger without needing to perform direct division operations.
Mathematical Problem Solving
Mathematical problem solving often requires an interdisciplinary approach combining insights from various methods. Clear reasoning and effective strategies distinguish efficient problem solvers from others.

To solve our fraction comparison problem efficiently, we used a blend of approximation and denominator analysis methods. Breaking down the problem into these small, manageable parts allowed us to leverage their strengths confidently.

Let's break this process down further:
  • First, simplify each fraction using insights from approximation methods to estimate their values as close to 1.
  • Then, utilize denominator analysis to understand the subtleties that even small changes in the denominator can have, providing a clearer pathway to compare accurately.
These methods streamline the path to a solution, ensuring clarity and reinforcing understanding of mathematical concepts. Problem-solving in this structured manner not only aids in finding the right answers but also enriches our mathematical intuition and skills.

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