Question

prove that root 5 is irrational​

Answer 1

Answer:

$let \sqrt{5} is \: not \: a \: irrational \: \: no. \\ \\ \sqrt{5} \: is \: a \: rational \: no. \\ \\ \: now \: let \: \sqrt{5} = x \: a \: rational \: no. \\ \\ {x}^{2} = ( \sqrt{5} ) {}^{2} \\ \\ {x}^{2} = \sqrt{25} \\ \\ {x}^{2} = 5 \\ \\ x = \sqrt{5} \\ \\ here. \\ \\ our \: consumptions \: is \: wrong \: \\ \\ therefore \: \\ \\ \: \sqrt{5} \: is \: a \: irrational \: no.$

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

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$\bf \huge \: Question \: \: \:$

prove that root 5 is irrational !

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$\bf \huge \: solution \: \: \:$

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Let's prove this by the method of contradiction-

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Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

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$\bf \: \sqrt{5} = \frac{p}{q} \\ \\ \bf</p><p></p><p>5= \frac{ {p}^{2} }{ {q}^{2} }$

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{Squaring both the sides}

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$\bf \: 5q^2=p^2\: (1)$

$\bf \: p^2 is\: a \: multiple \: of\: 5.$

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{Euclid's Division Lemma}

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p is also a multiple of 5.

{Fundamental Theorm of arithmetic}

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$\bf \: p=5m</p><p>$

$\bf \:p^2=25m^\: (2)$

From equations (1) and (2),

we get,

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$\bf \: 5q^2=25m^2$

$\bf \: q^2=5m^2$

$\bf \: q^2 \: is\: a \: multiple\: of \: 5.$

{Euclid's Division Lemma}

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q is a multiple of 5.

{Fundamental Theorm of Arithmetic}

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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.

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