Chapter 1: Rational Numbers

Solution:-

Rational number with denominator 4 in their lowest form are,

64/16 = 16/4, 36/-12 = 12/-4, 5/-4, 140/28 =20/4

Solution:-

First rearrange the rational numbers and add the numbers with same denominator.

= (4/7) + (3/7) – (4/9) – (13/9)

= ((4 + 3)/7) – ((4 + 13)/9)

= (7/7) – (17/9)

= 1 – (17/9)

= (9 – 17)/9

= -8/9

(b) -5 + (7/10) + (3/7) + (-3) + (5/45) + (-4/5)

Solution:-

= -5 + (-3) + (7/10) + (-4/5) + (3/7) + (5/14)

= 8 + [(7-8)/10] + [(6 + 5)/14]

= – 8 – (1/10) + (11/14)

LCM of 1, 10 and 14 is 70

= (-560 – 7 + 55)/70

= -512/70

= – 256/35

Solution:-

– x = -3/5

– (-x) = – (-3/5)

X = 3/5

(ii) x = -7/9

Solution:-

– x = – (-7/9)

-x = 7/9

– (-x) = – (7/9)

X = – 7/9

(iii) x = 13/-15

Solution:-

– x = – (-13/15)

-x = 13/15

– (-x) = – (13/15)

X = -13/15

Solution:-

Rational numbers are closed under addition:-

Example:- 5/4 + 1/2

The LCM of the denominators 4 and 2 is 4

∴ (5/4) = [(5×1)/ (4×1)] = (5/4)

and (1/2) = [(1×2)/ (4×2)] = (1/4)

Then,

= 5/4 + ¼

= (5 + 1)/ 4

= 6/4

= 3/2 is a rational number

Rational numbers are closed under subtraction:-

Example:- 5/4 – 1/2

The LCM of the denominators 4 and 2 is 4

∴ (5/4) = [(5×1)/ (4×1)] = (5/4)

and (1/2) = [(1×2)/ (4×2)] = (1/4)

Then,

= 5/4 – ¼

= (5 – 1)/ 4

= 4/4

= 1 is a rational number

Rational numbers are closed under addition:-

Example:- 5/4 × 1/2

= 5/8 is a rational number.

For any rational number x, x ÷ 0 is not defined,

Hence not all rational numbers are closed under division. We can say that except zero, all rational numbers are closed under division.

Example, ¾ ÷ 4/5

= ¾ × 5/4

= 15/16 is a rational number.

Solution:-

In the question is given to verify the property = x + y = y + x

Where, x = ½, y = ½

Then, ½ + ½ = ½ + ½

LHS = ½ + ½

= (1 + 1)/2

= 2/2

= 1

RHS = ½ + ½

= (1 + 1)/2

= 2/2

= 1

By comparing LHS and RHS

LHS = RHS

∴ 1 = 1

Hence x + y = y + x

(b) x = -2/3, y = -5/6

Solution:-

In the question is given to verify the property = x + y = y + x

Where, x = -2/3, y = -5/6

Then, -2/3 + (-5/6) = -5/6 + (-2/3)

LHS = -2/3 + (-5/6)

= -2/3 – 5/6

The LCM of the denominators 3 and 6 is 6

(-2/3) = [(-2×2)/ (3×2)] = (-4/6)

and (-5/6) = [(-5×1)/ (6×1)] = (-5/6)

Then,

= – 4/6 – 5/6

= (- 4 – 5)/ 6

= – 9/6

RHS = -5/6 + (-2/3)

= -5/6 – 2/3

The LCM of the denominators 6 and 3 is 6

(-5/6) = [(-5×1)/ (6×1)] = (-5/6)

and (-2/3) = [(-2×2)/ (3×2)] = (-4/6)

Then,

= – 5/6 – 4/6

= (- 5 – 4)/ 6

= – 9/6

By comparing LHS and RHS

LHS = RHS

∴ -9/6 = -9/6

Hence x + y = y + x

(c) x = -3/7, y = 20/21

Solution:-

In the question is given to verify the property = x + y = y + x

Where, x = -3/7, y = 20/21

Then, -3/7 + 20/21 = 20/21 + (-3/7)

LHS = -3/7 + 20/21

The LCM of the denominators 7 and 21 is 21

(-3/7) = [(-3×3)/ (7×3)] = (-9/21)

and (20/21) = [(20×1)/ (21×1)] = (20/21)

Then,

= – 9/21 + 20/21

= (- 9 + 20)/ 21

= 11/21

RHS = 20/21 + (-3/7)

The LCM of the denominators 21 and 7 is 21

(20/21) = [(20×1)/ (21×1)] = (20/21)

and (-3/7) = [(-3×3)/ (7×3)] = (-9/21)

Then,

= 20/21 – 9/21

= (20 – 9)/ 21

= 11/21

By comparing LHS and RHS

LHS = RHS

∴ 11/21 = 11/21

Hence x + y = y + x

(d) x = -2/5, y = – 9/10

Solution:-

In the question is given to verify the property = x + y = y + x

Where, x = -2/5, y = -9/10

Then, -2/5 + (-9/10) = -9/10 + (-2/5)

LHS = -2/5 + (-9/10)

= -2/5 – 9/10

The LCM of the denominators 5 and 10 is 10

(-2/5) = [(-2×2)/ (5×2)] = (-4/10)

and (-9/10) = [(-9×1)/ (10×1)] = (-9/10)

Then,

= – 4/10 – 9/10

= (- 4 – 9)/ 10

= – 13/10

RHS = -9/10 + (-2/5)

= -9/10 – 2/5

The LCM of the denominators 10 and 5 is 10

(-9/10) = [(-9×1)/ (10×1)] = (-9/10)

and (-2/5) = [(-2×2)/ (5×2)] = (-4/10)

Then,

= – 9/10 – 4/10

= (- 9 – 4)/ 10

= – 13/10

By comparing LHS and RHS

LHS = RHS

∴ -13/10 = -13/10

Hence x + y = y + x

Solution:-

The arrangement of the given rational number is as per the rule of distributive law over addition.

Now take out ½ as common.

Then,

= ½ [¼ + 6]

= ½ [(1 + 24)/4]

= ½ [25/24]

= ½ × (25/24)

= 25/8

(b) [(1/5) × (2/15)] – [(1/5) × (2/5)]

Solution:-

The arrangement of the given rational number is as per the rule of distributive law over subtraction.

Now take out 1/5 as common.

Then,

= 1/5 [(2/15) – (2/5)]

The LCM of the denominators 15 and 5 is 15

(2/15) = [(2×1)/ (15×1)] = (2/15)

and (2/5) = [(2×3)/ (5×3)] = (6/15)

= 1/5 [(2 – 6)/15]

= 1/5 [-4/15]

= (1/5) × (-4/15)

= -4/75

(c) (-3/5) × {(3/7) + (-5/6)}

Solution:-

The arrangement of the given rational number is as per the rule of distributive law over addition.

= (-3/5) × {(3/7) + (-5/6)}

The LCM of the denominators 7 and 6 is 42

(3/7) = [(3×6)/ (7×6)] = (18/42)

and (-5/6) = [(-5×7)/ (6×7)] = (-35/42)

= -3/5 [(18 – 35)/42]

= -3/5 [-17/42]

= (-3/5) × (-17/42)

= 51/210 … [divide both denominator and numerator by 3]

= 17/30

Solution:-

The arrangement of the given rational number is as per the rule of Associative property for Multiplication.

Solution:-

In the question is given to verify the property = x × y = y × x

Where, x = 7, y = ½

Then, 7 × ½ = ½ × 7

LHS = 7 × ½

= 7/2

RHS = ½ × 7

= 7/2

By comparing LHS and RHS

LHS = RHS

∴ 7/2 = 7/2

Hence x × y = y × x

(b) x = 2/3 and y = 9/4

Solution:-

In the question is given to verify the property = x × y = y × x

Where, x = 2/3, y = 9/4

Then, (2/3) × (9/4) = (9/4) × (2/3)

LHS = (2/3) × (9/4)

= (1/1) × (3/2)

= 3/2

RHS = (9/4) × (2/3)

= (3/2) × (1/1)

= 3/2

By comparing LHS and RHS

LHS = RHS

∴ 3/2 = 3/2

Hence x × y = y × x

(c) x = -5/7 and y = 14/15

Solution:-

In the question is given to verify the property = x × y = y × x

Where, x = -5/7, y = 14/15

Then, (-5/7) × (14/15) = (14/15) × (-5/7)

LHS = (-5/7) × (14/15)

= (-1/1) × (2/3)

= -2/3

RHS = (14/15) × (-5/7)

= (2/3) × (-1/1)

= -2/3

By comparing LHS and RHS

LHS = RHS

∴ -2/3 = -2/3

Hence x × y = y × x

(d) x = -3/8 and y = -4/9

Solution:-

In the question is given to verify the property = x × y = y × x

Where, x = -3/8, y = -4/9

Then, (-3/8) × (-4/9) = (-4/9) × (-3/8)

LHS = (-3/8) × (-4/9)

= (-1/2) × (-1/3)

= 1/6

RHS = (-4/9) × (-3/8)

= (-1/3) × (-1/2)

= 1/6

By comparing LHS and RHS

LHS = RHS

∴ 1/6 = 1/6

Hence x × y = y × x

Solution:-

In the question is given to verify the property x × (y × z) = (x × y) × z

The arrangement of the given rational number is as per the rule of associative property for multiplication.

Then, 1 × (-½ × ¼) = (1 × -½) × ¼

LHS = 1 × (-½ × ¼)

= 1 × (-1/8)

= -1/8

RHS = (1 × -½) × ¼

= (-½) × ¼

= -1/8

By comparing LHS and RHS

LHS = RHS

∴ -1/8 = -1/8

Hence x × (y × z) = (x × y) × z

(b) x = 2/3, y = -3/7 and z = ½

Solution:-

In the question is given to verify the property x × (y × z) = (x × y) × z

The arrangement of the given rational number is as per the rule of associative property for multiplication.

Then, (2/3) × (-3/7 × ½) = ((2/3) × (-3/7)) × ½

LHS = (2/3) × (-3/7 × ½)

= (2/3) × (-3/14)

= -6/42

RHS = ((2/3) × (-3/7)) × ½

= (-6/21) × ½

= -6/42

By comparing LHS and RHS

LHS = RHS

∴ -6/42 = -6/42

Hence x × (y × z) = (x × y) × z

(c) x = -2/7, y = -5/6 and z = ¼

Solution:-

In the question is given to verify the property x × (y × z) = (x × y) × z

The arrangement of the given rational number is as per the rule of associative property for multiplication.

Then, (-2/7) × (-5/6 × ¼) = ((-2/7) × (-5/6)) × ¼

LHS = (-2/7) × (-5/6 × ¼)

= (-2/7) × (-5/24)

= 10/168

RHS = ((-2/7) × (-5/6)) × ¼

= (10/42) × ¼

= 10/168

By comparing LHS and RHS

LHS = RHS

∴ 10/168 = 10/168

Hence x × (y × z) = (x × y) × z

Solution:-

In the question is given to verify the property x × (y + z) = x × y + x × z

The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.

Then, (-½) × (¾ + ¼) = (-½ × ¾) + (-½ × ¼)

LHS = (-½) × (¾ + ¼)

= (-½) × ((3 + 1)/4)

= -½ × (4/4)

= -½ × 1

= -½

RHS = (-½ × ¾) + (-½ × ¼)

= (-3/8) + (-1/8)

= (-3 – 1)/8

= -4/8

= -½

By comparing LHS and RHS

LHS = RHS

∴ -½ = -½

Hence x × (y + z) = x × y + x × z

(b) x = -½, y = 2/3, z = ¾

Solution:-

In the question is given to verify the property x × (y + z) = x × y + x × z

The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.

Then, (-½) × ((2/3) + ¾) = (-½ × (2/3)) + (-½ × ¾)

LHS = (-½) × ((2/3) + ¾)

= (-½) × ((8 + 9)/12)

= -½ × (17/12)

= -17/24

RHS = (-½ × (2/3)) + (-½ × ¾)

= (-1/3) + (-3/8)

= (-8 – 9)/24

= -17/24

By comparing LHS and RHS

LHS = RHS

∴ -17/24 = -17/24

Hence x × (y + z) = x × y + x × z

(c) x = -2/3, y = -4/6, z = -7/9

Solution:-

In the question is given to verify the property x × (y + z) = x × y + x × z

The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.

Then, (-2/3) × ((-4/6) + (-7/9)) = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))

LHS = (-2/3) × ((-4/6) + (-7/9))

= (-2/3) × ((-12 – 14)/18)

= – (2/3) × (-26/18)

= – (1/3) × (-26/9)

= 26/27

RHS = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))

= ((-1/3) × (-4/3)) + ((-2/3) × (-7/9))

= (4/9) + (14/27)

= (12 + 14)/27

= 26/27

By comparing LHS and RHS

LHS = RHS

∴ 26/27 = 26/27

Hence x × (y + z) = x × y + x × z

(d) x = -1/5, y = 2/15, z = -3/10

Solution:-

In the question is given to verify the property x × (y + z) = x × y + x × z

The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.

Then, (-1/5) × ((2/15) + (-3/10)) = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))

LHS = (-1/5) × ((2/15) + (-3/10))

= (-1/5) × ((4 – 9)/30)

= (-1/5) × (-5/30)

= (-1/1) × (-1/30)

= 1/30

RHS = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))

= (-2/75) + (3/50)

= (-4 + 9)/150

= 5/150

= 1/30

By comparing LHS and RHS

LHS = RHS

∴ 1/30 = 1/30

Hence x × (y + z) = x × y + x × z