Tag: maths

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

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

a³ = a2 x a is a rational number,

a⁴ = a³x a is a rational number,

......

......

∴ aⁿ = a^n-1 x a is a rational number.

So, the option (III) is true

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

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

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.