Tag: properties of irrational numbers

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Questions Related to properties of irrational numbers

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

State TRUE or FALSE 

${(2+\sqrt{3})}^{2}$ 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
Multiple choice maths number systems existence of irrational numbers irrational numbers properties of irrational numbers

$0.\overline{35}$ is equal to

  1. $\displaystyle\frac{35}{66}$
  2. $\displaystyle\frac{35}{77}$
  3. $\displaystyle\frac{35}{99}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$X=0.35353535$   -- i
Multiplying equation i with 100,

$100x=35.353535353$   --ii 
Subtracting equation i from ii 

$ 100x-x = 35.3535 - 0.3535 $
$99x=35$
$ x = \dfrac{35}{99}$
Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$3.\overline{25}$ is equal to

  1. $\displaystyle\frac{320}{99}$
  2. $\displaystyle\frac{321}{99}$
  3. $\displaystyle\frac{322}{99}$
  4. $\displaystyle\frac{323}{99}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,$3.\overline{25}$.


Let,

$x=3.\overline{25}$

 $x=3.252525.....$


Multiply by 100 both sides,

  $ 100x=100\times 3.252525..... $

 $ 100x=325.2525..... $

 $ 100x=322+3.2525..... $

 $ 100x=322+x $

 $ 99x=322 $

 $ x=\dfrac{322}{99} $


Hence, this is the answer.

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

$0.\overline{05}$ is equal to

  1. $\displaystyle\frac{3}{99}$
  2. $\displaystyle\frac{4}{99}$
  3. $\displaystyle\frac{5}{99}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,$0.\overline{05}$

Let,

  $ x=0.\overline{05} $

 $ x=0.05050505..... $

Multiply by $100$ both sides,

 $ 100x=100\times 0.05050505..... $

 $ 100x=5.050505..... $

 $ 100x=5+0.050505..... $

 $ 100x=5+x $

 $ 99x=5 $

 $ x=\dfrac{5}{99} $


Hence, this is the answer.