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(a) rational numbers are closed under addition.

Because, (3/8) + (-5/7)

Take the LCM of the denominators of the given rational numbers.

LCM of 8 and 7 is 56

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(3/8)= [(3×7)/ (8×7)] = (21/56)

(-5/7)= [(-5×8)/ (7×8)] = (-40/56)

Then,

= (21/56) + (-40/56) … [∵ denominator is same in both the rational numbers]

= (21 – 40)/56

= (-19/56)

6/13 x ( Reciprocal of -7/16) = 6/13 x -16/7 = -96/91

(-8/9) from (-3/5)

We have:

= (-3/5) – (-8/9)

= (-3/5) + (additive inverse of -8/9)

= (-3/5) + (8/9)

LCM of 5 and 9 is 45

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(-3×9)/ (5×9)] = (-27/45)

= [(8×5)/ (9×5)] = (40/45)

Then,

= (-27/45) + (40/45)

= (-27+40)/45

= (13/45)

(d) rational numbers are closed under division.

Because, rational numbers are closed under the operations of addition, subtraction and multiplication.

(i) -4/5 x 1 = 1 x-4/5 = -4/5

1 is the multiplicative identity.

(ii) Commuatativity

(iii) Multiplicative inverse

(-5/6) from (1/3)

We have:

= (1/3) – (-5/6)

= (1/3) + (additive inverse of -5/6)

= (1/3) + (5/6)

LCM of 3 and 6 is 6

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(1×2)/ (3×2)] = (2/6)

= [(5×1)/ (6×1)] = (5/6)

Then,

= (2/16) + (5/6)

= (2+5)/6

= (7/6)

(a) addition of rational numbers is commutative.

The arrangement of above rational numbers is in the form of Commutative law of addition [a + b=b + a]

2011 > 528 > 364 > 210 > 85 > 12 > – 58 > -1024 < -10,000 

2011, 528, 364, 210, 85, 12, -58, -1024, -10,000.

(a) [(2/3) × ((-6/7) × (3/5))] = [((2/3) × (-6/7)) × (3/5)]

Because, the arrangement of above rational numbers is in the form of Associative law of Multiplication [a × (b ×c)] = [(a× b) × c]

(3/4) from (1/3)

We have:

= (1/3) – (3/4)

= (1/3) + (additive inverse of 3/4)

= (1/3) + (-3/4)

LCM of 4 and 3 is 12

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(1×4)/ (3×4)] = (4/12)

= [(-3×3)/ (4×3)] = (-9/12)

Then,

= (4/12) + (-9/12)

= (4-9)/12

= (-5/12)

(d) Not defined

Reciprocal of 0 = 1/0

= not defined

(b) -1

Reciprocal of -1 = -1/1

= -1

(i) 5

Additive inverse of 5 is -5

(ii) -9

Additive inverse of -9 is 9

(iii) (3/14)

Additive inverse of (3/14) is (-3/14)

(iv) (-11/15)

Additive inverse of (-11/15) is (11/15)

(v) (15/-4)

First we write each of the given numbers with a positive denominator.

(15/-4) = [(15× (-1))/ (-4×-1)]

= (-15/4)

Then,

Additive inverse of (-15/4) is (15/4)

(vi) (-18/-13)

First we write each of the given numbers with a positive denominator.

(-18/-13) = [(-18× (-1))/ (-13×-1)]

= (18/13)

Then,

Additive inverse of (18/13) is (-18/13)

(vii) O

Additive inverse of 0 is 0

(viii) (1/-6)

First we write each of the given numbers with a positive denominator.

(1/-6) = [(1× (-1))/ (-6×-1)]

= (-1/6)

Then,

Additive inverse of (-1/6) is (1/6)

(a) 1

Reciprocal of 1 = 1/1

= 1

The correct option is (D) -2/8.

(a) Zero (0) is the identity for addition of rational numbers.

That means,

If a is a rational number.

Then, a+0=0+a = a

Note Zero (0) is also the additive identity for integers and whole number as well.

(c) One (1) is the identity for multiplication of rational numbers.

That means,

If a is  a rational number.

Then, a-1 = 1-a = a

Note One (1) is the multiplication identity for integers and whole number also.

(i)  2/8

Additive = -2/8

(ii) -5/9

Additive = 5/9

(ii) -6/5 = 6/5

Additive inverse = -6/5

(iv) 2/-9 = -2/9

Additive inverse = 2/9

(v) 19/-6 = -19/6

Additive inverse = 2/9

The correct option is (C) 19/56.

(a) x

= x + 0 = x [∵ identity for addition of rational numbers]

Additive inverse of (-7/19) is (b) (7/19)

The additive inverse of the rational number -a/b is a/b and vice-versa.

(d) 7/-8

= \(-1 \frac{1}{7}\)

= – 8/7

= 7/-8 [∵ reciprocal]

We know that integers between 2 and -4 are:
-4 < -3 < -2 < -1 < 0 < 1 < 2

∴ Five rational numbers between 2/5 and -4/5 will be -3/5, -2/5, -1/5, 0/5, 1/5.

(b) a negative rational number.

(-1/3) is a rational number so its multiplicative inverse is (-3/1) 

So that their multiplication will be,

= (-1/3) × (-3/1)

= – 1 × -1

= 1

Filling the blanks in the table,we get

The correct option is (A) -21/5.

(i) <

(ii) (0)

(iii) infinite

(iv) -11/18

(a) We know that, the sum of any rational number and zero (0) is the rational number itself.

Now, x + 0 = 0+ x= x, which is a rational number, then 0 is called identity for addition of rational numbers.

(b) -(-x) = x

Negative of negative rational number is equal to positive rational number.

L.H.S

1/2 + (-3/4 + -5/8)

= 1/2 + (-6 + (-5)/8)

= 1/2 - 11/8

= (4-11)/8

= -7/8

R.H.S

(1/2 + -3/4) + -5/8

= (2 + (-3)/4) + -5/8

= -1/4 + -5/8

= (-2 +(-5))/8

= -7/8

LHS = RHS

Yes, total of both sides are equal.

Filling the blanks in table,we get

(c) 21/8

Because,

= (8/21) × (21/8)

= (8 × 21) / (21 × 8)

= 168/168

= 1

108 = 2 × 2 × 3 × 3 × 3 = 2² × 3² × 3

If we multiply the factors by 3, then we get
2² × 3² × 3 × 3 ⇒ 2² × 3² × 3² = (2 × 3 × 3)²

Which is a perfect square.

∴ Again if we divide by 3 then we get 2² × 3² ⇒ (2 × 3 )², a perfect square.

∴ We have to multiply or divide 108 by 3 to get a perfect square.

Filling the blanks in table,we get

Correct option is (C) -5/9

\(\frac{2}{3}\times\frac{-5}{6}\) \(=\frac{2\times-5}{3\times6}=\frac{-10}{18}\) = \(\frac{-5}{9}\).

Correct option is  C) -5/9 

The correct option is (C) -4/3.

(i) -11/7 + 4/7

= --11/7 + 4/7

= (-11 +4)/7

= -7/7

= -1

(ii) 3/5+ (-2/5)

= 3/5 + (-2/5)

= (3+(-2))/5

= 1/5

(iii) -3/4 + (-5/4)

= -3/4 + (-5/4)

= (-3+(-5))/4

= -8/4

= -2

Product of two rational numbers = -9

One rational number = -12

Let the other rational number = x

Now,

According to the question,

-12 × x = -9

\(\Rightarrow\) x = \(\frac{-9}{-12}\)

\(\Rightarrow\) x = \(\frac{-9}{-12}= \frac{-9\times-1}{-12\times-1}=\frac{9}{12}\)

\(\Rightarrow\) x = \(\frac{9}{12} =\frac{9\div3}{12\div3}=\frac{3}{4}\)

Hence, the other rational number is \(\frac{3}{4}\)

Correct option is (D) Rs \(18\frac{3}{4}\)

\(\because\) Cost of \(5\frac{2}{5}\) litres of milk = Rs \(101\frac{1}{4}\) 

\(\Rightarrow\) Cost of \(\frac{27}{5}\) litres of milk = Rs \(\frac{405}4\)

\(\therefore\) Cost of 1 litres of milk = Rs \((\frac{405}4\div\frac{27}5)\)

= Rs \((\frac{405}4\times\frac5{27})\)

= Rs \((\frac{45}4\times\frac5{3})\)

= Rs \((\frac{15\times5}4)\) = Rs \(\frac{75}4\)

= Rs \(\frac{18\times4+3}4\) = Rs \((18+\frac34)\)

= Rs \(18\frac{3}{4}\).

Correct option is   D) ₹ 18\(\frac{3}{4}\)

Product of two rational numbers = \(\frac{-16}{9}\)

One rational number = \(\frac{-4}{3}\)

Let the other rational number = x 

Now, 

According to the question,

\(\frac{-4}{3}\times \text{x} = \frac{-16}{9}\) \(\)

\(\Rightarrow\) \(\text{x} =\frac{-16}{9}\div\frac{-4}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{-16}{9}\times\frac{3}{-4}\)

\(\Rightarrow\) \(\text{x}=\frac{-16\times3}{9\times-4}\)

\(\Rightarrow \) \(\text{x}=\frac{-48\times-1}{-36\times-1}=\frac{48}{36}\)

\(\Rightarrow \) \(\text{x}=\frac{48}{36}=\frac{48\div12}{36\div12}=\frac{4}{3}\)

Hence, the other rational number is \(\frac{4}{3}\)

Correct option is (A) 1 11/14

\(4\frac{2}{7}\div2\frac{2}{5}\) \(=\frac{4\times7+2}{7}\div\frac{2\times5+2}{5}\) \(=\frac{30}{7}\div\frac{12}{5}\)

\(=\frac{30}{7}\times\frac5{12}=\frac{5\times5}{7\times2}\) \(=\frac{25}{14}=\frac{14+11}{14}=1+\frac{11}{14}\)

\(=1\frac{11}{14}.\)

Correct option is   A)  \(1\frac{11}{14} \)

Let the required number be x. Then,

= (3/16) + x = (-3/8)

By sending (3/16) from left hand side to the right hand side it changes to – (3/16)

x = (-3/8) – (3/16)

LCM of 8 and 16 is 16

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(-3×2)/ (8×2)] = (-6/16)

= [(3×1)/ (16×1)] = (3/16)

Then,

= (-6/16) – (3/16)

= (-6-3)/16

= (-9/16)

Hence the required number is (-9/16)

Correct option is   B) \(\frac{91}{11}\)

i) According to commutative property.

= -4 × 7/9 = 79 × -4

ii) According to commutative property.

= 5/11 × -3/8 = -3/8 × 5/11

iii) According to commutative property

= -4/5 × (5/7 + -8/9) = (-4/5 × 5/7) + -4/5 × -8/9

Let the required number be x. Then,

= (-15/7) + x = (-3)

By sending (-15/7) from left hand side to the right hand side it changes to – (-15/7)

x = (-3) – (-15/7)

LCM of 1 and 7 is 7

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(-3×7)/ (1×7)] = (-21/7)

= [(-15×1)/ (7×1)] = (-15/7)

Then,

= (-21/7) – (-15/7)

= (-21+15)/7

= (-6/7)

Hence the required number is (-6/7)

(-7/10) by (-40/21)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

The above question can be written as (-7/10) × (-40/21)

We have,

= (-7×-40)/ (10×21)

On simplifying,

= (-1×4)/ (1×3)

= (4/3)

i) 
\(\frac{1}{\frac{1}{2}}=1\times2\\=2\)

ii) 5/-5/7 

 = 5 × 7/-5 

= -7

iii)  (-3/4) / (9/-16)

= (-3/4) × -16/9 

= 4/3

iv)  (2/3) / (-7/12)

= (2/3) × 12/-7 

= -8/7

v) 0 / (7/5) 

= 0

(25/-9) by (3/-10)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

First we write each of the given numbers with a positive denominator.

(25/-9) = [(25× (-1))/ (-9×-1)]

= (-25/9)

(3/-10) = [(3× (-1))/ (-10×-1)]

= (-3/10)

The above question can be written as (-25/9) × (-3/10)

We have,

= (-25×-3)/ (9×10)

On simplifying,

= (-5×-1)/ (3×2)

= (5/6)

Let the required number be x. Then,

= (-5/1) + x = (-4/3)

By sending (-5/1) from left hand side to the right hand side it changes to – (-5/1).

x = (-4/3) – (-5/1)

LCM of 3 and 1 is 3

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

= [(-4×1)/ (3×1)] = (-4/3)

= [(-5×3)/ (1×3)] = (-15/3)

Then,

= (-4/3) – (-15/3)

= (-4+15)/3

= (11/3)

Hence the required number is (11/3)