Tag: rational and irrational numbers

$\sqrt { 7 } =2.64575131106...\ The\quad decimal\quad representation\quad is\quad non\quad repeating\quad non\quad terminating.\ Hence,\quad \sqrt { 7 } is\quad an\quad irrational\quad number.\ \quad $

$\ { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )=4-2=2\ \  { 2 } is\quad a\quad rational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad terminating.\ Hence,\quad { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )\quad is\quad a\quad rational\quad number.\ \quad $

$\ { (\sqrt { 5 } -2) }\ \sqrt { 5 } =2.2360679775........\ \ \sqrt { 5 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Subtraction\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 5 } -2) }\quad is\quad an\quad irrational\quad number.\ \quad $

$\ { \frac { -2 }{ 5 } \sqrt { 8 }  }={ \frac { -4 }{ 5 } \sqrt { 2 }  }=-0.8*\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\frac { -2 }{ 5 } \sqrt { 8 } ) }\quad is\quad an\quad irrational\quad number.\ \quad $

$\displaystyle \frac { (2+\sqrt { 2 } )(3-\sqrt { 5 } ) }{ (3+\sqrt { 5 } )(2-\sqrt { 2 } ) } =\frac { { (2+\sqrt { 2 } ) }^{ 2 }{ (3-\sqrt { 5 } ) }^{ 2 } }{ (9-5)(4-2) } =\frac { (4+2+4\sqrt { 2 } )(9+5-6\sqrt { 5 } ) }{ 8 } \$

${ (2+\sqrt { 3 } ) }({ 2-\sqrt { 3 } ) }=4-3=1\ Hnece\quad rational.\ $

${ (2+\sqrt { 3 } ) }^{ 2 }=4+3+4\sqrt { 3 } =7+4\sqrt { 3 } \ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebric\quad equation\quad \quad \ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\ $

$\displaystyle 3.14=\frac{22}{7}=\pi(pi)$

Let $\sqrt3$ is a rational number
$\therefore \sqrt3 = \displaystyle \frac{a}{b}$ [Where a & b are co-primes]
$a^2=3b^2$ .......(i)
$\Rightarrow$ 3 divides $a^2$
$\Rightarrow$ 3 also divides a
$\Rightarrow$ a=3c
[Where c is any non-zero positive integer]
$\Rightarrow a^2 = 9c^2$
From equation (i)
$3b^2=9c^2$
$\Rightarrow b^2 = 3c^2  \Rightarrow$ 3 divides $b^2$
$\Rightarrow$ 3 also divides b
So, 3 is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means $\sqrt3$ is an irrational number.

$\cfrac{m}{n} = \sqrt{2} + \sqrt{3} $
Square both sides:
$\cfrac{m^2}{ n^2} = 5 + 2\sqrt{6} $