Tag: irrational numbers

To get the sum as irrational, the numbers need to have an irrational part as well which are different from each other.

Example, the pair of numbers $ \sqrt{3} + 2 $ and $ 5 + \sqrt {2} $ have the sum $ \sqrt{3} + 2 + 5 + \sqrt {2} = 7 + \sqrt {2} + \sqrt {3} $ which is an irrational number too.

Decimal form of $\sqrt { 3 } $ is non terminating and non repeating, So, it is irrational number.

$ \sqrt{2} \times \sqrt[3] {3} \times \sqrt[4]{4}$
$=2^{ \frac { 1 }{ 2 }  } \times 3^{ \frac { 1 }{ 3 }  }\times 2^{ \frac { 2 }{ 4 }  }$
$=2^{ \frac { 1 }{ 2 }  } \times 2^{ \frac { 1 }{ 2 }  }\times 3^{ \frac { 1 }{ 3 }  }$
$=2  \times3^{ \frac { 1 }{ 3 }  }$
$=2^{ \frac { 3 }{ 3 }  }\times3^{ \frac { 1 }{ 3 }  }  $
$=\sqrt [ 3 ]{ 2^{ 3 } }\times\sqrt[3]{3}$
$=\sqrt[3]{8\times3}$
$=\sqrt[3]{24}$

We will be considering the digits only after the decimal point.

An irrational number is any real number that cannot be expressed as a ratio of integers. Irrational numbers are those real numbers that cannot be represented as terminating or repeating decimals.
Among all the options only $(D)$ $0.123223222322223$...... is non terminating and non repeating decimal.Therefore, it is a irrational number.

Rational numbers are those numbers which can be expressed in the form $ \dfrac {p}{q} $, where p and q are integers and $ q \neq 0 $
Numbers which are not rational numbers are called irrational numbers.
Since, $ \sqrt {7} $ cannot be written in $ \dfrac {p}{q} $, where $p$ and $q$ are integers and $ q \neq 0 $; it is an irrational number.

Real number consists of collection of rationals and irrationals.

Hence, every irrational number is also a real number.

Example-2 is also real.

The statement is false since real numbers consists of both rational and irrational numbers. $5,65,8/9...$ are all real numbers which are rational.

$2$ is rational

Decimal representation of an irrational number is always non terminating non repeating.