Math, asked by itzsimransingh37, 9 months ago

prove that root 5 is irrational​

Answers

Answered by SugaryGenius
6

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Answered by XxRadhikaxX
9

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\sf{On\:Our\:contratry,\:let\:us\:assume\:that\:\sqrt{5}\:is\:a}

\sf{rational\:number.}

\therefore\:\:\sf{\sqrt{5}\:=\:\frac{p}{q}\:(q\:not\:equals\:to\:0)}\\

\sf{If\:p,\:q\:have\:a\:common\:factor,\:on\:cancelling}\\

\sf{the\:common\:factor\:let\:it\:be\:reduces}\\

\sf{to\:\frac{a}{b},\:where\:a,\:b\:are\:co-primes.}\\

\sf{Now\:\sqrt{5}\:=\:\frac{a}{b},\:where\:HCF\:(a,\:b)\:=\:1}\\

\sf{Squaring \:on\:both\:sides\:we\:get}\\

\sf{(\sqrt{5})^2 \: = \: (\:\frac{a}{b}\:)^2}\\

\sf{5\:=\:\frac{a^2}{b^2}\\

\longrightarrow\sf{5b^2 \: = \: a^2}\\

\longrightarrow\sf{5\:divides\:a^2\:and\:thereby\:5\:divides\:5}\\

\sf{\:\:\:Now,\:take\:a\:=\:5c}\\

\sf{\:\:\:then,\:a^2\:=\:25c^2}\\

\sf{I.e.,\:\:5b^2\:=\:25c^2}\\

\longrightarrow\sf{b^2\:=\:5c^2}\\

\longrightarrow\sf{5\:divides\:b^2\:and\:thereby\:b.}\\

\longrightarrow\sf{5\:divides\:both\:b\:and\:a.}\\

\sf{This\:contradicts\:that\:a\:and\:b\:are}

\sf{co-primes}\\

\sf{This\:contradiction \:arised\:due\:to\:our}

\sf{assumption\:that\:\sqrt{5}\:is\:a\:rational\:number.}\\

\sf{Hence\:our\:assumption\:is\:wrong}\\

\sf{i.e.,\:\sqrt{5}\:is\:an\:irrational \:number}\\\\

✍️ Ⓣhank Ⓨou ‼️

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