Number System

Yes, zero is a rational number and it can be represented in the form of as
or where z is an integer and z ≠0.

There are infinite rational numbers between 3 and 4.

So, we can represent 3 and 4 in terms of fractions as an example:

(Multiplying & dividing by 8)

NOTE: We can multiply & divide by other numbers as well.
We are multiplying and dividing at the same time because it is easier to write other fractions in between.
The decimal numbers between 3 & 4 can also be written for e.g. 3.1, 3.2, 3.3, etc.

Now, we can write the six rational numbers between 3 and 4 as:

Between and there exist infinite rational numbers & we know that rational numbers are fractions, whole numbers and decimals.

Trick: To find "n" rational numbers between any two numbers, multiply & divide both by "n+1".

Example: If we need to find 5 rational numbers between & , we multiply and divide both the numbers by 6 to obtain the set of rational numbers.

Hence, five rational numbers between and are:

Now its not necessary to only multiply and divide by n+1, you can do by any number greater than n.
For example if you multiply and divide by 10

So the rational numbers between 3 and 4 now will be

(i) True, since the collection of whole numbers contains all natural numbers.

Natural numbers are numbers starting from 1, i.e 1, 2, 3, 4, 5, 6, ..................
And whole numbers are numbers starting from 0. i.e, 0, 1, 2, 3, 4, 5.............

And you can see that all natural numbers are within whole numbers.

(ii) False, as integers may be negative or positive but whole numbers are always positive.

(iii) False, as rational numbers may be fractional but whole numbers are not fractional.

(i) True, since the real numbers are the collection of rational and irrational numbers.

For example √2 = 1.414... which can be represented in a number line.

All the numbers which can be represented in a number line are real.

(ii) False, since negative numbers cannot be expressed as the square of any other real number m.

for example: 4 = √2 but there is no number whose square root is -4.

(iii) False, as real numbers consist of both rational and irrational numbers. Therefore, every real number may not be an irrational number.

For example: √2 is irrational and 1/2 is rational and both are real numbers.

No, the square roots of all positive integers are not irrational. For example = 5.

Steps of construction:

Step 1: Let AB be a line of length 2 unit on the number line.

Step 2: At B, draw a perpendicular line BC of length 1 unit and then join CA.

Step 3: Now, ABC is a right angled triangle.

Using Pythagoras theorem,

AB² + BC² = CA²

2² + 1² = CA²

CA² = 5

CA = √5

Thus, CA is a line of length √5 units.

Step 4: Taking CA as a radius and A as the center, construct an arc touching the number line.

The number line gets intersected by the arc at a point which is at √5 distance from 0, as it is a radius of the circle with center A

Thus, √5 is represented on the number line as shown in the figure below.

Activity

For the square root spiral follow the given steps:-

Decimal Expansion which ends after the division is called terminating expansion. And when a fraction is divided such that the value repeats itself, it is a recurring expansion. When the expansion never ends it is called non terminating expansion.

For example: 0.2342516782943257 is neither repeating nor terminating expansion.


(i) = 0.36

(Since, decimal expansion has finite no. of figures. Hence, it is terminating).

(ii) = 0.09090909….

=

(Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion).

(iii)

= 4.125

(Since, decimal expansion has finite no. of figures. Hence, it is terminating).

(iv) = 0.230769230769

=

(Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion).

(v) = 0.1818181818….

=
(Since decimal expansion repeats itself so it is a non-terminating recurring decimal expansion)

(vi) = 0.8225

(Since, decimal expansion has finite no. of figures. Hence, it is terminating.)

Yes, we can do this by the following method:

2/7 = 2 x 1/7 = 2 x 0.͞1͞42͞8͞5͞7 = 0.2͞8͞5͞71͞4

3/7 = 3 x 1/7 = 3 x 0.1͞42͞8͞5͞7 = 0.4͞28͞5͞7͞1

4/7 = 4 x 1/7 = 4 x 0.1͞42͞8͞5͞7 = 0.͞57͞1͞4͞2͞8

5/7 = 5 x 1/7 = 5 x 0.1͞42͞8͞5͞7 = 0.7͞1͞4͞2͞8͞5

6/7 = 6 x 1/7 = 6 x 0.1͞42͞8͞5͞7 = 0.͞8͞5͞7͞1͞4͞2

(i) = 0.666…

Let x = 0.666666…

Multiplying both sides by 10

10 x = 6.666= 6 + 0.666

10 x = 6 + x

9 x = 6

(ii)

Let x = 0.4777777
Now multiplying both sides by 10,
10 x = 4.77777777........ eq(1)
Now multiplying eq(1) by 100 we get
100 x = 47.777777........ eq(2)
Now eq(2) - eq(1)
100 x - 10 x = 47 - 4
90 x = 43
x = 43/90

(iii)

Let x = 0.001001…eq(1)

Multiplying both sides by 1000

1000x = 1.001001… = 1 + 0.001001... eq(2)
we can see that,

1000x = 1 + x

999x = 1

Let x = 0.99999......... ......(i)

multiply both sides by 10

10x = 9.9999...... ....(2)

subtract (1) from (2).

9 x = 9

As x = 9/9 or x = 1.

Therefore, on converting 0.99999.... in the p/q form, we get the answer as1. .

The difference between 1 and 0.999999 is o.000001 which is negligible.

Hence, 0.999 is too much near to 1. Therefore, 1 as an answer can be justified

1/17 = 0.0͞5͞8͞8͞2͞3͞5͞2͞9͞4͞1͞1͞7͞6͞4͞7

There are 16 digits in the repeating numbers of the decimal expansion of 1/17.

Division check:

After 0.0588235294117647 digits will start repeating itself.

We observe that when q is 2, 4, 5, 8, 10… then the decimal expansion is terminating. For example:

, denominator q = 2¹

, denominator q = 2³

, denominator q = 5¹

It can be observed that the terminating decimal can be obtained in a condition where prime factorization of the denominator of the given fractions has the power of 2 only or 5 only or both.

Numbers with non-terminating decimal expression means that they never completely divided.

For example, when you divide 10 by 3, the answer you get keeps on dividing and no single value can be obtained but this number is recurring because 10/3 = 2.666666666 and in this value 6 is repeating or recurring.

So we have to write a number which is non terminating and non-repeating. Write any number which does not end and does not repeat.

(0.123123456345876....... or 2.123231245627549.... there can be infinite such numbers)

Three numbers whose decimal expansions are non-terminating non-recurring are:

0.304004000400003…

0.80506000900005…

0.7205200820008200008200000…

OR

All irrational numbers are non-terminating and non-repeating.

Example: √2 ,√3 and √5

Irrational Numbers: Numbers that cannot be expressed in p/q form are called irrational number.
Any number that is non-repeating and non terminating is an irrational number.
To find irrational numbers first we find a decimal expansion of given fractions.

Now, for irrational numbers, we write any non-terminating and non-repeating decimal number between given numbers.

Three different irrational numbers can be:

0.74074007400074000074…

0.75075007500075000075…

0.78078007800078000078…

Rational NUmber: A number that can be expressed in the form p/q, where q ≠ 0.

Irrational Number: A number which can not be expressed in the form p/q. The numbers that are non-repeating and non-terminating are called irrational.

(i) = 4.79583152331…

Since the decimal expansion is non-terminating and non-recurring hence, it is an irrational number

(ii) = 15 = 15/1

It is a rational number as it can be represented in the form of p/q

(iii) 0.3796

Since the decimal expansion is terminating number.

Therefore, it is a rational number

(iv) 7.478478 = 7.4͞7͞8

Since the decimal expansion is a non-terminating recurring number.

Therefore, it is a rational number

(v) 1.101001000100001…

Since the number has decimal expansion as non-terminating and non-repeating.

Hence, it is an irrational number

First, we will divide the number line between points 3 and 4. And then we will divide the points between 3.7 and 3.8 as the number is between both of them.
The number to visualize is 3.765 so now visualize the points between 3.76 and 3.77.

4.2͞6 = 4.2626

(i) 2 - = 2 – 2.2360679…

= -0.2360679…

Since the number is in non-terminating non-recurring, therefore, it is an irrational number

(ii) (3 + ) - = 3 + -

= 3

= 3/1

A number is a rational number as it can be represented in the form of p/q.

(iii) = 2/7

The number is a rational number as it can be represented in the form of p/q.

(iv) =

= 0.7071067811.....

Since the number is a non-terminating and non-recurring number.

Hence, it is an irrational number

(v) 2π = 2 * 3.1415…

= 6.2830......

Since the number is a non-terminating and non-recurring number.

Hence, it is an irrational number

(i) (3 + √3) (2 + √2)

= 3 (2 + √2) + √3 (2 + √2)

= 3 × 2 + 3√2 + 2√3 + √2 x √3

= 6 + 3√2 + 2√3 + √6

(ii) (3 + √3) (3 - √3) [Since, (a + b) (a – b) = a² – b²]

= 3² – (√3)²

= 9 – 3

= 6

(iii) (√5 + √2)² [Since, (a + b)² = a² + b² + 2ab]

= (√5)² + (√2)² + 2 × √5×√2

= 5 + 2 + 2 ×

= 7 + 2√10

(iv) (√5 - √2) (√5 + √2) [Since, (a + b) (a – b) = a² – b²]

= (√5)² – (√2)²

= 5 – 2

= 3

There is no contradiction in the statement.

When we measure any value with a scale, we only obtain an approximate value.

We never obtain an accurate value.

Hence, we cannot say that either c or d is irrational.

The value of π is almost equal to 22/7 or 3.142857…

Step 1: Draw a line segment AB of 9.3 unit. Then, extend it to C so that BC = 1 unit.

Step 2: Now, AC = 10.3 units. Find the center of AC and mark it as O

Step 3: Draw a semi-circle with radius OC and center O.

Step 4: Draw a perpendicular line BD to AC at point B intersecting the semi-circle at D. And then, join OD

Step 5: Now, OBD is a right angled triangle

Here, OD (Radius of semi-circle)

OC

BC = 1

Then, OB = OC – BC

Using Pythagoras theorem,

OD² = BD² + OB²

()² = BD² + ()²

BD² = ()² - ()²

BD² = ( – ) ( + )

BD² = 9.3

BD=

Thus, the length of BD is

Step 6: Taking BD as radius and B as the center, construct an arc which touches the line segment.

Now, the point where it touches the line segment is at a distance of from O as shown in the figure below

Rationalizing the denominator means that we have to remove the irrational component from the denominator of the fraction.



(ii)

(iii) =

=

=

=

(iv) =

=

=

=

(i) (64)^1/2

Writing Prime factors of 64 we get that : 64 = 8 x 8

So, (64)^1/2 = (2⁶)^1/2

(64 )^1/2 = 2³
= 8

(ii) (32)^1/5

Writing Prime Factors of 32 we get that: 32 = 2 x 2 x 2 x 2 x 2

So, 32 = 2⁵

(iii)(125)^1/3

Writing Prime Factors of 125 we get that: 125 = 5 x 5 x 5

So, 125 = 5 ³


(125)^1/3 = (5³)^1/3

(125) ^1/3 = 5

= 5

(i) 9^3/2 = (3²)^3/2

= 3^2*3/2

= 3³

= 27

(ii) 32^2/5 = (2⁵)^2/5

= (2)^5*2/5

= 2²

= 4

(iii) (16)^3/4 = (2⁴)^3/4

= 2³

= 8

(iv) (125)^-1/3 =

=

(i) a^m . aⁿ = a^{(m + n)}

(ii) (a^m)ⁿ = a^mn

(iii)

(iv) a^m . b^m = (a b)^m