Tag: proof of irrationality of numbers

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

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

  1. $c>a>b$
  2. $a>b>c$
  3. $a>c>b$
  4. $b>a>c$
Reveal answer Fill a bubble to check yourself
B Correct answer
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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Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

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

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

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

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

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

Every irrational number is

  1. a surd

  2. a prime number

  3. not a surd

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

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

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

  1. A rational number

  2. An irrational number

  3. An integer

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
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.