Class 9 Operations on Real Numbers

### Topic covered

Operations on Real Numbers

### Operations on Real Numbers

You have learnt that rational numbers satisfy the commutative, associative and distributive laws for addition and multiplication.

Moreover, if we add, subtract, multiply or divide (except by zero) two rational numbers, we still get a rational number (that is, rational numbers are ‘closed’ with respect to addition, subtraction, multiplication and division).

It turns out that irrational numbers also satisfy the commutative, associative and distributive laws for addition and multiplication.

However, the sum, difference, quotients and products of irrational numbers are not always irrational. For example, color{orange}((sqrt6)+(-sqrt6),(sqrt(2)-(sqrt2),(sqrt3).(sqrt3)) and color{orange}((sqrt17)/(sqrt17)) are rationals.

Let us look at what happens when we add and multiply a rational number with an. irrational number. For example sqrt3 is irrational. What about color{blue}(2 + sqrt3 and 2sqrt(3)?).

Since color{green}(2 + sqrt3) and color{blue}(2 + sqrt3) Therefore, both color{blue}(2 + sqrt3) and color{blue}(2 sqrt3) are also irrational numbers.
Q 3210645519

Check whether 7 sqrt5, 7/(sqrt5) , sqrt2 +21, pi -2 are irrational numbers or not.
Class 9 Chapter 1 Example 12
Solution:

sqrt5 = 2.235., sqrt2 = 1.4142....pi=3.1415...

Then 7 sqrt5 = 15.652 ....., 7/(sqrt5) = (7sqrt5)/(sqrt5 sqrt5) = (7sqrt5)/5 = 3.1304...

sqrt2 + 21 = 22.4142..., pi – 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.
Q 3210745610

Add 2sqrt2 +5sqrt3 and sqrt2-3sqrt3
Class 9 Chapter 1 Example 13
Solution:

(2sqrt2+5)+(sqrt2-3sqrt3)= (2sqrt+sqrt2)+(5 sqrt3-3sqrt3)
=(2+1)sqrt+(5-3)sqrt3=3 sqrt2+2sqrt3
Q 3220745611

Multiply 6 sqrt(5) by 2sqrt5
Class 9 Chapter 1 Example 14
Solution:

6sqrt(5)xx2sqrt(5)= 6xx2xxsqrt5xxsqrt5= 12 xx5 = 60
Q 3220167011

Divide 8sqrt15 by 2 sqrt3
Class 9 Chapter 1 Example 15
Solution:

8sqrt15 ÷ 2 sqrt3= (8sqrt3xx sqrt5)/(2sqrt3)=4sqrt5

### Note

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.

We now turn our attention to the operation of taking square roots of real numbers.
Recall that, if a is a natural number, then sqrta=b means b^2=a and b>a he same definition can be extended for positive real numbers.

Let a > 0 be a real number. Then sqrta = b menas b^2=a b > 0.
In Section 1.2, we saw how to represent sqrtn for any positive integer n on the number line. We now show how to find sqrtx for any given positive real number x geometrically.

For example, let us find it for x = 3.5, i.e., we find sqrt(3.5) geometrically.

Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such that AB = 3.5 units (see Fig. 1.15). From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.
Then, BD = sqrt(3.5)

More generally, to find sqrtx for any positive real number x, we mark B so that AB = x units, and, as in Fig. 1.16, mark C so that BC = 1 unit. Then, as we have done for the case x = 3.5, we find BD = sqrtx
(see Fig. 1.16).

We can prove this result using the Pythagoras Theorem.

Notice that, in Fig. 1.16, Delta OBD is a right-angled triangle. Also, the radius of the circle is (x+1)/2
Therefore, OC = OD = OA = (x+1)/2

Now, OB = x - ((x+1)/2) = (x-1)/2

So, by the Pythagoras Theorem, we have
BD^2 = OD^2 –OB^2 = ((x+1)/2)^2 - (x-1)/2 = (4x)/4 =x
Now, OB = x- ((x+1)/2) = (x-1)/2

So, by the Pythagoras Theorem, we have

BD^2 = OD^2 – OB^2 = ((x+1)/2)^2 - ((x-1)/2)^2 = (4x)/4=x
This shows that BD = sqrtx

This construction gives us a visual, and geometric way of showing that sqrtx exists for
all real numbers x > 0.

If you want to know the position of sqrtx on the number line, then let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with centre B and radius BD, which intersects the number line in E (see Fig. 1.17). Then, E represents sqrtx

We would like to now extend the idea of square roots to cube roots, fourth roots, and in general nth roots, where n is a positive integer. Recall your understanding of square roots and cube roots from earlier classes.

What is 8sqrt8 ? Well, we know it has to be some positive number whose cube is 8, and you must have guessed 8sqrt8 =2 . Let us try 5sqrt(243) Do you know some number b such that b^2=243? The answer is 3. Therefore, 5sqrt(243)=3

From these examples, can you define nsqrta for a real number a > 0 and a positive integer n?

Let a > 0 be a real number and n be a positive integer. Then 'sqrt' used in sqrt2, 3sqrt8,nsqrta etc. is called the radical sign.
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)=x^2-y^2 for any real numbers x and y. Let a and b be positive real numbers. Then

(i) sqrt(ab) = sqrtasqrtb
(ii) sqrt(a/b)=sqrt(a)/sqrt(b)
(iii) (sqrta+sqrtb)(sqrta-sqrtb)=a-b
(iv)(a+sqrtb)(a-sqrtb)=a2-b
(v)(sqrta+sqrtb)(sqrtc+sqrtd)=sqrt(ac)+sqrt(ad)+sqrt(bd)
(vi) (sqrta+sqrtb)^2=a+2sqrt(Ab)+b
Q 3230167012

Simplify the following expressions:

(i) (5+sqrt7)(2+sqrt(5)
(ii)(5+sqrt5)(5-sqrt5)
(iii)(sqrt3+sqrt7)^2
(iv)(sqrt11-sqrt7)(sqrt11+sqrt7)
Class 9 Chapter 1 Example 16
Solution:

(i) (5+sqrt7)(2+sqrt(5) = 10 +5sqrt5 +2 sqrt7 +sqrt35
(ii)(5+sqrt5)(5-sqrt5)= 5^2- (sqrt5)^2=25 -5 = 20
(iii)(sqrt3+sqrt7)^2 = (sqrt3)^2 + 2sqrt3 sqrt7 + (sqrt7)^2=3 2sqrt(21) +7 =1 0+2 sqrt21
(iv)(sqrt11-sqrt7)(sqrt11+sqrt7) = (sqrt11)^2-(sqrt7)^2=11-7=4

"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 end this section by considering the following problem. Look at 1/(sqrt2) Can you tell where it shows up on the number line? You know that it is irrational. May be it is easier to handle if the denominator is a rational number. Let us see, if we can ‘rationalise’ the denominator, that is, to make the denominator into a rational number. To do so, we need the identities involving square roots. Let us see how.
Q 3240167013

Rationalise the denominator of 1/(sqrt2)
Class 9 Chapter 1 Example 17
Solution:

We want to write 1/(sqrt2) as an equivalent expression in which the denominator is a rational number. We know that sqrt2.sqrt2 is rational. We also know that multiplying 1/sqrt2 by (sqrt2)/(sqrt2) will give us an equivalent expression, since (sqrt2)/(sqrt2) =1 So, we put these two facts together to get

1/sqrt(2)=1/(sqrt2)xx(sqrt2)/(sqrt2)=(sqrt2)/2
In this form, it is easy to locate 1/(sqrt2) on the number line. It is half way between 0 and sqrt2!
Q 3250167014

Rationalise the denominator of 1/(2+sqrt3)
Class 9 Chapter 1 Example 18
Solution:

We use the Identity (iv) given earlier. Multiply and divide 1/(2+sqrt3) by
2-sqrt3 to get 1/(2+sqrt3)xx(2-sqrt3)/(2-sqrt3)= (2-sqrt3)/(4-3) = 2- sqrt3
Q 3214167959

Rationalise the denominator of 5/(sqrt3 - sqrt5)
Class Chapter 1 Example 19
Solution:

Here we use the Identity (iii) given earlier.
So, 5/(sqrt3 - sqrt5) = 5/(sqrt3 - sqrt5)*(sqrt3 + sqrt5)/(sqrt3 + sqrt5)

=(5(sqrt3 + sqrt5))/2 = (-5/2)((sqrt3 + sqrt5))
Q 3260167015

Rationalise the denominator of 1/(7+3sqrt2)
Class 9 Chapter 1 Example 20
Solution:

1/(7+3sqrt2) = 1/(7+3sqrt2) xx((7-3sqrt2)/(7-3sqrt2)) = (7-3sqrt2)/(49-18) = (7-3sqrt2)/(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 rationalising the denominator.