Tag: irrational numbers

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 $

$\ \sqrt { 3 } =1.73205080757......\ \sqrt { 3 } is\quad an\quad irrational\quad number,\quad since\quad it's\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad a\quad rational\quad and\quad irrational\quad number\quad is\quad irrational.\ Hence,\quad 2+\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\ \sqrt { 3 } =1.73205080757......\ Also\quad \sqrt { 2 } =1.41421356237........\ \sqrt { 3 } and\quad \sqrt { 2 } are\quad irrational\quad numbers,\quad since\quad their\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad two\quad irrational\quad numbers\quad is\quad irrational.\ Hence,\quad \sqrt { 2 } +\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\ \sqrt { 3 } =1.73205080757......\ Also\quad \sqrt { 5 } =2.2360679775........\ \sqrt { 3 } and\quad \sqrt { 5 } are\quad irrational\quad numbers,\quad since\quad their\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ And\quad addition\quad of\quad two\quad irrational\quad numbers\quad is\quad irrational.\ Hence,\quad \sqrt { 5 } +\sqrt { 3 } \quad is\quad an\quad irrational\quad number.\ \quad $

$\sqrt { 7 } =2.64575131106...\ The\quad decimal\quad representation\quad is\quad non\quad repeating\quad non\quad terminating.\ Hence,\quad \sqrt { 7 } is\quad an\quad irrational\quad number.\ \quad $

$\ { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )=4-2=2\ \  { 2 } is\quad a\quad rational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad terminating.\ Hence,\quad { (2-\sqrt { 2 } ) }(2+\sqrt { 2 } )\quad is\quad a\quad rational\quad number.\ \quad $