Contents
Real Numbers
Question 1:
Exercise 1A
The numbers of the form
Ten examples of rational numbers are:
, , , , , , , , 1,
Question 2:
- 5
- -3
(v) 1.3
(vii)
Question 3:
Therefore, we have
Therefore, three rational numbers between
, and
Question 5:
Let
Then, x < y because
Or we can say that,
We know that, 8 < 9 < 10 < 11 < 12 < 13 < 14 < 15.
Therefore, we have,
, , , and
Question 6:
Let x = 3 and y = 4
Then, x < y, because 3 < 4 We can say that,
We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28.
Therefore, we have,
, , , and
Question 7:
Let x = 2.1 and y = 2.2
Then, x < y because 2.1 < 2.2 Or we can say that,
That is, we have,
We know that, 2100 < 2105 < 2110 < 2115 < 2120 < 2125 < 2130 < 2135 < 2140 <
2145 < 2150 < 2155 < 2160 < 2165 < 2170 < 2175 < 2180 < 2185 < 2190 < 2195 <
2200
Therefore, 16 rational numbers between, 2.1 and 2.2 are:
So, 16 rational numbers between 2.1 and 2.2 are:
2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17,
2.175, 2.18
Exercise 1B
Question 1:
If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.
Since, 80 has prime factors 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since 12 has prime factors 2 and 3 and 3 is different from 2 and 5,
If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.
Since 35 has prime factors 5 and 7, and 7 is different from 2 and 5,
If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.
Since 125 has prime factor 5 only
Question 2:
Question 3:
-
Let x =
i.e x = 0.333 …. (i)
⇒ 10x = 3.333 …. (ii) Subtracting (i) from (ii), we get 9x = 3
-
Let x = 1.
i.e x = 1.333 …. (i)
⇒10x = 13.333 …. (ii)
Subtracting (i) from (ii) we get; 9x = 12
⇒ x =
-
Let x = 0.
i.e x = 0.3434 …. (i)
⇒ 100x = 34.3434 …. (ii)
Subtracting (i) from (ii), we get 99x = 34
⇒ x =
-
Let x = 3.
i.e x = 3.1414 …. (i)
⇒ 100x = 314.1414 …. (ii)
Subtracting (i) from (ii), we get 99x = 311
⇒ x =
-
Let x = 0.
i.e. x = 0.324324 ….(i)
⇒ 1000x = 324.324324….(ii)
Subtracting (i) from (ii), we get 999x = 324
⇒ x =
-
Let x = 0.
i.e. x = 0.177 …. (i)
⇒ 10x = 1.777 …. (ii)
and 100x = 17.777…. (iii) Subtracting (ii) from (iii), we get 90x = 16
⇒ x =
-
Let x = 0.
i.e. x = 0.544 …. (i)
⇒ 10 x = 5.44 …. (ii) and 100x = 54.44 ….(iii)
Subtracting (ii) from (iii), we get 90x = 49
⇒ x =
⇒ 10x = 1.6363 …. (ii)
and 1000 x = 163.6363 …. (iii) Subtracting (ii) from (iii), we get
990x = 162
Question 4:
- True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers
- False. Since 0 is whole number but it is not a natural number.
- True. Every integer can be represented in a fraction form with denominator 1.
-
False. Since division of whole numbers is not closed under division, the value of
, p and q are integers and q ≠ 0, may not be a whole number. - True. The prime factors of the denominator of the fraction form of terminating
decimal contains 2 and/or 5, which are integers and are not equal to zero.
- True. The prime factors of the denominator of the fraction form of repeating decimal contains integers, which are not equal to zero.
-
True. 0 can considered as a fraction
, which is a rational number.
Question 1:
Exercise 1C
Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot
be expressed in the fraction form,
For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.
Also,
Question 2:
(i)
We know that, if n is a perfect square, then
(ii)
We know that, if n is a perfect square, then
(iii)
We know that, if n is a not a perfect square, then
(iv)
We know that, if n is a not a perfect square, then
(v)
Theorem: The sum of a rational number and an irrational number is an irrational number.
So by the above theorem, the sum,
(vi)
Theorem: The sum of a rational number and an irrational number is an irrational number.
So by the above theorem, the sum,
So,
(vii)
Theorem: The product of a non-zero rational number and an irrational number is an irrational number.
Thus, by the above theorem,
So,
-
0.
Every rational number can be expressed either in the terminating form or in the non- terminating, recurring decimal form.
Therefore, 0.
Question 3:
Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.
Take OA = 1 unit and draw BA ⊥ OA such that AB = 1 unit, join OB. Then,
With O as centre and OB as radius, drawn an arc, meeting OX at P. Then, OP = OB =
Thus the point P represents
Now draw BC ⊥ OB such that BC = 1 units Join OC. Then,
With O as centre and OC as radius, draw an arc, meeting OX at Q. The, OQ = OC =
Thus, the point Q represents
Now draw DE ⊥ OD such that DE = 1 units Join OE. Then,
With O as centre and OE as radius draw an arc, meeting OX at R. Then, OR = OE =
Thus, the point R represents
Question 4:
Draw horizontal line X’OX taken as the x-axis Take O as the origin to represent 0.
Let OA = 2 units and let AB ⊥ OA such that AB = 1 units
With O as centre and OB as radius draw an arc meeting OX at P. Then, OP = OB =
Now draw BC ⊥ OB and set off BC = 1 unit Join OC. Then,
With O as centre and OC as radius, draw an arc, meeting OX at Q. Then, OQ = OC =
Thus, Q represents
With O as centre and OD as radius, draw an arc, meeting OX at R. Then OR = OD =
Thus, R represents
Question 5:
(i)
Since 4 is a rational number and
So,
(ii)
Since – 3 is a rational number and
So,
(iii)
Since 5 is a rational number and
So,
(iv)
Since -3 is a rational number and
So,
Question 6:
- True
- False
- True
- False
- True
- False
- False
- True
- True
Exercise 1D
Question 1:
We have:
We have:
Question 2:
Question 3:
(ii)
Question 4:
Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 units. Find the midpoint O of AC.
With O as centre and OA as radius, draw a semicircle. Now, draw BD AC, intersecting the semicircle at D. Then, BD =
With B as centre and BD as radius, draw an arc meeting AC produced at E. Then, BE = BD =
Question 6:
Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC.
With O as centre and OA as radius, draw a semicircle.
Now, draw BD AC, intersecting the semicircle at D. Then, BD =
With D as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD =
Question 7:
Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for al real numbers a, b, c.
Commutative Law: a + b = b + a, for all real numbers a and b.
Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (-a) such that
a + (-a) = (-a) + a= 0
a and (-a) are called the additive inverse of each other. Existence of inverse of multiplication:
a and
Question 1:
Exercise 1E
Question 2:
Question 3:
Question 5:
Question 9:
Question 12:
Question 14:
Question 15:
Question 17:
Exercise 1F
Question 1:
(ii)
Question 2:
(iii)
Question 3:
(ii)
Question 4:
(ii)
Question 5:
(ii)
Question 6:
(ii)
(iii)
Question 7:
(ii)