NCERT Solutions for Class 9 Maths Exercise 1.3 Question 6

NCERT Solutions for Class 9 Maths Exercise 1.3 Question 5

Understanding the Question 🧐

This question has two parts. First, it asks for the theoretical maximum number of digits in the repeating part of the decimal expansion of &&\frac{1}{17}&&. Second, it asks us to verify this by actually performing the long division. This problem helps us understand the properties of rational numbers and their decimal forms. These ncert solutions will guide you through both parts.

What can the maximum number of digits be in the repeating block of digits in the decimal expansion of &&\frac{1}{17}&&? Perform the division to check your answer.

Step-by-Step Solution 📝

Part 1: Finding the Maximum Number of Repeating Digits

For any rational number in the form &&\frac{p}{q}&&, where &&p&& and &&q&& are integers and &&q \neq 0&&, the maximum number of digits in the repeating block of its decimal expansion is always less than the denominator, &&q&&. The maximum possible length is &&q-1&&.

In our case, the fraction is &&\frac{1}{17}&&.
The denominator, &&q&&, is &&17&&.
Therefore, the maximum number of digits in the repeating block is &&17 – 1 = 16&&.

So, theoretically, the repeating block in the decimal expansion of &&\frac{1}{17}&& can have at most 16 digits.

Part 2: Performing the Long Division to Verify

Now, let’s perform the long division of &&1 \div 17&& to find the actual repeating block and check our theory.

1 ÷ 17 = 0.0588235294117647… 100 (Remainder 15) 150 (Remainder 14) 140 (Remainder 4) 40 (Remainder 6) 60 (Remainder 9) 90 (Remainder 5) 50 (Remainder 16) 160 (Remainder 7) 70 (Remainder 2) 20 (Remainder 3) 30 (Remainder 13) 130 (Remainder 11) 110 (Remainder 8) 80 (Remainder 12) 120 (Remainder 1)

We started with the dividend &&1&&. After &&16&& steps of division, the remainder is &&1&& again. This means the entire sequence of division will now repeat, and so will the digits in our quotient.

The quotient we obtained is &&0.0588235294117647…&&
The repeating block of digits is 0588235294117647.
Let’s count the number of digits in this block: There are exactly 16 digits.

Conclusion and Key Points ✅

Our long division confirms the theoretical prediction. The decimal expansion of &&\frac{1}{17}&& is a non-terminating, repeating decimal.

Final Answer: &&\frac{1}{17} = 0.\overline{0588235294117647}&&. The maximum number of digits in the repeating block is 16.

Trick: To quickly find the maximum possible length of a repeating decimal for a fraction &&1/q&& (where q is a prime number), just calculate &&q-1&&. This gives you the upper limit without doing any division!

Points to Remember:

A rational number (a number of the form &&\frac{p}{q}&&) will have either a terminating decimal expansion or a non-terminating repeating decimal expansion.
The maximum length of the repeating block of digits for &&\frac{p}{q}&& is always less than &&q&&.
The long division process repeats as soon as a remainder is repeated.

FAQ

Q: How can you determine the maximum length of the repeating block for &&\frac{1}{17}&& without division?

A: For any rational number of the form &&\frac{p}{q}&&, the maximum number of digits in the repeating block of its decimal expansion is always less than the denominator, &&q&&. For &&\frac{1}{17}&&, the denominator is &&17&&, so the maximum number of repeating digits is &&17 – 1 = 16&&.

Q: What does a non-terminating repeating decimal mean?

A: A non-terminating repeating decimal is a decimal number that continues infinitely with a specific sequence of digits repeating over and over. For &&\frac{1}{17}&&, the block of digits &&0588235294117647&& repeats endlessly.

Q: Why does the division for &&\frac{1}{17}&& result in a repeating block of 16 digits?

A: During the long division of &&1&& by &&17&&, the remainders can be any integer from &&1&& to &&16&&. Once a remainder repeats (in this case, the remainder becomes &&1&& again), the sequence of digits in the quotient starts repeating. Since there are &&16&& possible non-zero remainders, the block repeats after a maximum of &&16&& steps.

Q: What is the final decimal expansion of &&\frac{1}{17}&&?

A: The decimal expansion of &&\frac{1}{17}&& is &&0.\overline{0588235294117647}&&, where the bar over the digits indicates that this block of &&16&& digits repeats infinitely.

Q: Is &&\frac{1}{17}&& a rational or irrational number?

A: &&\frac{1}{17}&& is a rational number because it can be expressed in the form &&\frac{p}{q}&&, where &&p&& and &&q&& are integers and &&q&& is not zero. Its decimal expansion is non-terminating but repeating, which is a characteristic of rational numbers.