Tag: existence of irrational numbers

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.

No. Not all square roots of integers are irrational. Examples are $\sqrt{4}=2$, $\sqrt{1}=1$, $\sqrt{9}=3$, etc.

$\sqrt{2} = 1.4142136...$
The decimal expansion of the number $\sqrt{2}$ is Non-terminating non recurring

$\ { (\sqrt { 2 } -2) }^{ 2 }=2+4-4\sqrt { 2 } =6-4\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad and\quad subtration\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 2 } -2) }^{ 2 }\quad is\quad an\quad irrational\quad number.\ \quad $

$\ { (\sqrt { 2 } +\sqrt { 3 } ) }^{ 2 }=2+3+2\sqrt { 6 } =5+2\sqrt { 6 } \ \sqrt { 6 } =2.44948974278........\ \ \sqrt { 6 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad ans\quad addition\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 2 } +\sqrt { 3 } ) }^{ 2 }\quad is\quad an\quad irrational\quad number.\ \quad $