Number System: 9 Math

NCERT Exercise-1.3(part1) Decimal Expansions: 9th math

Real Numbers and their Decimal Expansions

Rational and Irrational Numbers can be differentiated on the basis of their decimal expression.

Decimal Expansion of a Rational Number

A Rational Number can have either Terminating Decimal expansion or Non-Terminating Recurring Decimal expansion.

Thus, a Rational Number Can be divided into two types:

(a) Rational Number having Terminating Decimal Expansion

Rational numbers which remainder becomes zero after dividing the p by q, then these are called Rational Numbers with Terminating Decimal Expansion.

Example

`7/2, 1/2, 8/10, 5/8`, etc. are examples of some rational number numbers which have Terminating Decimal Expansion.

`7/2 = 3.5` (Terminating decimal expansion)

`1/2=0.5` (Terminating decimal expansion)

`8/10 = 0.8` (Terminating decimal expansion)

`5/8 =0.625` (Terminating decimal expansion)

(b) Rational Number having Non-Terminating Recurring (Repeating) Decimal Expansion.

Rational Numbers of whose remainder never becomes zero, but remainder repeat after a certain stage forcing the decimal expansion to go on forever then such numbers are called Rational Number having Non-Terminating Recurring (Repeating) Decimal Expansion.

Example:

`1/3` = 0.33333. . . . . . (Non-terminating recurring (repeating) decimal expansion)

`1/7` = 0.14285714285714 . . . . (Non-terminating recurring (repeating) decimal expansion)

The usual way of showing that 3 repeats in the quotient of `1/3` is to write it as `0.bar(3)` or `0.dot(3)`

Similarly, the block of digits 142857 repeats in the quotient of `1/7` this is written as `0.bar(142857)`

A bar over the number after decimal means, that very number or that block of numbers will be repeated endlessly.

Thus, 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.

Decimal Expansion of Irrational Number

A number whose decimal expansion is non-terminating non-recurring is called Irrational Number.

The decimal Expansion of an Irrational number is non-terminating non-recurring.

Example

(a) The decimal expansion of `sqrt2` = 1.414213562373 . . . . . . . .

Since, decimal expansion of `sqrt2` is non terminating non recurring, thus, `sqrt2` is an Irrational Number.

(b) `pi` is an Irrational number, because decimal expansion of `pi` = 3.141592653589 . . . . , which is non terminating non recurring.

(c) Similarly, `22/7`, `sqrt5, sqrt3, sqrt7`, etc. are examples of Irrational Numbers.

Definition of Irrational Number

Irrational Number can be defined in three ways:

(a) Numbers which cannot be expressed in the form of `p/q` where p and q are integers and `q!=0`, are called Irrational Number.

(b) Numbers whose decimal expansion is non-terminating non-recurring are called Irrational Numbers.

(c) A non-terminating non-recurring number is called Irrational Number.

Solution of NCERT Exercise 1.3 (class ninth mathematics)

Question (1) Write the following in decimal form and say what kind of decimal expansion each has:

(i) `36/100`

Solution:

Given, `36/100`

=0.36

Thus, decimal expansion of given number = terminating Answer

(ii) `1/11`

Solution

Here, since 09 is getting repeated after decimal.

Thus, decimal expansion of given number = non-terminating recurring Answer

(iii) `4\ 1/8`

Solution

Given, `4\1/8 =33/8`

Thus, decimal expansion of given number = terminating Answer

(iv) `3/13`

Solution

Given, `3/13`

Thus, decimal expansion of given number = non-terminating recurring Answer

(v) `2/11`

Solution

Given, `2/11`

Thus, decimal expansion of given number = non-terminating recurring Answer

(vi) `329/400`

Solution

Thus, decimal expansion of given number = terminating Answer

Question (2) You know that `1/7=0.bar(142857)`. 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.]

Solution

Here, given, `1/7=0.bar(142857)`

Thus, `2/7=1/7xx2`

`=0.bar(142857)xx2`

`=0.bar(285714)`

In similar way,

`3/7=1/7xx3`

`=0.bar(142857)xx3`

`=0.bar(428571)`

Similarly,

`4/7=1/7xx4`

`=0.bar(142857)xx4`

`=0.bar(571428)`

Similarly,

`5/7=1/7xx5`

`=0.bar(142857)xx5`

`=0.bar(714285)`

Similarly,

`6/7=1/7xx6`

`=0.bar(142857)xx6`

`=0.bar(857142)`

Question (3) Express the following in the form `p/q`, where p and q are integers and `q!=0`.

(i) `0.bar(6)`

Solution:

Given, `0.bar(6)`

This means, 6 is repeating digit after decimal,

That is, `0.bar(6)`= 0.6666 . . . .

Now, Let, m = 0.6666 . . . . ----------(i)

Now, multiply with 10 to both sides for one repeating digit

⇒ 10 m = 10 × (0.666 . . . )

⇒ 10 m = 6.666 . . . .

⇒ 10 m = 6 + 0.666. . . .

⇒ 10 m = 6 + m

[∵ m = 0.666. . .]

⇒ 10 m – m = 6

⇒ 9 m = 6

`=> m = 6/9 = 2/3`

Therefore, `0.bar(6) = 2/3` Answer

Alternate Method

Given, `0.bar(6)`

Now, here, only digit is to be repeated.

Thus, place one 9 as denominator and omit decimal and recurring sing.

`:. 0.bar(6) = 6/9`

`=(3xx2)/(3xx3) = 2/3`

`:. 0.bar(6) = 2/3` Answer

(ii) `0.4bar(7)`

Solution

Given, `0.4bar(7)`

= 0.4777 . . . .

Let, m = 0.4777 . . . .

Here only one digit is repeating, thus, multiply both sides by 10

⇒ 10 × m = 10 × 0.4777 . . .

⇒ 10 × m = 4.777 . . .

⇒ 10 × m = 4.3 + 0.4777 . . .

[∵ 4.3 + 0.4777 . . . = 4.777. . . ]

⇒ 10 × m = 4.3 + m

[m = 0.4777. . . as assumed earlier]

⇒ 10 m – m = 4.3

⇒ 9 m = 4.3

`=>m = 4.3/9 = 43/90`

Therefore, `0.4bar(7) = 43/90` Answer

Alternate Method

Given, `0.4bar(7)`

Here only one digit is repeating, and one digit is non-repeating thus take 90 as denominator.

Now, take (47–4) as numerator.

Thus, `0.4bar(7) = (47-7)/90`

`= 43/90`

Thus, `0.4bar(7) = 43/90` Answer

(iii) `0.bar(001)`

Solution

Given, `0.bar(001)`

= 0.001001001. . . .

Let, m = 0.001001001. . . .

Here, since there are three repeating numbers, thus multiply both sides by 1000

⇒ 1000 m = 1000 × 0.001001001. . . .

⇒ 1000 m = 1.001001 . . .

⇒ 1000 m = 1 + 0.001001 . . . .

⇒ 1000 m = 1 + m

[∵ m = 0.001001001. . . . as supposed]

⇒ 1000 m – m = 1

⇒ 999 m = 1

`=> m = 1/999`

Thus, `0.bar(001)=1/999` Answer

Alternate Method

Given, `0.bar(001)`

Here, there are three repeating numbers, thus take 999 as denominator and given number after omitting recurring and decimal sign as numerator.

Thus, `0.bar(001)=1/999` Answer

Question (4) Express 0.99999 . . . . in the form `p/q`. Are you surprise b your answer? With your teacher and classmates discuss why the answer makes sense.

Solution

Given, 0.99999 . . . .

Let, m = 0.9999 . . .

Here, one digit is repeating, thus multiply both sides by 10

⇒ 10 m = 10 × 0.9999 . . .

⇒ 10 m = 9.999 . . .

⇒ 10 m = 9 + m

[∵ m = 0.9999. . . .]

⇒ 10 m – m = 9

⇒ 9 m = 9

`:. m = 9/9 = 1`

Thus, 0.9999 . . . . = 1 Answer

Alternate method

Given, 0.9999 . . . .

`=0.bar(9)`

Here, since only one digit is repeating, thus 9 is taking as denominator

Thus, `=0.bar(9) = 9/9=1`

Hence, 0.9999 . . . . = 1 Answer

Question (5) What can the maximum number of digits be in the repeating block of digits in the decimal expansion of `1/17`? Perform division to check your answer.

Solution

`=0.bar(0588235294117647)`

Thus, maximum number of digits in the repeating block = 17 Answer

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