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 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.
Example : Find the decimal expansions of `10/3,7/8, and 1/7.
Remainders : 1, 1, 1, 1, 1... Remainders : 6, 4, 0 Remainders : 3, 2, 6, 4, 5, 1,
Divisor : 3 Divisor : 8 3, 2, 6, 4, 5, 1,...
Divisor : 7
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
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.
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 `1/2`= 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 terminating.
Case (ii) : The remainder never 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, `1/3` = 0.3333... and `1/7` = 0.142857142857142857...
The usual way of showing that 3 repeats in the quotient of `1/3` is to write it as 0.3 .
Similarly, since the block of digits 142857 repeats in the quotient of `1/7`, we write `1/7` as 0.142857 , where the bar above the digits indicates the block of digits that repeats.
Also 3.57272... can be written as 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, 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
Example : Show that 3.142678 is a rational number. In other words, express 3.142678 in the form `p/q`, where p and q are integers and q ≠ 0.
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.
Example : Show that 0.3333... = 03. can be expressed in the form `p/q`, where p and q are integers and q ≠ 0.
Solution : Since we do not know what 03. is , let us call it ‘x’ and so x = 0.3333...
Now here is where the trick comes in. Look at 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 : Show that 1.272727... = 1.27 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
100 x = 127.2727...
So, 100 x = 126 + 1.272727... = 126 + x
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`.
Example : Show that 0.2353535... = 0.235 can be expressed in the form `p/q`, where p and q are integers and q ≠ 0.
Solution : Let x = 0.235 . 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`= 0.2`bar35` .
So, every number with a non-terminating recurring decimal expansion can be expressed in the form `p/q` (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 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
The decimal expansion of an irrational number is non-terminating non-recurring.
Moreover, a number whose decimal expansion is non-terminating non-recurring
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 irrationals similar to s.
What about the famous irrationals √2 and p? Here are their decimal expansions upto a certain stage.
√2 = 1.4142135623730950488016887242096...
π = 3.14159265358979323846264338327950...
(Note that, we often take `22/7`as an approximate value for π, but π ≠ `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 (1/4xx 1/3)-( 1/34xx1/4xx1/3) `= 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 p is very interesting.
The Greek genius Archimedes was the first to compute digits in the decimal expansion of π. He showed 3.140845
< π < 3.142857. Aryabhatta (476 – 550 C.E.), 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, π has been
computed to over 1.24 trillion decimal places!
Archimedes (287 BCE – 212 BCE)
Now, let us see how to obtain irrational numbers.
Example : Find an irrational number between `1/7` and `2/7`.
Solution : We saw that `1/7= 0.bar142857` . So, you can easily calculate `2/7 = 0.bar285714.`
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, you can find infinitely many such numbers.
An example of such a number is 0.150150015000150000...
Shaalaa.com | Decimal Expansion of Real Numbers
You know that `1/7=0.bar142857.` 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.]
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?
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.