NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

In this section, we are going to study rational and irrational numbers from a different point of view. We will look at the decimal expansions of real numbers and see if we can use the expansions to distinguish between rational and irrational numbers. We will also explain how to visualise the representation of real numbers on the number line using their decimal expansions. Since we are more familier with rational numbers, let us start with them. Let us take three examples : 10÷3, 7÷8, 1÷7.

Pay special attention to the remainders and see if you can find any pattern.

Example 5

Find the decimal expansions of 10÷3,7÷8 and 1÷7.

Solution :

(i) Let us look at the division of 10 by 3.

We can see for ourselves that as we divide 10 by 3 we get a remainder of 1 and we write a zero , and again divide it by 3 and get a remainder of 1. Likewise, as we keep dividing we will keep getting a remainder of 1 and each time we will get a quotient of 3.

(ii) Let us divide 7 by 8. We will find that the remainders are 6,4,0 and the quotient is 0.875.

(iii ) Let us divide 1 by 7, We observe that the remainders are 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, 1,.. and the quotient is 0.142857….

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What have we noticed? We should have noticed at least three things:
(i) The remainders either become 0 after a certain stage, or start repeating itself.
(ii) The number of entries in the repeating string of remainders is less than the divisor itself; for example, in 10÷3, only one number repeats itself and the divisor is 3, in 1÷7, there are six entries 326451 in the repeating string of remainders and 7 is the divisor.
(iii) If the remainders repeat, then we get a repeating block of digits in the quotient; for example, for 10÷3, 3 repeats in the quotient and for 1÷7, we get the repeating block 142857 in the quotient.

Although, we have noticed this pattern using only the examples above, it is true for all rational numbers of the form p ÷ q, where, q ≠ 0. On division of p by q, two main things happen -
either the remainder becomes zero or
never becomes zero and we get a repeating string of remainders.
Let us look at each case separately.

Case (i) : The remainder becomes zero

In the example of 7÷8 , we found that the remainder becomes zero after some steps and the decimal expansion of 7÷8 = 0.875. Other examples are ½ = 0.5, 639÷250 = 2.556. In all these cases, the decimal expansion terminates or ends after a finite number of steps. We call the decimal expansion of such numbers as terminating.

Case (ii) : The remainder never becomes zero

In the examples of 10÷3 and 1÷7 , we notice that the remainders repeat after a certain stage forcing the decimal expansion to go on for ever. In other words, we have a repeating block of digits in the quotient. We say that this expansion is non-terminating recurring. For example,
10÷3 = 3.3333... and
1÷7 = 0.142857142857142857...

The usual way of showing that 3 repeats in the quotient of 10÷3 is to write it as 3.3 with a bar above the repeating number 3. Similarly, since the block of digits 142857 repeats in the quotient of 1÷7, we write 1÷7 as 0.142857 with a bar over the repeating block 142857, where the bar above the digits indicates the block of digits that repeats.

Also 3.57272... can be written as 3.572 with bar over 72.

So, all these examples give us non-terminating recurring (repeating) decimal expansions.

Thus, we see that the decimal expansion of rational numbers have only two choices: either they are terminating or non-terminating recurring.

Now suppose, on the other hand, on your walk on the number line, you come across a number like, 3.142678 whose decimal expansion is terminating or a number like 1.272727... that is, 1.27 with bar over 27, whose decimal expansion is non-terminating recurring, can you conclude that it is a rational number? The answer is yes!

We will not prove it but illustrate this fact with a few examples. The terminating cases are easy.

Example 6

Show that 3.142678 is a rational number. In other words, express 3.142678 in the form of p/q , where p and q are integers and q ≠ 0.

Solution :

We have 3.142678 = 3142678 ÷1000000 , and hence is a rational number.

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Now, let us consider the case when the decimal expansion is non-terminating recurring.

Example 7

Show that 0.3333... = 0.3 bar can be expressed in the form p ÷q , where p and q are integers and q ≠ 0.

Solution :

Since we do not know what 0.3 bar is , let us call it ‘x’, and so

x = 0.3333...

Now here is where the trick comes in. Look at

Multiply both sides with 10. We get,
10 x = 10 × (0.333...) = 3.333...

Now, 3.3333... = 3 + x, since x = 0.3333...

Therefore, 10 x = 3 + x

Solving for x, we get

9x = 3, i.e., x = 1÷3

Example 8

Show that 1.272727... = 1.27 with bar over 27 can be expressed in the form p/q , where p and q are integers and q ≠ 0.

Solution :

Let x = 1.272727...

Since two digits are repeating, we multiply x by 100 to get
100 x = 127.2727...

So, 100 x = 126 + 1.272727... = 126 + x

Therefore, 100 x − x = 126, i.e., 99 x = 126

Hence, x = 126÷99 = 14÷11

You may check the reverse that 14÷11 = 1.27 bar over 27.

Example 9

Show that 0.2353535... = 0.235 with bar on 35 can be expressed in the form p/q , where p and q are integers and q ≠ 0.

Solution :

Let x = 0.235 bar on 35.

Over here, note that 2 does not repeat, but the block 35 repeats. Since two digits are repeating, we multiply x by 100 to get
100 x = 23.53535...

So, 100 x = 23.3 + 0.23535... = 23.3 + x

Therefore, 99 x = 23.3

Hence, 99 x = 233 ÷10 , which gives x = 233 ÷ 990

You may also check the reverse that 233 ÷990 = 0.235 with bar on 35.

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So, every number with a non-terminating recurring decimal expansion can be expressed in the form p/q, where q ≠ 0, where p and q are integers.

Let us summarise our results in the following form:

The decimal expansion of a rational number is either terminating or non terminating recurring. Moreover, a number whose decimal expansion is terminating or non-terminating recurring is rational.

Important technique for writing repeating decimal numbers into p ÷q form by Dr T K Bansal:

Follow the following steps,
(i) Observe the repeating number carefully.
(ii) Find the number of repeating digits in the given repeating number. For example in the number 0.2343434…, there are 2 digits that are repeating, namely 3, and 4.
(iii) Find the numbers that are not repeating. For example, in the number 0.234, 0.2 is the non repeating.
(iv) Shift the decimal point to the right by as many digits as the number of repeating numbers in the given number. For example, in 0.234, shift the decimal to the right by 2 digits and the number becomes 23.4.
(v) Now subtract the non repeating number from this number. For example 0.2 in this case and the number becomes 23.2.
(vi) Last step, now divide this number by as many nines as is the number of repeating digits in the given number. For example 2 in the present case, and the number becomes 23.2÷99 or 232÷990.
End of the technique.

Let us try to work back. Please check yourself that 232÷990 = 0.2343434..

From the technique, it appears that it is a difficult one. However, it is very easy to work, and can be done verbally. Let us try some examples quickly;

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Rewrite the following repeating decimal numbers into fractions;
(i) 2.333.. .. = (23 − 2) ÷9 = 21 ÷9 = 7÷3.
(ii) 2.6666.. .. = (26 − 2) ÷9 = 24 −9 = 8 ÷3.
(iii) 4.32323232.. .. = (432 − 4) ÷99 = 428÷99

So you can see for yourself, that you can solve the problems without writing anything on a notebook. Which is very important for visually impaired students.

Eventually, now we know what the decimal expansion of a rational number can be.
What about the decimal expansion of irrational numbers?
Because of the property above, we can conclude that their decimal expansions are non-terminating non-recurring.

So, the property for irrational numbers, similar to the property stated above for rational numbers, is

The decimal expansion of an irrational number is non-terminating non-recurring. Moreover, a number whose decimal expansion is non-terminating non-recurring is irrational.

Recall s = 0.10110111011110... from the previous section. Notice that it is non terminating and non-recurring. Therefore, from the property above, it is irrational. Moreover, notice that you can generate infinitely many irrational numbers similar to s.

What about the famous irrational numbers √2 and π? Here are their decimal expansions up to a certain stage.

√2 = 1.4142135623730950488016887242096...

π = 3.14159265358979323846264338327950...

(Note that, we often take 22÷7 as an approximate value for π, but π is not equal to 22÷7.)

Over the years, mathematicians have developed various techniques to produce more and more digits in the decimal expansions of irrational numbers. For example, you might have learnt to find digits in the decimal expansion of √2 by the division method. Interestingly, in the Sulbasutras (rules of chord), a mathematical treatise of the Vedic period (800 BC - 500 BC), you find an approximation of √ 2 as follows:

√2 = 1 +1÷3 + (1÷4 × 1÷3) - (1÷34 × ¼ × 1÷3) = 1.4142156

Notice that it is the same as the one given above for the first five decimal places. The history of the hunt for digits in the decimal expansion of pi is very interesting.

Start of yellow box:

The Greek genius Archimedes was the first to compute digits in the decimal expansion of π. He showed that

3.140845 < π < 3.142857.

Aryabhatta (476 - 550 AD), the great Indian mathematician and astronomer, found the value of π correct to four decimal places (3.1416).

Using high speed computers and advanced algorithms, pi has been computed to over 1.24 trillion decimal places!

 

Archimedes (287 BCE - 212 BCE)]

End of yellow box.

Now, let us see how to obtain irrational numbers.

Example 10

Find an irrational number between 1/7 and 2/7.

Solution :

We saw that 1/7 = 0.142857 with bar over 142857. So, you can easily calculate 2/7 = 0.285714 with bar over 285714.

To find an irrational number between 1÷7 and 2÷7, we find a number which is non-terminating non-recurring lying between them. Of course, we can find infinitely many such numbers.

An example of such a number is 0.150150015000150000...

EXERCISE 1.3

Q1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) 36/100
(ii) 1/11
(iii) 4 × (1/8)
(iv) 3/13
(v) 2/11
(vi) 329/400

A1.
(i) 0.36, terminating.
(ii) 0.09 (bar on 09), non-terminating repeating.
(iii) 4.125, terminating.
(iv) 0.230769 (bar on 230769), non-terminating repeating.
(v) 0.18 (bar on 18) non-terminating repeating.
(vi) 0.8225 terminating.

Q2. You know that 1÷7 = 0.142857 with bar over 142857. Can you predict what the decimal expansions of 2÷7, 3÷7, 4÷7, 5÷7, 6÷7 are, without actually doing the long division? If so, how?
[Hint : Study the remainders while finding the value of 1÷7 carefully.]

A2. 2÷7 = 2 × 1÷7 = 0.285714 (bar on 285714),
3÷7 = 3 × 1÷7 = 0.428571 (bar on 428571),
4÷7 = 4 × 1÷7 = 0.571428 (bar on 571428),
5÷7 = 5 × 1÷7 = 0.714285 (bar on 714285),
6÷7 = 6 × 1÷7 = 0.857142 (bar on 857142)

Q3. Express the following in the form p / q , where p and q are integers and q ≠ 0.
(i) 0.6 with bar on 6
(ii) 0.47 with bar on 47
(iii) 0.001 with bar on 001

A3.
(i)2÷3 [Let x = 0.666. . . So 10x = 6.666. . . or, 10x = 6 + x or , x = 6÷9 = 2÷3]
(ii) 43÷90
(iii) 1÷999

Q4. Express 0.99999 .... in the form p ÷ q. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

A4. 1 [Let x = 0.9999. . . So 10 x = 9.999. . . or, 10 x = 9 + x or, x = 1]

Q5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1÷17? Perform the division to check your answer.

A5. 0.0588235294117647 (bar on 0588235294117647)

Q6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

A6. The prime factorisation of q has only powers of 2 or powers of 5 or both.

Q7. Write three numbers whose decimal expansions are non-terminating non-recurring.

A7. 0.01001000100001. . ., 0.202002000200002. . ., 0.003000300003. . .

Q8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

A8. 0.75075007500075000075. . ., 0.767076700767000767. . ., 0.808008000800008. . .

Q9. Classify the following numbers as rational or irrational :
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478...
(v) 1.101001000100001...

A9.
(i) and (v) irrational;
(ii), (iii) and (iv) rational.