Tag: proofs of irrationality

Questions Related to proofs of irrationality

If $a=\sqrt{11}+\sqrt{3}, b =\sqrt{12}+\sqrt{2}, c=\sqrt{6}+\sqrt{4}$, then which of the following holds true ?

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $c>a>b$

  2. $a>b>c$

  3. $a>c>b$

  4. $b>a>c$

Show answer
Correct Option: B
Explanation:

$a=\sqrt{11}+\sqrt{3}$

$a^{2}=11+3+2\sqrt{33}=14+2\sqrt{33}$

$b=\sqrt{12}+\sqrt{2}$

$b^{2}=14+2\sqrt{24}$

As $\sqrt{33} > \sqrt{24}, a^{2} > b^{2}, a>b$

$c=\sqrt{6}+\sqrt{4}$

$c^{2}=10+2\sqrt{24}$

As $14>10, b^{2} > c^{2}, b>c$

Hence, $a>b>c$.
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State whether the following statement is true or not:
$\left( 3+\sqrt { 5 }  \right) $ is an irrational number. 

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
Let us suppose $3+\sqrt 5$ is rational.

$=>3+\sqrt 5$ is in the form of $\dfrac pq$ where $p$ and $q$ are integers and $q\neq0$

$=>\sqrt5=\dfrac pq-3$

​$=>\sqrt5=\dfrac{p-3q}{q}$

as $p, q$ and $3$ are integers $\dfrac{p-3q}{q}$ is a rational number.

$=>\sqrt 5$ is a rational number.

But we know that $\sqrt 5$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that $3+\sqrt 5$ is a rational number.

Hence $3+ \sqrt5$  is an irrational number.

A rational number equivalent to  $ \displaystyle \frac{-5}{-3}  $ is -

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $ \displaystyle \frac{25}{15} $

  2. $ \displaystyle \frac{-15}{25} $

  3. $ \displaystyle \frac{-25}{15} $

  4. None of these

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Correct Option: A
Explanation:

 $ \displaystyle  \because  \frac{-5}{-3} $= $ \displaystyle  \frac{-5}{-3} $X $ \displaystyle  \frac{-5}{-5} $= $ \displaystyle  \frac{25}{15} $

Every irrational number is

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. a surd

  2. a prime number

  3. not a surd

  4. none

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Correct Option: C
Explanation:

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.


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 C.

Which of the following are not a surd?

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $\sqrt{3+2\sqrt{5}}$

  2. $\sqrt [ 4 ]{ 3 } $

  3. $\sqrt [ 3 ]{ \sqrt{3} } $

  4. $\sqrt{343}$

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Correct Option: A,C
Explanation:

Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals 

What is the square of $(2 + \sqrt {2})$?

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. A rational number

  2. An irrational number

  3. A natural number

  4. A whole number

Show answer
Correct Option: B
Explanation:

Square of (2+√2) is (4+2.2.√2+2) = 6+4√2 which is an irrational number

State whether the following statement is True or False.
3.54672 is an irrational number.

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$3.54672$ can be written in simple fraction.$\Rightarrow $ Rational number.

State the following statement is True or False
 35.251252253...is an irrational number

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$35.251522253...\Rightarrow$ It is a non-terminating decimal. This number cannot be written as a simple fraction.

Therefore, it's an irrational number.

For three irrational numbers $p,q$ and $r$ then $p.(q+r)$ can be 

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. A rational number

  2. An irrational number

  3. An integer

  4. All of the above

Show answer
Correct Option: D
Explanation:

$p,q$ and $r$ are all irrational 
Let $p=q=r=\sqrt2$
$p(q+r)=p.q+p.r$
$\Rightarrow \sqrt { 2 } (\sqrt { 2 } +\sqrt { 2 } )=\sqrt { 2 } .\sqrt { 2 } +\sqrt { 2 } .\sqrt { 2 } =2+2=4$
which is rational as well as integer
Let us take another case in which $p=\sqrt2$ and $q=r=\sqrt3$
$\Rightarrow \sqrt { 2 } (\sqrt { 3 } +\sqrt { 3 } )=\sqrt { 2 } .\sqrt { 3 } +\sqrt { 2 } .\sqrt { 3 } =\sqrt { 6 } +\sqrt { 6 } =2\sqrt { 6 } $
which is an irrational number.
So on applying distributive property on three irrational numbers we can get an integer,a rational as well as an irrational number .
So option $D$ is correct.

Which of the following irrational number lies between $\dfrac{3}{5}$ and $\dfrac{9}{10}$

maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $\dfrac{\sqrt80}{10}$

  2. $\dfrac{\sqrt85}{10}$

  3. $\dfrac{\sqrt82}{10}$

  4. $\dfrac{\sqrt83}{10}$

Show answer
Correct Option: A
Explanation:
$\sqrt36<\sqrt{80}<\sqrt{81}$
On dividing with 10, we get
$\dfrac{6}{10}<\dfrac{\sqrt{80}}{10}<\dfrac{9}{10}$