Tag: irrational numbers

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Questions Related to irrational numbers

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.


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

Every surd is

  1. a natural number

  2. an irrational number

  3. a whole number

  4. a rational number
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
When a number cannot be simplified further to remove a square root then it is a surd.  
A surd is an irrational number.

For. eg: square root of 2 cannot be simplified. thus it is a surd.

By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt2$ is a surd since 2 is rational and $\sqrt 2$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand, $\sqrtπ$ is not a surd even though $\sqrtπ$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd.

The answer is Option B