Number System

Yes, zero is a rational number,
It can be written in the form of where q ≠ 0 such as , , , etc.

Given: to find five rational numbers between 1 and 2, we multiply & divide both the numbers by 6.

Trick: To find "n" rational numbers between any two numbers "a" & "b", just multiply & divide the numbers "a" & "b" by "n+1".

Example,

To find five rational numbers between 1 and 2, we multiply & divide both the numbers by 6, as shown:

Therefore, five rational numbers between 1 and 2 are:

Given, to find six rational numbers between 3 and 4.

We have,

3 * = and 4 * =

We know that,

21 < 22 < 23 < 24 < 25 < 26 < 27 < 28

required rational numbers = < < < < < < <

= 3 < < < < < < < 4

Hence, 6 rational numbers between 3 and 4 are:

, , , , ,

Given, to find 5 rational numbers between and

We have, × = and × =

We know that,

18 < 19 < 20 < 21 < 22 < 23 < 24

= < < < < < <

= < < < < < <

= < < < < < <

Hence, 5 rational numbers between and are:

, , , ,
Note: You can multiply and divide with any number you want to find the rational numbers.

(i) False: As whole numbers include zero, whereas natural numbers doesn’t include zero.

(ii) True: As integers are a part of rational numbers.

(iii) False: As integers are a part of natural numbers.

(iv) True: As whole numbers include all the natural numbers.

(v) False: As whole numbers are a part of integers.

(vi) False: As rational numbers include all the whole numbers.

(i) By long division

∴

(ii) By long division method, we have

(iii) By long division method, we have

∴

(i) By long division method, we have

∴

(ii) By long division method, we have

∴

(iii) By long division method, we have

∴

(iv) By long division method, we have

∴

(v) By long division method, we have

∴

(vi) By long division method, we have

∴

A rational number is a terminating decimal only, when prime factors of q are 2 and 5 only. Therefore, is a terminating decimal only, when prime factorisation of q must have only powers of 2 or 5 or both.

To convert decimal into fraction count no of decimal places in decimal number.
Let it be x. Then multiply and divide the decimal number with 10^x

(i) We have,

0.39 =

(ii) We have,

0.750 = =

=

(iii) We have,

2.15 = = =

(iv) We have,

7.010 = =

=

(v) We have,

9.90 = =

=

(vi) We have,

1.0001 =

(i) Let x =

Now, x = 0.4̅ = 0.444… (1)

Multiplying both sides of equation (1) by 10, we get,

10x = 4.444… (2)

Subtracting equation (1) by (2)

10x – x = 4.444… - 0.444…

9x = 4

x =

Hence, =

(ii) Let x =

Now, x = 0.3737…. (1)

Multiplying equation (1) by 10

10x = 3.737…. (2)

Multiplying equation (2) by 10

100x = 37.3737…. (3)

100x – x = 37

99x = 37

x =

Hence, =

(iii) Now x =

= 0.5454… (i)

Multiplying both sides of equation (i) by 100, we get

100x = 54.5454…. (ii)

Subtracting (i) by (ii), we get

100x – x = 54.5454…. – 0.5454….

99x = 54

x =

(iv) Now x =

= 0.621621…. (i)

Multiplying both sides by 1000, we get

1000x = 621.621621…. (ii)

Subtracting (i) by (ii), we get

1000x – x = 621.621621…. – 0.621621….

999x = 621

x = =

(v) Now x =

= 125.3333…. (i)

Multiplying both sides of equation (i) by 10, we get

10x = 1253.3333…. (ii)

Subtracting (i) by (ii), we get

10x – x = 1253.3333…. – 125.3333….

9x = 1128

x = 1128 / 9 = 376/3

(vi) Now x =

= 4.7777…. (i)

Multiplying both sides of equation (i) by 10, we get

10x = 47.7777…. (ii)

10x – x = 47.7777…. – 4.7777….

9x = 43

x =

(vii) Now, x =

= 0.47777….

Multiplying both sides by 10, we get

10x = 4.7777…. (i)

Multiplying both sides of equation (i) by 10, we get

100x = 47.7777…. (ii)

Subtracting (i) from (ii), we get

100x – 10x = 47.7777…. – 4.7777….

90x = 43

x =

A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number. For example,

1.01001000100001…

A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number. For example,

0.33033003300033…

On the other hand, every rational number is expressible either as a terminating decimal or as a repeating decimal. For example, 3.24̅ and 6.2876 are rational numbers.

(i) is not a perfect square root, so it is an irrational number.

(ii) We have,

= 2 =

can be expressed in the form of , so it is a rational number.

(iii) 2 ia a rational number, whereas is an irrational number.

Because, sum of a rational number and an irrational number is an irrational number, so 2 + is an irrational number

(iv) is an irrational number. Also is an irrational number. The sum of two irrational numbers is irrational.

Therefore, is an irrational number.

(v) is an irrational number. Also, is an irrational number. The sum of two irrational numbers is irrational.

Therefore, is an irrational number.

(vi) We have,

()² = ()² – 2 * * 2 + (2)²

= 2 - 4 + 4

= 6 - 4

Now 6 is a rational number, whereas 4 is an irrational number

The difference of a rational number and an irrational number is an irrational number.

So, it is an irrational number.

(vii) We have,

(2 - ) (2 + ) = (2)² – ()² [Therefore, (a – b) (a + b) = a² – b²]

= 4 – 2 = 2 =

Since 2 is a rational number

Therefore, (2 - ) (2 + ) is a rational number

(viii) We have,

()² = ()² + 2 × × + ()²

= 2 + 2 + 3

= 5 + 2

The sum of a rational number and an irrational number is irrational number. Therefore, it is an irrational number.

(ix) The difference of a rational number and an irrational number is an irrational number.

Therefore, 5 - is an irrational number.

(x) = 4.79583152331….

Therefore, it is an irrational number

(xi) = 15 =

Therefore, it is a rational number as it is represented in the form of , where q ≠ 0

(xii) 0.3796, as a decimal expansion of this number is terminating, so it is an irrational number.

(xiii) 7.478478…. = 7.4̅7̅8̅

As, decimal expansion of this number is non – terminating recurring so it is a rational number

(xiv) 1.101001000100001…..

It is an irrational number

(i) = 2 =

can be written in the form of , so it is a rational number.

Its decimal expansion is 2.0

(ii) 3 = 3

= 3 * 3

= 9

Since, the product of a rational and an irrational is an irrational number.

Therefore, 9 is an irrational;

3 is an irrational number

(iii) We have,

=

= 1.2

Every terminating decimal is a rational number, so 1.2 is a rational number.

(iv) we have,

= =

=

Quotient of a rational and an irrational number is irrational number. Therefore, it is an irrational number.

(v) - = -

= - 8 = -8/1

AS it can be expressed in the form of , so it is a rational number.

(vi) = 10 =

Thus it can be expressed in the form of, so it is a rational number.

(i) We have,

x² = 5

Taking square root on both sides,

= ² =

= x =

is not a perfect square root, so it is an irrational number.

(ii) We have,

y² = 9

y =

= 3 =

can be expressed in the form of , so it is a rational number.

(iii) We have,

z² = 0.04

Taking square root on both the sides, we get,

² =

z =

= 0.2 =

=

z can be expressed in the form of , so it is a rational number.

(iv) We have,

u² =

Taking square root on both the sides, we get

² =

u =

Quotient of an rational number is irrational, so u is an irrational number.

(v) We have,

v² = 3

Taking square roots on both the sides, we get,

² =

v =

is not a perfect square root, so v is an irrational number.

(vi) We have,

w² = 27

Taking square roots on both the sides, we get,

² =

w = = 3

Product of a rational number and an irrational number is irrational number. So, it is an irrational number.

(vii) We have,

t² = 0.4

Taking square roots on both the sides, we get,

² = =

=

Since, quotient of a rational number and an irrational number is irrational number, so t is an irrational number.

(i) is an irrational number.

Now, () – () = 0

0 is the rational number.

(ii) Let two irrational numbers are 5 and

Now, (5) – () = 4

4 is an irrational number.

(iii) Let two irrational numbers be and -

Now, () + (-) = 0

0 is a rational number

(iv) Let two irrational numbers are 4 and

Now, (4) + ( = 5

5 is an irrational number.

(v) Let two irrational numbers are 2 and

Now, 2 * = 2 * 3

= 6

6 is a rational number.

(vi) Let two irrational numbers are and

Now, * =

(vii) Let two irrational numbers are 3 and

Now, = 3

# is a rational number.

(viii) Let two irrational numbers are and

Now, =

=

is an irrational number.

Let a = 0.212112111211112

And, b = 0.232332333233332….

Clearly a < b because in the second decimal place a has digit 1 and b has digit 3

If we consider rational numbers in which decimal place has the digit 2, then they will lie between a and b

Let,

x = 0.22

y = 0.22112211…

Then,

a < x < y < b

Hence, x and y are required rational numbers.

Let a = 0.515115111511115….

b = 0.5353353335….

We observe that the second decimal place a has digit 1 and b has digit 3, therefore, a < b. So if we consider rational numbers

x = 0.52

y = 0.52052052

Then,

a < x < y < b

Hence, x and y are required rational numbers.

Let a = 0.2101

And, b = 0.2222….

We observe that the second decimal place a has digit 1 and b has digit 2, therefore, a < b. In the third decimal place a has digit 0. So, if we consider irrational number

x = 0.211011001100011…

We find that,

a < x < b

Hence, x is required irrational number.

Let a = 0.3010010001….

And, b = 0.3030030003…..

We observe that the second decimal place a has digit 1 and b has digit 3, therefore, a < b. In the third decimal place a has digit 1. So, if we consider rational and irrational numbers

x = 0.302

y = 0.302002000200002…

We find that,

a < x < b

And, a < y < b

Hence, x and y are required rational and irrational numbers respectively.

Let a = 0.5 = 0.50

And, b = 0.55

We observe that in the second decimal place a has digit 0 and b has digit 5. Therefore a < b. So, if we consider irrational numbers

x = 0.51051005100051…

y = 0.5305343055353530…

We find that,

a < x < y < b

Hence, x and y are required irrational numbers.

Let, a = 0.1 = 0.10

And, b = 0.12

We observe that in the second decimal place a has digit 0 and b has digit 2. Therefore a < b. So, if we consider irrational numbers

x = 0.11011001100011…

y = 0.111011110111110…

We find that,

a < x < y < b

Hence, x and y are required irrational numbers.

If possible, let be a rational number equal to x. Then,

x =

x² = ()²

= ()² + ()² + 2 * *

= 3 + 5 + 2

= 8 + 2

x² – 8 = 2

=

Now, x is rational

x² is rational

is rational

is rational

But, is irrational

Thus, we arrive at a contradiction. So, our supposition that + is rational is wrong.

Hence, + is an irrational number.

3 irrational numbers are:

0.73073007300073000073…..

0.75075007500075000075….

0.79079007900079000079….

(i) Every point on the number line corresponds to a REAL number which many be either RATIONAL or IRRATIONAL

(ii) The decimal from of an irrational number is neither TERMINATING Nor REPEATING.

(iii) The decimal representation of a rational number is either TERMINATING or NON TERMINATING

(iv) Every real number is either RATIONAL number or IRRATIONAL Number

Draw a number line and mark point O, representing zero, on it.

To represent on number line follow the following steps:
Step 1- Draw a line and mark a point A on it.
Step 2- Mark a point B on the line such that AB=3.5 units.
Step 3- Mark a point C on AB produced such that BC=1 unit.
Step 4- Find the mid-point of AC. Let it be O.
Step 5- Taking O as the centre and OC=OA as radius draw a semi circle. Also draw a line passing trough B perpendicular to OB. Suppose it cuts the semi circle at D.
Step 6- Taking B as centre and BD as radius draw an arc cutting OC produced at E. Point E so obtain represent

Similarly, represnt other square root numbers on number line.

(i) True: As we know that rational and irrational numbers taken form the set of real numbers.

(ii) True: As, is ratio of the circumference of a circle to its diameter, it is an irrational number.

=

(iii) False: Irrational numbers can be represented by point on the number line.

The following steps for successive magnification to visualise 2.665 are:

(1) We observe that 2.665 is located somewhere between 2 and 3 on the number line. So, let us look at the portion of the number line between 2 and 3.

(2) We divide this portion into 10 equal parts and mark each point of division. The first mark to the right of 2 will represent 2.1, the next 2.2 and soon. Again we observe that 2.665 lies between 2.6 And 2.7.

(3) We mark these points A₁ and A₂ respectively. The first mark on the right side of A₁, will represent 2.61, the number 2.62, and soon. We observe 2.665 lies between 2.66 and 2.67.

(4) Let us mark 2.66 as B₁ and 2.67 as B₂. Again divide the B₁B₂ into ten equal parts. The first mark on the right side of B₁ will represent 2.661, then next 2.662, and so on.

Clearly, fifth point will represent 2.665.

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

D)

The decimal expansion of an irrational number never repeats or terminates (essentially, that is repeating zeroes), unlike any rational number does.

If the irrational parts on adding forms a rational term then the whole number will be rational and if the irrational parts on adding gives again an irrational term then the complete number will be irrational.

Let the rational number be of the form , where p Z, while the rational number be r. If r + is a rational then we have that,

r + = for some a Z and b Z \ {0}. This means that r = - = where aq – bp Z and this contradicts the facts that r is irrational. Hence, our assumption that r + is a rational is false. Hence, it is an irrational number.

Example: Take any rational number other than Zero let’s take 2 as rational number and​√2 as irrational number than product as:

= 2 ​ × √2 = 2 √2

2 √2 is irrational number so if we can say that if rational number other than zero product with any irrational number the result is also irrational.

Proof: let us assume that √7 be rational.

then it must in the form of p / q [q ≠ 0] [p and q are co-prime]

√7 = p / q

√7 x q = p

squaring on both sides

7q²= p² (i)

p² is divisible by 7

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p² = 49 c² (ii)

Substitute p² in eq (i), we get,

7q² = 49 c²

q² = 7c²

q is divisible by 7

Thus q and p have a common factor 7.

There is a contradiction

As our assumption p & q are co - prime but it has a common factor.

So that √7 is an irrational.

Since, it is non - terminating and non - repeating decimal.

Since it is in the form of p/q , and where q ≠ 0.

Since it is a non - terminating repeating decimal, hence it is a rational number.

If n can be written in the form of p/q, where q≠0, then it is a rational number else irrational.

Since, it can be represented as 0.27272727... which is a non - terminating repeating decimal.

A real number is a value that represents a quantity along a line.

Since, it is non - terminating and non - repeating decimal.

Since, after dividing 14 from 11, we get that number.

Since, among the following only the division of 1 by 3 gives that specified number.

Since, among them 3/10 is the only number which when multiplied, then its decimal expansion terminates after one place of decimal.

Let x =

10x = (i)

100x = (ii)

Now, subtracting (i) from (ii), we get

100x – 10x = -

90x = 29

x =

x = _(i)

100x = 2343. (ii)

Subtracting (i) from (ii), we get

100x – x = 2343. - 23. .

99x = 2320

x =

x = (i)

(ii)

Subtracting (i) from (ii), we get

999x = 1

x =

= 0.232323….

0.2̅2̅ = 0.222222….

Now, = 0.23232323…. + 0.22222222….

= 0.45454545….

=