Tag: existence of irrational numbers

Quizzes (0) Questions (127)

Questions Related to existence of irrational numbers

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.

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

Which statement is true?

  1. $ \displaystyle \frac{-8}{12} $= $ \displaystyle \frac{10}{-15} $
  2. $ \displaystyle \sqrt{3} $ is not a real number

  3. Additive identity of 5 is -5
  4. $ \displaystyle \frac{2}{5} $>$ \displaystyle \frac{4}{5} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Option A is correct because both fractions simplify to -2/3. Option B is false as sqrt(3) is a real number. Option C is false because the additive identity is 0, while -5 is the additive inverse of 5. Option D is false because 2/5 is less than 4/5.

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

Irrational number is defined as 

  1. a real number that cannot be made by dividing two integers.

  2. a real number that can be made by dividing two integer.

  3. a number that can be made derived after multiplying two integers.

  4. a real number that can be written as whole number.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An irrational is any real number that cannot be expressed as a ratio of integers.

Therefore, $A$ is the correct answer.