Tag: rational numbers between two rational numbers

Questions Related to rational numbers between two rational numbers

 ________ are rational numbers between $\displaystyle -\dfrac{3}{4}$ and $\displaystyle \dfrac{1}{2}.$

  1. $\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{9}{16}$

  2. $\dfrac{-15}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$

  3. $\dfrac{-7}{16}, \dfrac{-1}{8}, \dfrac{3}{16}$

  4. none of the above


Correct Option: C
Explanation:
A rational number between two numbers $ a $ and $ b = \dfrac {(a + b)}{2} $ 

So,
a rational number between $ - \dfrac {3}{4} $ and $ \dfrac {1}{2} $
$= \dfrac {-\dfrac {3}{4} + \dfrac {1}{2}}{2} = - \dfrac {1}{8} $

Now,
another rational number between $ - \dfrac {3}{4} $ and $ - \dfrac {1}{8} $
$=\dfrac {- \dfrac {3}{4} - \dfrac {1}{8}}{2} = - \dfrac {7}{16} $ 

Another rational number between $ - \dfrac {1}{8} $ and $ \dfrac {1}{2} =$

$\dfrac { - \dfrac {1}{8} + \dfrac {1}{2}} {2} =  \dfrac {3}{16} $ 

Hence, required three rational numbers between $ - \dfrac {3}{4} $ and $  \dfrac {1}{2} $ are $ - \dfrac {3}{4}, - \dfrac {7}{16},  - \dfrac {1}{8}, \dfrac {3}{16}, \dfrac {1}{2} $

The rational number lying between the numbers $\displaystyle \frac{1}{3}$ and $\displaystyle \frac{3}{4}$ are

  1. $\displaystyle \frac{97}{300}$,$\displaystyle \frac{299}{500}$

  2. $\displaystyle \frac{99}{300}$,$\displaystyle \frac{301}{400}$

  3. $\displaystyle \frac{95}{300}$,$\displaystyle \frac{301}{400}$

  4. $\displaystyle \frac{117}{300}$,$\displaystyle \frac{287}{400}$


Correct Option: D
Explanation:

To insert rational numbers between $2$ numbers, we will arrange the options and check if they are in ascending order.
$\dfrac { 1 }{ 3 } { ? }\dfrac { 117 }{ 300 } \ \Longrightarrow 300<351\ \qquad \dfrac { 3 }{ 4 } { ? }\dfrac { 287 }{ 400 } \ \Longrightarrow 1200>1148$
They are in ascending order, i.e., $\dfrac { 1 }{ 3 } ,\dfrac { 117 }{ 300 } ,\dfrac { 287 }{ 400 } ,\dfrac { 3 }{ 4 } $
From the given options only option $D$ satisfies this condition. Hence, $D$ is the answer.

Let a, b, c be positive integers such that $\frac {a\sqrt 2+b}{b\sqrt 2+c}$ is a rational number, then which of the following is always an integers?

  1. $\frac {2a^2+b^2}{2b^2+c^2}$

  2. $\frac {a^2+b^2-c^2}{a+b-c}$

  3. $\frac {a^2 _2b^2}{b^2+2c^2}$

  4. $\frac {a^2+b^2+c^2}{a+c-b}$


Correct Option: D

Let $x\;\in\;Q,\;y\;\in\;Q^c$, which of the following statement is always WRONG ?

  1. $xy\;\in\;Q^c$

  2. $y/x\;\in\;Q$, whenever defined

  3. $\sqrt{2}x+y\;\in\;Q$

  4. $x/y\;\in\;Q^c$, whenever defined


Correct Option: B
Explanation:

Let $x=1,\;y=\sqrt{2}$
Then $xy=\sqrt{2}\;\in\;Q^c$
Obvious
$x=-1,\;y=\sqrt{2}$ then $\sqrt{2}x+y=0\;\in\;Q$
$x=1,\;y=\sqrt{2}$ then $x/y=\displaystyle\frac{1}{\sqrt{2}}\;\in\;Q^c$

Which of these is true?
$(I)$ $5\sqrt {3}$ is not a rational number
$(II)$ $1$ is not the cube of a rational number
$(III)$ If a is rational and $n$ is an integer greater than $1$, then ${a}^{n}$ is rational.

  1. $I$ and $II$

  2. $II$ and $III$

  3. $III$ and $I$

  4. all three


Correct Option: C
Explanation:

(I) In $5\sqrt{3}$

5 ia s rational number and $\sqrt{3}$ is an Irrational number
As we know, The product of a rational and irrational number is an irrational number.
So, $5\sqrt{3}$ is not a rational number.
Hence, the option (I) is true
(II) 1 is a rational number 
and cube of 1 is 1 only, which is a rational number
Hence the option (II) is False
(III) We know that product of two rational number is always a rational number
Hence if a is a rational number and n is greater than one 
Then,
a2 = a x a is a rational number.

a3 = a2 x a is a rational number,

a4 = a3x a is a rational number,

......

......

 an = an-1 x a is a rational number.

So, the option (III) is true

Which of the following numbers lies between $\dfrac {5}{24}$ and $\dfrac {3}{8}$?

  1. $\dfrac {7}{2}$

  2. $1$

  3. $\dfrac {7}{24}$

  4. $0$


Correct Option: C
Explanation:

Mean $= \dfrac {\dfrac {3}{8} + \dfrac {5}{24}}{2} = \dfrac {\dfrac {9 + 5}{24}}{2} = \dfrac{\left (\dfrac {14}{24}\right )}{2}$


$= \dfrac {7}{12}\times \dfrac {1}{2}$

$= \dfrac {7}{24}$

Mean of two numbers lies between the two numbers.  
So, $ \dfrac {7}{24}$ lies between $\dfrac {3}{8}$ and $\dfrac {5}{24}.$

Which of the following numbers lies between $-1$ and $-2$?

  1. $\dfrac {-1}{2}$

  2. $\dfrac {-3}{2}$

  3. $\dfrac {1}{2}$

  4. $\dfrac {3}{2}$


Correct Option: B
Explanation:

Mean $= \dfrac {(-1) + (-2)}{2} = \dfrac {-1 -2}{2} = \dfrac {-3}{2}$.

Mean of two numbers always lies between the two numbers.
So, answer is option $B.$

Which of the following represents a rational number between $-6$ and $-7$?

  1. $\dfrac {-6 - 7}{2}$

  2. $\dfrac {-6 + 7}{2}$

  3. $\dfrac {6 + 7}{2}$

  4. $-6 - 7$


Correct Option: A
Explanation:

Mean $= \dfrac {(-6) + (-7)}{2} = \dfrac {-6 -7}{2}$.

Mean of two numbers always lies between the two numbers.
So, answer is option $A.$

A rational number between $\dfrac {-9}{10}$ and $\dfrac {4}{5}$ is:

  1. $\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times \dfrac {1}{2}$

  2. $\left (\dfrac {-9}{10} - \dfrac {4}{5}\right ) + \dfrac {1}{2}$

  3. $\left (\dfrac {-9}{10} + \dfrac {4}{5}\right ) \times 2$

  4. All above are correct


Correct Option: A
Explanation:
Mean of two numbers always lies between the two numbers.

Mean 

$= \dfrac{\left (\dfrac {-9}{10} + \dfrac {4}{5}\right )}{2}$

$= \left (\dfrac {-9}{10} + \dfrac {4}{5}\right )\times \dfrac {1}{2}$.

So, answer is option $A.$

Which of the following rational numbers lies between $\dfrac {3}{4}$ and $\dfrac {13}{8}$?

  1. $\dfrac {11}{16}$

  2. $\dfrac {12}{16}$

  3. $\dfrac {19}{16}$

  4. $\dfrac {26}{16}$


Correct Option: C
Explanation:

$Mean = \dfrac{\left (\dfrac {3}{4} + \dfrac {13}{8}\right )}{2} = \dfrac {6 + 13}{8} \times \dfrac {1}{2} = \dfrac {19}{16}$.

Mean of two numbers always lies between the two numbers.
So, answer is option $C.$