Class 9 NUMBER SYSTEMS FOR CBSE NCERT
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Class 9 Chapter - 1 Number Systems

NUMBER SYSTEMS FOR CBSE NCERT

Topic Covered

Numbers and their Decimal Expansions

Numbers and their Decimal Expansions

Here, 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 rationals and irrationals.

We will also explain how to visualise the representation of real numbers on the number line using their decimal expansions. Since rationals are more familiar to us, 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.
Q 3210445310

Find the decimal expansions of `(10)/3,7/8` and `1/7`
Class 9 Chapter 1 Example 5
Solution:

What have you noticed? You should have noticed at least three things:
(i) The remainders either become 0 after a certain stage, or start repeating themselves (ii) The number of entries in the repeating string of remainders is less than the divisor (in `1/3` 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 `1/3, 3` repeats in the quotient and for `1/7` we get the repeating block 142857 in the quotient).

Note

Although we have noticed this pattern using only the examples above, it is true for all rationals of the form `p/q \ \ (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.

`color{fuchsia}color({fuchsia}("Case (i) : The remainder becomes zero")`
In the examples of `1/3` 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 `color{green}(1/3 = 0.3333..)` and `color{green}(1/7 = 0.142857142857142857...)`
The usual way of showing that 3 repeats in the quotient of `1/3` is to write it as `bar0.3` Similarly, since the block of digits 142857 repeats in the quotient of `1/7` we write `1/7` as `bar(0.142857)` where the bar above the digits indicates the block of digit Also 3.57272... can be written as `bar(3.572)` 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, `bar(1.27)` , 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.
Q 3240445313

Show that 3.142678 is a rational number. In other words, express 3.142678 in the from `p/q` where `p` and `q` are integers and `q ne 0.`
Class 9 Chapter 1 Example 6
Solution:

We have `3.142678 = (3142678)/(1000000)` and hence is a rational number.
Now, let us consider the case when the decimal expansion is non-terminating recurring.
Q 3270445316

Show that `0.3333... = 0.bar3` . can be expressed in the form `p/q`, where `p` and `q` are integers and `q ne 0.`
Class 9 Chapter 1 Example 7
Solution:

Since we do not know what `0.bar3` . is , let us call it `‘x’` and so

`x = 0.3333...`
Now here is where the trick comes in. Look at
Now, `10 x = 10 × (0.333...) = 3.333...`
Therefore, `10 x = 3 + x`

Solving for `x,` we get

`9x = 3, i.e., x =1/3`
Q 3280445317

Show that `1.272727... = 1.27` can be expressed in the for `p/q` where `p` and q are integers and `q ne 0.`
Class 9 Chapter 1 Example 8
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`
i.e., `x = (126)/99= (14)/(11)`
You can check the reverse that `(14)/(11) = 1.bar27`
Q 3270545416

Show that `0.2353535... = 0.235` can be expressed in the form `p/q` where `p` and `q` are integers and `q ne 0.`
Class 9 Chapter 1 Example 9
Solution:

Let `x = 0.bar235` . 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`
i.e., `99 x = (233)/10` , which gives `x = (233)/(990)`
You can also check the reverse that (233)/(990) = `0.bar235`
So, every number with a non-terminating recurring decimal expansion can be expressed in the form `p/q (q ne 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 nonterminating recurring. Moreover, a number whose decimal expansion is terminating or non-terminating recurring is rational.
So, 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 nonterminating and non-recurring. Therefore, from the property above, it is irrational. Moreover, notice that you can generate infinitely many irrationals similar to s.

What about the famous irrationals `sqrt2` and `pi?` Here are their decimal expansions up to a certain stage.

`sqrt2` = 1.4142135623730950488016887242096...
`pi` = 3.14159265358979323846264338327950...
(Note that, we often take `(22)/7` as an approximate value for `pi` but `pi ne 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 `sqrt2` 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 `sqrt2` as follows:

`sqrt2 = 1+1/3 + (1/4xx1/3) - (1/34xx1/4 xx1/3) = 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.
Now, let us see how to obtain irrational numbers.
Q 3230645512

Find an irrational number between `1/7` and `2/7`
Class Chapter 1 Example 10
Solution:

We saw that `1/7 = 0.bar(142857)` So, you can easily calculate `2/7=bar(285714)` To find an irrational number betwe en `1/7` and `2/7` we find a number which is non-terminating non-recurring lying between them. Of course, you can find infinitely
many such numbers.

An example of such a number is `0.150150015000150000...`
 
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