Math, asked by aanchalbang06, 7 months ago

Prove that √7-√5 is an irrational number

Answers

Answered by gnanasekarsathya546

Answer:

√7-√5=0.409683334..

Step-by-step explanation:

√7=2.64575131..

√5=2.23606798...

Therefore it is an irrational number

Answered by XxMissPaglixX

${\huge{\mathtt{\pink{AnSwEr:-}}}}$

$<body bgcolor="black"><font color="green">$

$let \: ( \sqrt{7} - \sqrt{5} ) \: be \: a \: rational \: no. \\ so( \sqrt{7} - \sqrt{5} ) = \frac{p}{q} \\ {( \sqrt{7 } - \sqrt{5}) }^{2} = { (\frac{p}{q}) }^{2} \\ { (\sqrt{7 } \: ) }^{2} + { (\sqrt{5} \: )}^{2} - 2 \sqrt{7} . \sqrt{5} = \frac{ {p}^{2} }{ {q}^{2} } \\ 7 + 5 - 2 \sqrt{35} = \frac{ {p}^{2} }{ {q}^{2} } \\ 12 - \frac{ {p}^{2} }{ {q}^{2} } = 2 \sqrt{35} \\ \sqrt{35} = \frac{12 {q}^{2} - {p}^{2} }{2 {q}^{2} }$

Now,

$\sqrt{35} \: is\: a \: irrational \: no \\ but \: \frac{12 {q}^{2} - {p}^{2} }{2 {q}^{2} } \: is \: a \: rational \: no.$

So,

$\sqrt{35} \: not \: = \frac{12 {q}^{2} - {p}^{2} }{2 {q}^{2} } \\ which \: contradicts \: our \: supposition \\t hat \:( \sqrt{7} - \sqrt{5} ) \:is \: rational$

$hence \: it \: is \: irrational$

Hence Proved

${\huge{\mathtt{\pink{Thank\:You}}}}$

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Prove that √7-√5 is an irrational number​

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Now,

So,

Hence Proved

Prove that √7-√5 is an irrational number