Tag: maths

Questions Related to maths

If the product of two irrational numbers is rational, then which of the following can be concluded?

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

  1. The ratio of the greater and the smaller numbers is an integer.

  2. The sum of the numbers must be rational.

  3. The excess of the greater irrational number over the irrational number must be rational.

  4. None of the above

Show answer
Correct Option: A

 $\frac { 2 } { 2 + \sqrt { 3 } }$ 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

State whether the statement is true/false.
$\sqrt{72}$ is irrational

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:

72 is not an Irrational number because it can be expressed as the quotient of two integers: 72 ÷ 1

So, given statement is true.

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, $a>b>c$.
'

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

Show answer
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

Show answer
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. $\sqrt{343}$

Show answer
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.