Rational numbers between two rational numbers - class-IX

5/16 = 0.3125 and 1/2 = 0.5. Comparing options: 3/8 = 0.375 (in range), 7/16 = 0.4375 (in range), 1/4 = 0.25 (not in range), 13/32 = 0.40625 (in range).

$2 < 2.1=(2+0.1) < 2.2=(2.1+0.1) < 2.3=(2.2+0.1) < 2.4=(2.3+0.1) < ... < 2.9=(2.8+0.1) < 3=(2.9+0.1)$

We know that, $ \sqrt {2} = 1.414

$ and $ \sqrt {3} = 1.732 $

Hence two rational numbers between $ 1.414 $ and $ 1.732

$  can be $ 1.5 (= \frac {3}{2}) $ and $ 1.6 (= \frac {16}{10}= \frac {8}{5}) $

We know that, $ \sqrt {3} = 1.732$ and $ \sqrt {5} = 2.236 $

Hence three rational numbers between $ 1.732 $ and $ 2.236$  can be $ 1.8 \left(= \dfrac {18}{10}\right)  $ , $ 2 $ and $ 2.2 \left(= \dfrac {22}{10}\right) $.

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

As $\sqrt 6$ is irrational therefore option A is wrong.

In option B,

The rational number between $\dfrac12$ and $\sqrt1$ :

We know $\dfrac{5}{16} =0.3125$

We know $\dfrac{1}{4} = 0.25$

Between any two rational numbers we can find infinitely many rational numbers. 

rational number always lies on the line.

$\dfrac{-3}{2},-1,3,0,\dfrac{1}{2}$

There are infinite rational numbers between any two rational numbers 

All the options have denominator $75$. Hence, let us convert into equivalent fractions having denominator $75$. 
$\dfrac{3}{5} $ $=\dfrac{3\times 15}{5\times 15} $ $=\dfrac{45}{75}$

$\dfrac{2}{3}$ $=\dfrac{2\times 25}{3\times 25}$ $=\dfrac{50}{75}$

here, $\dfrac {5}{\sqrt 3-\sqrt 5}$

Converting the fractions to decimals or common denominators helps locate a rational number between 3/7 (approx 0.428) and 2/3 (approx 0.667). Option B, 4/7, is approximately 0.571, which correctly lies between these two values.

A rational number between two numbers $ a $ and $ b = \dfrac {(a +

b)}{2} $
So, a rational number between $\dfrac {3}{8} $ and $ \dfrac {7}{12}$

A

rational number between two numbers $ a $ and $ b = \frac {(a +

b)}{2} $

So, a

rational number between $\frac {1}{3} $ and $ \frac {1}{4} = \frac {(\frac {1}{3} + \frac {1}{4})}{2} = \frac {7}{24} $

Now, another rational number between $ \frac {7}{24} $ and $ \frac {1}{4} = \frac {(\frac {7}{24} + \frac {1}{4})}{2} = \frac {13}{48} $



Hence, required two rational numbers between $\frac {1}{3} $ and $ \frac {1}{4} $ are $\frac {1}{3} ,\frac {7}{24}, \frac {13}{48}, \frac {1}{4}$

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.

For (asqrt(2)+b)/(bsqrt(2)+c) to be rational, the irrational parts must cancel. This implies a/b = b/c, so b^2 = ac. Testing option D: (a^2+b^2+c^2)/(a+c-b) is a standard algebraic identity related to this condition.

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 $x/y=\displaystyle\frac{1}{\sqrt{2}};\in;Q^c$

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

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

$Mean = \dfrac{\left (\dfrac {4}{9} + \dfrac {4}{5}\right )}{2} = \left (\dfrac {20 + 36}{45}\right ) \times \dfrac {1}{2} = \dfrac {56}{45}\times \dfrac {1}{2}$
$= \dfrac {28}{45}$

Consider $3 \times 4 = 12$, so 
$\dfrac 23 = \dfrac{8}{12}$

The given rational numbers $-\dfrac { 2 }{ 3 }$ and $-\dfrac { 1 }{ 5 }$ are negative rational numbers because the numerator and denominator of both the rational numbers are of opposite signs that is the numerator of both the integers is negative while the denominators are positive.

3/8 = 0.375 and 7/12 = 0.5833. Option A: 41/96 = 0.427 and 23/48 = 0.479. Both are between 0.375 and 0.5833.