Question

prove that √5 is irational number​

Answer 1

"√5 is an “irrational number”.

Given:

√5

To prove:

√5 is a rational number

Solution:

Let us consider that √5 is a “rational number”.

We were told that the rational numbers will be in the “form” of \frac {p}{q}

q

p

form Where “p, q” are integers.

So, \sqrt { 5 } = \frac {p}{q}

5

=

q

p

p = \sqrt { 5 } \times qp=

5

×q

we know that 'p' is a “rational number”. So 5 \times q should be normal as it is equal to p

But it did not happens with √5 because it is “not an integer”

Therefore, p ≠ √5q

This denies that √5 is an “irrational number”

So, our consideration is false and √5 is an “irrational number”."

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Answer 2

Let us assume that √5 is rational.

$\therefore \: \sqrt{5} = \frac{p}{q} \: \: (where \: p \: and \: q \: are \: coprimes) \\ = > {( \sqrt{5} )}^{2} = {( \frac{p}{q} )}^{2} \\ = > 5 = \frac{ {p}^{2} }{ {q}^{2} } \\ = > 5 {q}^{2} = {p}^{2} \\ = > 5 \: is \: a \: factor \: of \: {p}^{2} \\ = > 5 \: is \: the \: factor \: of \: p \\ \\ Let \:5c = p, where\:c \:is\: some\: integer.\\ 5 {q}^{2} = {(5c)}^{2} \\ = > 5 {q}^{2} = 25 {c}^{2} \\ = > {q}^{2} = \frac{25 {c}^{2} }{5} \\ = > {q}^{2} = 5 {p}^{2} \\ = > 5 \: is \: the \: factor \: of \: {q}^{2} \\ = > 5 \: is \: the \: factor \: of \: q$

.°. 5 is the factor of both p and q.

But, this contradicts the fact that p and q are co-primes.

So, our assumption is wrong.

Hence, √5 is irrational.