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CBSE Questions for Class 8 Maths Rational Numbers Quiz 1 - MCQExams.com
CBSE
Class 8 Maths
Rational Numbers
Quiz 1
State whether the statements are true (T) or false (F).
The rational numbers $$\dfrac{1}{2}$$ and $$-\dfrac{5}{2}$$ are on the opposite sides of $$0$$ on the number line.
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True
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False
State whether the statements are true (T) or false (F).
The rational numbers can be represented on the number line.
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True
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False
Solve following equation-
$$\dfrac { 3 }{ 7 } +\left( \dfrac { -6 }{ 11 } \right) +\left( \dfrac { -8 }{ 21 } \right) +\left( \dfrac { 5 }{ 22 } \right)$$
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$$\dfrac { -125 }{ 462 }$$
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$$\dfrac { 125 }{ 462 }$$
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$$\dfrac { 462 }{ 125 }$$
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$$\dfrac { -462 }{ 125 }$$
Find multiplicative inverse of the following-
$$\dfrac { -5 }{ 8 } \times \dfrac { -3 }{ 7 }$$
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$$\dfrac { 15 }{ 56 }$$
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$$\dfrac { 56 }{ 15 }$$
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$$\dfrac { -15 }{ 56 }$$
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$$\dfrac { -56 }{ 15 }$$
Name which property is used in following operation-
$$\dfrac { 1 }{ 2 } \times \left( 5\times \dfrac { 2 }{ 5 } \right) =\left( \dfrac { 1 }{ 2 } \times 5 \right) \times \dfrac { 2 }{ 5 }$$
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Associative
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Reciprocal
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Commutative
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Additive inverse
Multiply $$\dfrac { 6 }{ 13 }$$ by the reciprocal of $$\dfrac { -7 }{ 16 }$$
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$$\dfrac { -91 }{ 96 }$$
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$$\dfrac { 96 }{ 91 }$$
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$$\dfrac { 91 }{ 96 }$$
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$$\dfrac { -96 }{ 91 }$$
$$\displaystyle \frac{-7}{5} + \left(\displaystyle \frac{2}{-11} + \frac{-13}{25} \right) = \left(\displaystyle \dfrac{-7}{5} + \frac{2}{-11} \right) + \frac{-13}{25}$$
This property is
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closure
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commutative
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associative
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identity
Explanation
Since a + (b + c) = (a + b) + C is associative property.
$$x$$ is additive identity of
rational numbers
.Then the value of x is
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0
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1
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-1
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None
Explanation
Since $$ 5 + 0 = 5 $$
$$\therefore$$ 0 is the additive identity of rational numbers.
For any rational number $$\dfrac {a}{b}$$, its additive inverse is ..........
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$$\dfrac {a}{b}$$
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$$\dfrac {b}{a}$$
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$$\dfrac {-b}{a}$$
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$$\dfrac {-a}{b}$$
Explanation
If $$\dfrac {a}{b}$$ is a rational number, then its negative $$\dfrac {-a}{b}$$ is called the additive inverse of $$\dfrac {a}{b}$$.
State true or false.
The three rational number between
$$5$$ and $$6$$ are
$$ \displaystyle\frac{21}{4},\frac{22}{4},\frac{23}{4}$$.
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True
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False
Explanation
$$5=\frac { 20 }{ 4 } \quad and\quad 6=\frac { 24 }{ 4 } ,\\ \frac { 21 }{ 4 } ,\frac { 22 }{ 4 } \quad and \quad \frac { 23 }{ 4 } \quad are\quad three\quad rational\quad numbers\quad lying\quad between\quad 5\quad and\quad 6.\quad \quad \\ $$
Choose the correct option for the following statement.
The property allows you to compute $$\displaystyle \frac{1}{3}\times \left ( 6 \times\frac{4}{3} \right )as \left ( \frac{1}{3} \times 6\right )\times \frac{4}{3}$$ is Associativity.
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The given statement is true.
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The given statement is false.
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Incomplete information
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None of these
Explanation
For any element $$a,b,c\in A$$, associative property states that
$$a(bc)=(ab)c$$
Take $$a=\dfrac{1}{3}, b=6, c=\dfrac{4}{3}$$, we get
$$\dfrac{1}{3}\times \left(6\times \dfrac{4}{3}\right)=\left(\dfrac{1}{3}\times 6\right)\times \dfrac{4}{3}$$
Hence, the statement is true.
State the following statement is True or False
The reciprocal of $$0$$ lie on the number
line.
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True
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False
Explanation
$$0$$ is the only rational number that does not have a reciprocal as $$\dfrac{1}{0}$$ is not defined, Hence cannot be represented on a number line .
Therefore the statement is false.
The property under multiplication used is:
$$-\dfrac{13}{17}\times\dfrac{-2}{7}=\dfrac{-2}{7}\times \dfrac{-13}{17}$$
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Associative Property
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Commutative Property
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Distributive Property
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Identity Property
Explanation
Commutative property says that if $$a$$ and $$b$$ are two number then
$$a\times{b}=b\times{a}$$
So correct answer will be Option B
Two rational numbers between $$\dfrac{2}{3}$$ and $$\dfrac{5}{3}$$ are :
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$$\dfrac{1}{6}$$ and $$\dfrac{2}{6}$$
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$$\dfrac{1}{2}$$ and $$\dfrac{2}{1}$$
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$$\dfrac{5}{6}$$ and $$\dfrac{7}{6}$$
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$$\dfrac{2}{3}$$ and $$\dfrac{4}{3}$$
Explanation
Changing the denominators of both numbers to 6, we get
$$\dfrac { 2 }{ 3 } =\dfrac { 4 }{ 6 } \quad \& \quad \dfrac { 5 }{ 3 } =\dfrac { 10 }{ 6 }$$
Numbers between the given rational numbers from the options are
$$\dfrac { 5 }{ 6 } \quad \& \quad \dfrac { 7 }{ 6 } $$
So, correct answer is option C.
Between any two rational numbers,
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there is no rational number
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there is exactly one rational number
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there are infinitely many rational numbers
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there are only rational numbers and no irrational numbers
Explanation
Re
call that to find a rational number between $$r$$ and $$s,$$ you can add
$$r$$ and $$s$$ and divide the sum by $$2,$$ that is $$\displaystyle \frac { r+s }{ 2 }$$ lies between r and s.
For example, $$\displaystyle \frac { 5 }{ 2 }$$ is a number between $$2$$ and
$$3.$$ We can proceed in this manner to find many more rational numbers between $$2$$ and $$3.$$
Hence, we can conclude that there are infinitely many rational numbers between any two given rational numbers.
The property under multiplication used in each of the following.
$$\dfrac{-4}{5}\times 1 = 1 \times\dfrac{-4}{5}= -\dfrac{4}{5}$$
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Commutative Property
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Associative Property
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Distributive Property
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Identity Property
Explanation
According to commutative property if $$a$$ and $$b$$ are real numbers then
$$a\times b=b\times a$$
According Identity if a is any real number then
$$a\times 1=a$$
So correct answer will be option $$A$$ and $$D$$
The property under multiplication used in each of the following is the
$$\dfrac{-19}{29}\times\dfrac{29}{-19}=1$$
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Commutative Property
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Multiplicative inverse
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Associative Property
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Indentity Property
Explanation
The multiplicative inverse for any number $$n$$ is $$\dfrac {1}{n}$$ where $$n\ne0.$$ The product of a number and its
multiplicative
inverse is $$1$$.
Let us take a number $$n=\dfrac {-19}{29}$$, then according to the definition, the multiplicative inverse of $$n$$ is
$$\dfrac { 1 }{ n } =\dfrac { 29 }{ -19 }$$
Now consider the product of $$n$$ and
$$\dfrac {1}{n}$$ as follows:
$$\dfrac { -19 }{ 29 } \times \dfrac { 29 }{ -19 } =1$$
Hence, multiplicative inverse property is used in
$$\dfrac { -19 }{ 29 } \times \dfrac { 29 }{ -19 } =1$$
A rational number between $$\displaystyle \frac{1}{4}$$ and $$\displaystyle \frac{1}{3}$$ is
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$$\displaystyle \frac{7}{24}$$
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$$\displaystyle \frac{16}{48}$$
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$$\displaystyle \frac{13}{48}$$
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All the above
Explanation
$$\dfrac{1}{4} = \dfrac{6}{24} = \dfrac{12}{48} $$
$$\dfrac{1}{3} = \dfrac{8}{24} = \dfrac{16}{48}$$
From this, we can see $$\dfrac{7}{24} , \dfrac{13}{48}, \dfrac{16}{48}$$ all lie between $$\dfrac{1}{4}$$ and $$\dfrac{1}{3}$$
Find the five rational numbers between
$$\displaystyle \frac{1}{2}$$ and $$\displaystyle \frac{3}{2}$$
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$$0.5 < 0.6 < 0.7 < 0.8 ... < 1.1 < ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 1.7 < 3.8 ... < 1.8< ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 0.7 < 2.8 ... < 1.1 < ... < 1.15 < 1.50$$
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$$0.5 < 0.6 < 0.7 < 0.8 ... < 3.1 < ... < 1.15 < 1.50$$
State true or false:
The product of two rational numbers is always a rational number.
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True
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False
Explanation
According to the properties of rational numbers, the product of two rational numbers is also a rational number.
Hence, the given statement is true.
If D be subset of `the set of all rational numbers, which can be expressed as terminating decimals, then D is closed under the binary operations of
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addition, subtraction and division.
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addition, multiplication and division.
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addition, subtraction and multiplication.
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subtraction, multiplication and division.
Explanation
The sum, difference or product of two rational numbers will be a number that can be expressed as a terminating decimal.
The ratio of two rational numbers need not be always expressible as a terminating decimal.
So, the answer is option C.
Choose the rational number which lie between rational numbers $$-\dfrac{2}{5}$$ and $$-\dfrac{1}{5}$$.
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$$-\dfrac{1}{4}$$
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$$-\dfrac{3}{10}$$
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$$\dfrac{3}{10}$$
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$$-\dfrac{7}{20}$$
Explanation
Let $$p$$ be a rational number between $$-\dfrac {2}{5} $$ and $$ -\dfrac {1}{5}$$
$$ p= \dfrac {-\dfrac {2}{5} + (-\dfrac {1}{5})}{2} \\= -\dfrac {3}{10} $$
Let $$m$$ be an another rational number between $$ -\dfrac {2}{5} $$ and $$ -\dfrac {3}{10} $$
$$m= \dfrac {-\dfrac {2}{5} + (-\dfrac {3}{10})}{2} \\= -\dfrac {7}{20} $$
Let $$q$$ be an another rational number between $$ -\dfrac {1}{5} $$ and $$ -\dfrac {3}{10} $$
$$q= \dfrac {-\dfrac {1}{5} + (-\dfrac {3}{10})}{2} \\= -\dfrac {1}{4} $$
Hence, required rational numbers between $$(-\dfrac {2}{5}) $$ and $$ (-\dfrac {1}{5}) $$ are $$(-\dfrac {3}{10}) ,(-\dfrac {7}{20}), (-\dfrac {1}{4})$$
For a rational number to lie between $$\dfrac{-2}{5}$$ and $$\dfrac{-1}{5}$$,it should be less than $$\dfrac{-1}{5}$$ and greater than $$\dfrac{-2}{5}$$.
Now,$$\dfrac{3}{10}$$ is not less than $$\dfrac{-1}{5}$$.
So,$$\dfrac{3}{10}$$ does not lie between $$\dfrac{-1}{5}$$ and $$\dfrac{-2}{5}$$.
The additive inverse of $$\displaystyle \frac{-a}{b}$$ is
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$$\displaystyle \frac{a}{b}$$
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$$\displaystyle \frac{b}{a}$$
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$$\displaystyle \frac{-b}{a}$$
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$$\displaystyle \frac{-a}{b}$$
Explanation
$$\therefore$$ p + (-P) = (-P) + P = 0 is the additive inverse property.
So additive inverse of $$\displaystyle \frac{-a}{b}$$ is $$\displaystyle \frac{a}{b}$$
Which one of the following is the rational number lying between $$\displaystyle \frac{6}{7} \ and \ \frac{7}{8}?$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{99}{122}$$
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$$\displaystyle \frac{95}{112}$$
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$$\displaystyle \frac{97}{112}$$
Explanation
Required rational number $$\displaystyle =\frac{1}{2}\left ( \frac{6}{7}+\frac{7}{8} \right )=\frac{1}{2}\left ( \frac{48+49}{56} \right )=\frac{97}{112}$$
Hence option (d) is correct
Multiplicative inverse of '0' is
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$$\displaystyle \frac{1}{0}$$
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0
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Does not exist
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$$\displaystyle \frac{0}{0}$$
Explanation
Does not exist since division by zero is not defined.
Write the additive inverse of each of the following: $$\dfrac{2}{8}$$ and $$-\dfrac{5}{9}$$
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$$-\dfrac{2}{8}\;;\; \dfrac{5}{9}$$.
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$$-\dfrac{1}{5};\dfrac{3}{5}$$
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$$\dfrac{1}{8};\dfrac{5}{9}$$
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$$\dfrac17;\dfrac89$$
Explanation
The additive inverse of $$\displaystyle\frac{2}{8}$$ is $$\begin{pmatrix}\displaystyle\frac{-2}{8}\end{pmatrix}=\displaystyle\frac{-2}{8}$$.
The additive inverse of $$\displaystyle\frac{-5}{9}$$ is $$-\begin{pmatrix}\displaystyle\frac{-5}{9}\end{pmatrix}=\displaystyle\frac{5}{9}$$.
Write the additive inverse of each of the following rational numbers:
$$\displaystyle\frac{4}{9}$$; $$\displaystyle\frac{-13}{7}$$; $$\displaystyle\frac{5}{-11}$$; $$\displaystyle\frac{-11}{-14}$$
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$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-13}{14}$$
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$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{1}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$
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$$\displaystyle\frac{-4}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$
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$$\displaystyle\frac{-5}{9};\;\displaystyle\frac{13}{7};\;\displaystyle\frac{5}{11};\;\displaystyle\frac{-11}{14}$$
Explanation
The additive inverse of $$\displaystyle\frac{4}{9}$$ is $$-\displaystyle\frac{4}{9}=\displaystyle\frac{-4}{9}$$
The additive inverse of $$\displaystyle\frac{-13}{7}$$ is $$-\begin{pmatrix}\displaystyle\frac{-13}{7}\end{pmatrix}=-\begin{pmatrix}-\displaystyle\frac{13}{7}\end{pmatrix}=\displaystyle\frac{13}{7}$$
We have, $$\displaystyle\frac{5}{-11}=\displaystyle\frac{-5}{11}$$
The additive inverse of $$\displaystyle\frac{-5}{11}$$ is $$-\begin{pmatrix}\displaystyle\frac{-5}{11}\end{pmatrix}$$
$$=-\begin{pmatrix}-\displaystyle\frac{5}{11}\end{pmatrix}=\displaystyle\frac{5}{11}$$
We have, $$\displaystyle\frac{-11}{-14}=\displaystyle\frac{11}{14}$$
The additive inverse of $$\displaystyle\frac{11}{14}$$ is $$-\begin{pmatrix}\displaystyle\frac{11}{14}\end{pmatrix}$$ $$=\displaystyle\frac{-11}{14}$$
Which of the following statement is
true about a rational number $$\displaystyle \frac{-2}{3}$$ ?
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It lies to the left side of $$'0'$$ on the number line.
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It lies to the right side of $$'0'$$ on the number line.
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It is not possible to represent on the number line.
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It cannot be determined on which side the number lies.
Explanation
We know that every positive number occupies space right to the origin while the negative numbers are present on the left side of the origin. Let us learn to represent a rational number on a number line with few steps.
(i) Draw a line and locate the point having the coordinate as $$0$$. This point is known as the origin.
If the given number is positive, locate it on the right side of the origin. If it is a negative number, locate it on the left side of zero.
Divide each unit into the values which are equal to the denominator of the fraction. For example: for representing $$\dfrac {4}{5}$$ on the number line you need to divide each unit into $$5$$ subunits.
Similarly,
$$\dfrac {-2}{3}$$
is a negative rational number and it is known that
$$\dfrac {-2}{3}$$
is greater than $$-1$$ and less than $$0$$. Therefore,
$$\dfrac {-2}{3}$$
lies between $$0$$ and
$$-1$$
and on the left side of
'0'
on
the number line.
Hence,
$$\dfrac {-2}{3}$$
lies to the left side of '0' on the number line.
Rational numbers are closed under substraction.
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True
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False
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Cannot be determined
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None
Explanation
Rational numbers are closed under subtraction.Because if we subtract a rational number with another
rational number we will get
rational number.
For example $$ 45 - 35 = 10 $$, which is also a rational number.
What is the additive inverse of $$\displaystyle\frac{a}{b}$$?
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$$\dfrac{-a}{b}$$
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$$\dfrac{b}{a}$$
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$$\dfrac{-b}{a}$$
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$$\dfrac{a}{b}$$
Explanation
The additive inverse of a number is what you add to a number to create the sum of zero.
So in other words, the additive inverse of $$x$$ is another number, $$y$$, as long as the sum of $$x + y$$ equals zero.
The additive inverse of $$x$$ is equal and opposite in sign to it (so, $$y = -x$$ or vice versa).
Let the additive inverse of $$\cfrac{a}{b}$$ be $$z$$. Then,
$$\cfrac{a}{b}+z=0$$
$$=>z=-\cfrac{a}{b}$$
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