# Number System – Class IX [CBSE]

Rational Number:

A number ‘r’ is called a rational number, if it can be written in the form p/q ,where p and q are integers and q ≠ 0. The collection of rational numbers is denoted by Q.

(Why do we insist that q ≠ 0?)

Notice that all the numbers now in the bag can be written in the form p/q, where p and q are integers and q ≠ 0. For example, –25 can be written as –25/1 here p = –25 and q = 1. Therefore, the rational numbers also include the natural numbers, whole numbers and integers.

Example 1 : Are the following statements true or false? Give reasons for your answers.

(i) Every whole number is a natural number.

(ii) Every integer is a rational number.

(iii) Every rational number is an integer.

Solution :

(i) False, because zero is a whole number but not a natural number.

(ii) True, because every integer m can be expressed in the form m/1, and so it is a rational number.

(iii) False, because 3/5 is not an integer.

Example 2 : Find five rational numbers between 1 and 2.

We can approach this problem in at least two ways.

Solution:

Recall that to find a rational number between r and s, you can add r and s and divide the sum by 2, that is r + s/2 lies between r and s. So, 3/2 is a number between 1 and 2. You can proceed in this manner to find four more rational numbers between 1 and 2. These four numbers are 5/4 ,11/8 ,13/8 and 7/4 .

Remark : Notice that in Example 2, you were asked to find five rational numbers between 1 and 2. But, you must have realized that in fact there are infinitely many rational numbers between 1 and 2. In general, there are infinitely many rational numbers between any two given rational numbers.

There are infinitely many rational numbers between any two given rational numbers.

### Number System – Exercise 1.1

Irrational Numbers:

A number ‘s’ is called irrational, if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.

Some examples are: √2 , √3, √15, π, 0.10110111011110…

Remark : Recall that when we use the symbol √, we assume that it is the positive square root of the number. So √4 = 2, though both 2 and –2 are square roots of 4.

Example 3 : Locate √2 on the number line.

Solution : It is easy to see how the Greeks might have discovered √2 . Consider a unit square OABC, with each side 1 unit in length (see Fig 1.6).Then you can see by the Pythagoras theorem that OB = √(12 + 12) = √2 . How do we represent √2 on the number line?

This is easy. Transfer Fig. 1.6 onto the number line making sure that the vertex O coincides with zero (see Fig. 1.7). We have just seen that OB = √2 . Using a compass with centre O and radius OB, draw an arc intersecting the number line at the point P. Then P corresponds to √2 on the number line.

Example 4 : Locate √3 on the number line. Construct BD of unit length perpendicular to OB (as in Fig. 1.8). Then using the Pythagoras theorem, we see that OD = √(√2)2+12) = √3 . Using a compass, with centre O and radius OD, draw an arc which intersects the number line at the point Q. Then Q corresponds to √3 . In the same way, you can locate n for any positive integer √n, after √(n – 1) has been located.

Number System – Exercise 1.2

Real Numbers and their Decimal Expansions:

In this section, we are going to study rational and irrational numbers from a different point of view.

Example 5: Find the decimal expansion of 10/3 , 7/8 and 1/7Solution:

3)       10      (3.333…

9

——————————

10

9

——————————–
10

9
——————————–
10

9

——————————-
1

Remainder 1, 1, 1, 1, 1 ; Divisor :3

8)       7.0     (0.875

64

———————————-
60
56
———————————
40
40
———————————
0

Remainders 6, 4, 0; Divisor :8

7)       1.0     (0.142857

7

————————————-
30

28

————————————-

20
14
————————————
60
56

————————————-
40
35
————————————–
50
49
————————————-
1
Remainder 3, 2, 6, 4, 5, 1, 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 the quotient).

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.

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 forever. 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 .

Number System – Exercise 1.3

Representing Real Numbers on the Number Line:

Example 11: Visualize the representation of 5.377777…. on the number line  up to 5 decimal places, that is, up to 5.37777…

Solution:

Once again we proceed by successive magnification, and successively decrease the lengths of the portions of the number line in which 5.377777……. is located. First, we see that 5.37 is located between 5 and 6. In the next step, we locate 5.37 between 5.3 and 5.4. To get a more accurate visualization of the representation, we divide this portion of the number line into 10 equal parts and use a magnifying glass to visualize that 5.37 lies between 5.37 and 5.38. To visualize 5.37 more accurately, we again divide the portion between 5.37 and 5.38 into ten equal parts and use a magnifying glass to visualize that 5.37 lies between 5.377 and 5.378. Now to visualize 5.37 still more accurately, we divide the portion between 5.377 an 5.378 into 10 equal parts, and visualize the representation of 5.37 as in Fig. 1.14 (iv). Notice that 5.37 is located closer to 5.3778 than to 5.3777

Number System – Exercise 1.4

Operation on Real Numbers:

Example 12: Check whether 7√5, 7/√5. √2+21 , π – 2 are irrational numbers or not.

Solution:

√5 = 2.236 …,

√2 = 1.4142…,

π = 3.1415…,

Then, 7√5 = 15.652…,

7/5 = 75/55 = 75/5 = 3.1304…,

√2 + 21 = 22.4142…,

π – 2 = 1.1415…,

All these are non-terminating non-recurring decimals. So, all these are irrational numbers. Now, let us see what generally happens if we add, subtract, multiply, divide, take square roots and even nth roots of these irrational numbers, where n is any natural number. Let us look at some examples.

Example 13: Add 2√2 + 5√3 and √2 – 3√3

Solution:

(2√2 + 5√3) + (√2 – 3√3) = (2√2 +√2) + (5√3 – 3√3)

=  (2 + 1)√2 + (5 – 3)√3

= 3√2 + 2√3

Example 14: Multiply 6√5 by 2√5

Solution:

6√5 x 2√5 = 6 x 2 x √5 x √5 = 12 x 5 = 60

Example 15: Divide 8√15 by 2√3

Solution:

8√15 ÷ 2√3 = 83 x 5/2√3 = 4√5

These examples may lead you to expect the following facts, which are true:

(i) The sum or difference of a rational number and an irrational number is irrational.

(ii) The product or quotient of a non-zero rational number with an irrational number is irrational.

(iii) If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational.

Example 16: Simplify the following expressions:

(i) (5 + √7) (2 + √5)

(ii) (5 + √5)(5 – √5)

(iii) (√3 + √7)2

(iv) (√11 – √7)( √11 + √7)

Solution:

(i) (5 + √7) (2 + √5) = 10 + 5√5 + 2√7 + √35

(ii) (5 + √5)(5 – √5) = 25 – 5 = 20

(iii) (√3 + √7)2 = (3 + 7 + 2√3√7) = 10 + 2√21

(iv) (√11 – √7)( √11 + √7) = (√11)2 – (√7)2 = 11 – 7 = 4

Remark : Note that ‘simplify’ in the example above has been used to mean that the expression should be written as the sum of a rational and an irrational number.

We now list some identities relating to square roots, which are useful in various ways. You are already familiar with some of these from your earlier classes. The remaining ones follow from the distributive law of multiplication over addition of real numbers, and from the identity (x + y) (x – y) = x2 – y2 , for any real numbers x and y. Let a and b be positive real numbers. Then

(i) √(ab) = √a. √b

(ii) √(a/b) = √a/√b

(iii) (√a + √b) (√a – √b) = a – b

(iv) (a + √b) (a – √b) = a2 – b

(v) (√a + √b) (√c + √d) = √(ac) + √(ad)+ √(bc)+ √(bd)

(vi) (√a + √b)2 = a + 2√(ab) + b

Example 17 : Rationalize the denominator of 1/√2

Solution : We want to write 1/√2 as an equivalent expression in which the denominator is a rational number. We know that √2 . √2 is rational.

We also know that multiplying 1/√2 by √2/√2 will give us an equivalent expression, since √2/√2  = 1.

So, we put these two facts together to get,

1/√2 = 1/√2 x √2/√2 = √2/2

In this form, it is easy to locate 1/√2 on the number line. It is half way between 0 and √2

Example 18: Rationalize the denominator of 1/2 + √3

Solution:

We use the identity (iv) given earlier. Multiply and divide 1/2+√3 by 2 – √3 to get 1/2+√3 x 2+√3/2-√3 = 2 – √3/4 – 3 = 2 – √3

Example 19: Rationalize the denominator of 5/√3-√5

Solution:

Here we use the identity (iii) given earlier

So, 5/√3-√5 x √3+√5/√3+√5 = 5(√3+√5)/3 – 5 = (-5/2)( √3+√5)

Example 20: Rationalize the denominator of 1/7+3√2

Solution:

1/7+3√2 = 1/7+3√2 x 7-3√2/7-3√2 = 7-3√2/49-18 = 7-3√2/31

So, when the denominator of an expression contains a term with a square root (or a number under a radical sign), the process of converting it to an equivalent expression whose denominator is a rational number is called rationalizing the denominator.

Number system – Exercise 1.5

Law of Exponents for Real Numbers:

Example 21: Simplify

(i) 22/3.21/3

(ii)(31/5)4

(iii) 71/5/71/3

(iv) 131/5.171/5

Solution:

(i) 22/3.21/3

= 2(2/3 x 1/3)

= 23/3

= 21

= 2

(ii)(31/5)4

= 3 (1/5)x4

= 34/5

(iii) 71/5/71/3

= 7 1/5 – 1/3

= 7 3-5/15

= 7-2/15

(iv) 131/5.171/5

= (13 x 17)1/5

= 2211/5

Number System – Exercise 1.6

Summary:

1. A number r is called a rational number, if it can be written in the form p/q , where p and q are integers and q ≠ 0.
2. A number s is called a irrational number, if it cannot be written in the form p/q , where p and q are integers and q ≠ 0.
3. 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.
4. 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.
5. All the rational and irrational numbers make up the collection of real numbers.
6. There is a unique real number corresponding to every point on the number line. Also, corresponding to each real number, there is a unique point on the number line.
7. If r is rational and s is irrational, then r + s and r – s are irrational numbers, and rs and r/s are irrational numbers, r ≠ 0.
8. For positive real numbers a and b, the following identities hold:

(i) √(ab) = √a. √b

(ii) √(a/b) = √a/√b

(iii) (√a + √b) (√a – √b) = a – b

(iv) (a + √b) (a – √b) = a2 – b

(v) (√a + √b) (√c + √d) = √(ac) + √(ad)+ √(bc)+ √(bd)

(vi) (√a + √b)2 = a + 2√(ab) + b

1. To rationalise the denominator of 1/√a + b we multiply this by , √a – b/√a – b , where a and b integers.
2. Let a > 0 be a real number and p and q be rational numbers. Then

(i) ap + aq = ap + q

(ii) (ap)q = apq

(iii)ap/aq = ap-q

(iv)apbp = (ab)p