Tag: maths

$\sqrt 7 $ is a rational number

We know that difference of a rational and irrational number is also irrational, 

when we divide $100$ with $99$ we will get remainder $1$ and hence next we divide 99 with $1$ which gives us remainder $0$.

Since we know that ratio of areas of two similar triangles is equal to the square of the ratio of their altitude
therefore
Ratio of their altitude=$\sqrt {\dfrac{{200}}{{128}}} $
$ = \sqrt {\dfrac{{100}}{{64}}} $
$ = \dfrac{{10}}{8}$
$ = \dfrac{5}{4}$
$ = 5:4$

The series is 3 - 1 + 1/3 - 1/9 + ... which can be split into 3 + (-1 + 1/3 - 1/9 + ...). The part in parentheses is a geometric series with first term a = -1 and common ratio r = -1/3. The sum is a / (1 - r) = -1 / (1 - (-1/3)) = -1 / (4/3) = -3/4. Adding the initial 3 gives 3 - 3/4 = 9/4.

Above in an infinite A.G.S. with $a = 1, d = 1$ for A.P., $b = \dfrac {1}{3}, r = \dfrac {1}{3}$ for G.P.
$\therefore S _{\infty} = \dfrac {ab}{1 - r} + \dfrac {dbr}{(1 - r)^{2}} = \dfrac {\dfrac {1}{3}}{1 - \dfrac {1}{3}} + \dfrac {1 . \dfrac {1}{3} . \dfrac {1}{3}}{\left (1 - \dfrac {1}{3}\right )^{2}} = \dfrac {1}{2} + \dfrac {1}{4} = \dfrac {3}{4}$
$\therefore P = 3^{S} = 3^{3/4} \therefore P^{1/3} = 3^{1/4}$.

$ar = 2, \dfrac {a}{1 - r} = \dfrac {1}{8}$. Eliminate $r$ and we get
$a^{2} - 8a + 16 = 0 \therefore (a - 4)^{2} = 0$ or $a = 4$

$(9)^{S _{\infty}} = 9^{1/2} = 3\because S _{\infty} = \dfrac {1/3}{1 - (1/3)} = \dfrac {1}{2}$.

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $