Question

prove that
$\sqrt{5}$
is irrational.


​

Answer 1

Step-by-step explanation:

$Let's \: prove \: this \: by \: the \: method \: of \: contradiction- </p><p></p><p>Say, √5 \: is \: a \: rational \: number. ∴ It \: can \: be \: expressed \: in \: the \: form \: p/q \: where \: p,q \: are \: co-prime \: integers. \: </p><p></p><p>⇒√5=p/q</p><p></p><p> \: ⇒5=p²/q²  \: {Squaring \: both \: the \: sides} \: </p><p></p><p>⇒5q²=p²  \: (1) \: </p><p></p><p>⇒p² \: is \: a \: multiple \: of \: 5. \: {Euclid's Division Lemma} \: </p><p></p><p>⇒ \: p \: is \: also \: a \: multiple \: of \: 5. \: {Fundamental \: Theorem \: of \: arithmetic} \: </p><p></p><p> \: ⇒p=5m \: </p><p></p><p>⇒p²=25m²    \: (2) \: </p><p></p><p> \: From \: equations \: (1) \: and \: (2), \: we \: get, \: </p><p></p><p>5q²=25m²</p><p></p><p> \: ⇒q²=5m²</p><p></p><p> \: ⇒ \: q² \: is \: a \: multiple \: of \: 5. \: {Euclid's \: Division \: Lemma} \: </p><p></p><p>⇒ \: q \: is \: a \: multiple \: of \: 5. \: {Fundamental \: \: Theorem \: of \: Arithmetic} \: </p><p></p><p> \: Hence, \: p,q \: have \: a \: common \: factor \: 5. \: this \: contradicts \: that \: they \: are \: co-primes. \: Therefore, \: p/q \: is \: not \: a \: rational \: number. \: This \: proves \: that \: √5 \: is \: an \: irrational \: number. </p><p></p><p> \: For \: the \: second \: query, \: as \: we've \: proved \: √5 \: irrational. \: Therefore \: 2-√5 \: is \: also \: irrational \: because \: difference \: of \: a \: rational \: and \: an \: irrational \: number \: is \: always \: an \: irrational \: number.</p><p></p><p>$