NCERT Solutions for Class 9 Maths Exercise 1.3 Question 3
Understanding the Question 🧐
This question is the reverse of what we did in Question 1. Here, we are given decimals that have repeating digits and our task is to convert them back into their rational number form, which is &&\frac{p}{q}&& (where p and q are integers and &&q \neq 0&&). This process proves that any decimal with a repeating pattern is a rational number.
Express the following in the form &&\frac{p}{q}&&, where p and q are integers and &&q \neq 0&&.
Part (i): Express &&0.\overline{6}&& in the form &&\frac{p}{q}&& 📝
- Let the number be &&x&&.
&&x = 0.\overline{6} = 0.666…&& —(1)
- Since only one digit is repeating, we multiply both sides by 10 to shift the decimal point one place to the right.
&&10x = 6.666…&& —(2)
- Now, subtract equation (1) from equation (2) to eliminate the repeating decimal part.
&&10x – x = (6.666…) – (0.666…)||
&&9x = 6&&
- Solve for &&x&& and simplify the fraction.
&&x = \frac{6}{9} = \frac{2}{3}&&
Answer: Thus, &&0.\overline{6} = \frac{2}{3}&&.
Part (ii): Express &&0.4\overline{7}&& in the form &&\frac{p}{q}&& 📝
This case is slightly different because one digit (‘4’) does not repeat, while the next digit (‘7’) does.
- Let the number be &&x&&.
&&x = 0.4\overline{7} = 0.4777…&& —(1)
- First, multiply by 10 to move the non-repeating part to the left of the decimal.
&&10x = 4.777…&& —(2)
- Now, multiply equation (2) by 10 again to shift one repeating digit to the left.
&&100x = 47.777…&& —(3)
- Subtract equation (2) from equation (3) to cancel the repeating part.
&&100x – 10x = (47.777…) – (4.777…)||
&&90x = 43&&
- Solve for &&x&&.
&&x = \frac{43}{90}&&
Answer: Thus, &&0.4\overline{7} = \frac{43}{90}&&.
Part (iii): Express &&0.\overline{001}&& in the form &&\frac{p}{q}&& 📝
- Let the number be &&x&&.
&&x = 0.\overline{001} = 0.001001001…&& —(1)
- Since there are three repeating digits, we multiply both sides by 1000 to shift the decimal point three places to the right.
&&1000x = 1.001001001…&& —(2)
- Subtract equation (1) from equation (2).
&&1000x – x = (1.001001…) – (0.001001…)||
&&999x = 1&&
- Solve for &&x&&.
&&x = \frac{1}{999}&&
Answer: Thus, &&0.\overline{001} = \frac{1}{999}&&.
Conclusion and Key Points ✅
As these ncert solutions demonstrate, we can use a simple algebraic method to convert any non-terminating repeating decimal into a fraction of the form &&\frac{p}{q}&&. The key is to create two equations where the repeating part after the decimal is identical, allowing it to be eliminated through subtraction.
- For &&0.\overline{a}&&, the fraction is &&\frac{a}{9}&&. (e.g., &&0.\overline{6} = \frac{6}{9} = \frac{2}{3}&&)
- For &&0.\overline{ab}&&, the fraction is &&\frac{ab}{99}&&.
- For &&0.\overline{abc}&&, the fraction is &&\frac{abc}{999}&&. (e.g., &&0.\overline{001} = \frac{1}{999}&&)
- Let the decimal equal &&x&&.
- Multiply by a power of 10 to shift the repeating block.
- Subtract the original equation from the new one.
- Solve for &&x&& and simplify.
FAQ
Q: What is the first step to convert a repeating decimal to a fraction?
A: The first step is to set the repeating decimal equal to a variable, usually ‘x’. For example, let &&x = 0.666…&&.
Q: Why do we multiply by powers of 10 in this method?
A: We multiply by powers of 10 (like 10, 100, 1000) to shift the decimal point. The goal is to create two equations where the infinite repeating part after the decimal point is identical, so it can be eliminated through subtraction.
Q: How do you decide whether to multiply by 10, 100, or 1000?
A: The power of 10 depends on the number of digits in the repeating block. If one digit repeats, you multiply by 10. If two digits repeat, you multiply by 100. If three digits repeat, you multiply by 1000, and so on.
Q: What do you do if there’s a non-repeating part, like in &&0.4\overline{7}&&?
A: If there’s a non-repeating part, you create two equations. First, multiply by a power of 10 to move the decimal just past the non-repeating part (e.g., &&10x = 4.777…&&). Then, multiply again to shift one full repeating block (e.g., &&100x = 47.777…&&). Finally, you subtract these two new equations.
Q: Does this algebraic method work for all repeating decimals?
A: Yes, this method is a reliable way to convert any repeating decimal into its equivalent fraction in the form &&\frac{p}{q}&&, which proves that all repeating decimals are rational numbers.
Further Reading
To practice more problems and solidify your understanding of rational numbers, you can refer to the official NCERT textbooks.
