Question

Prove that √5 is irrational.​

Answer 1

$\orange{\small\underline\bold{ solution }}$

$\pink{\hookrightarrow \: prove \: \sqrt{5} \: is \: irrational \: }$

$\green{ \mapsto \: if \: we \: have \: to \: prove \: that \: \sqrt{5} \: is \: irrational \: } \\ \blue{\leadsto \: we \: have \: to \: suppose \: \sqrt{5} \: is \: a \: rational \: number \: }$

$\color{red}and \: as \: we \: know \: - \\ \color{purple}they \: must \: be \: in \: the \: form \: of \: \frac{p}{q} \: where \: q \: should \: not \: be \: equal \: to \: 0 \: (zero)$

$\color{purple} \hookrightarrow \: and \: p \: and \: q \: should \: be \: co - prime \: numbers$

$\color{plum}\sqrt{5} = \frac{p}{q} \\ \color{pink}\sqrt{5} \times q = p$

Squaring both sides -

$\color{blue} {5 \: q}^{2} = {p}^{2} \: \: \: \: - eq \: \fbox{ \: 1}$

$\pink{{p}^{2} \: is \: divisible \: by \: 5}$

$\pink{so \: (p) \: \: is \:divisible \: by \: 5}$

$\color{blue}p = 5c$

Squaring on both sides -

$\color{blue}{p}^{2} ={25 \: c}^{2} \: \: \: - eq\:\fbox{2}$

$\green{ \dagger \: putting \: {p}^{2} \: in \: equation \: \fbox{1}}$

$\color{blue}{5 \: q}^{2} = {25(c)}^{2}$

$\color{blue} {q}^{2} = {5 \: c}^{2}$

$\pink{so \: q \: is \: divisible \: is \: 5}$

Thus P and Q have common factor of 5

$\color{plum} \circ \: so \: there \: is \: contradiction \:$

$\color{red}\longmapsto \: we \: have \: assumed \: p \: and \: q \: are \: co - prime \: but \: here \: they \: are \: common \: factor \: of \: 5$

$\blue{ \dagger \: the \: above \: statement \: contradicts \: our \: statement \: }$

$therefore \:$

$\color{purple} \leadsto \: it \: is \: proved \: \sqrt{5} \: is \: an \: \: irrational \:number \:$

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