Question

show that root 6 is an irrational numbers​

Answer 1

Question:

Show that √6 is an irrational number.

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Step-by-step explanation:

$let \: \sqrt{6} \: be \: rational \: number.$

$we \: can \: write \: it \: in \: the \: form \: \\ of \: \frac{p}{q} ..where \: q ≠ 0 \\also \: p&q \: are \: co - prime \: numbers \\ \sqrt{6} = \frac{p}{q} \\ squaring \: on \: both \: sides. \\ ({6})^{2} = {( \frac{p}{q} })^{2} \\ 36 = \frac{ {p}^{2} }{ {q}^{2} } \\ {6q}^{2} = {p}^{2}$

Here p is divisible by 6.

>p=6k where k is +ve integer.

again in squaring both sides.

${(p)}^{2} = ( {6k}^{2}) \\ {p}^{2} = {36k}^{2}$

substituting 6q²=p²

$= > {6q}^{2} = {36k}^{2} \\ {q}^{2} = {6k}^{2} \\ q \: is \: divisible \: by \: 6.$

From this we can say that p&q have a common factor of 6.

> it shows that our assumption is wrong p&q are not co-prime.

Hence, √6 is an irrational no.

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