Tag: rational and irrational numbers

Quizzes (6) Questions (185)

Questions Related to rational and irrational numbers

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State True or False.

$(2-\sqrt{2})(2+\sqrt{2})$ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\ { (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 $

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State True or False.

$\sqrt{5}-2$ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\ { (\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 $

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State True or False.

$-\displaystyle\frac{2}{5}\sqrt{8}$ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\ { \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 $

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State TRUE or FALSE
$\displaystyle\frac{(2+\sqrt{2})(3-\sqrt{5})}{(3+\sqrt{5})(2-\sqrt{2})}$ is Rational.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\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 } \$


$\displaystyle =\frac { (6+4\sqrt { 2 } )(14-6\sqrt { 5 } ) }{ 8 } =\frac { (3+2\sqrt { 2 } )(7-3\sqrt { 5 } ) }{ 2 } =\frac { (21-9\sqrt { 5 } +14\sqrt { 2 } -6\sqrt { 10 } ) }{ 2 } \ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebraic\quad equation\quad in\quad numerator\quad \ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\ $
The given statement is false.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State TRUE or FALSE 

${(2+\sqrt{3})}^{2}$ is Irrational

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ (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.\ $

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

 $\sqrt3$ is 

  1. rational number

  2. irrational number

  3. natural number

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

 $\sqrt2 + \sqrt3$ is 

  1. irrational

  2. rational

  3. natural

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

"Solve" for $\sqrt{6}$
$\sqrt{6} = \cfrac{\left(m^{2} - 5n^{2}\right)}{\left(2n^{2}\right)} $
so if  $\sqrt{2} + \sqrt{3} $ is  rational,  then  so  is $ \sqrt{6}$
Let a and b be the integers with gcd(a,b) = 1 such that
$\cfrac{a}{b} = \sqrt{6}$
Square both sides and multiply by $b^2$:
$a^2 = 6b^2 $
Now, the right side is divisible by 2, so $a^2$ is divisible by 2, which
then implies that a is divisible by 2 (since 2 is prime).
Therefore we  can write a=2k for some integer k:
$4k^{2} = \left(2k \right)^{2} = 6b^{2} $
Divide by 2:
$2k^{2} = 3b^{2} $
Now the left side is divisible by 2, so $3b^{2}$ is divisible by 2, from which it follows that b is divisible by 2.
However, this would mean that 2 divides gcd(a,b) = 1. Contradiction.
$\therefore  \sqrt{6} $ is  irrational
,  and  $\therefore \sqrt{2} + \sqrt{3} $ is  also irrational.