Question

Prove that Under root 5 is irratitional​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Steps to prove:

Let √5 be rational number

√5=p/q where p and q are integers q!=0 and p and q have no common factors (except 1)

Squaring both the sides

$5 = \frac{ {p}^{2} }{ {q}^{2} }$

$5 {q}^{2} = {p}^{2}$

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

$As \: \: 5 \: \: divides \: \: 5 {q}^{2} \: so \: \: 5 \: \: divides \: \: {p}^{2} \: \: but \: \: 5 \: \: is \: \: prime$

5 divides p

Let p=5k where k is an integer.

Substituting the value of p in (1)

$( {5k})^{2} = 5 {q}^{2}$

$25 {k}^{2} = 5 {q}^{2}$

${q}^{2} = 5 {k}^{2}$

$As \: \: 5 \: \: divides \: \: 5 {k}^{2} \: so \: \: 5 \: \: divides \: \: {q}^{2} \: \: but \: \: 5 \: is \: prime$

5 divides q

Thus p and q have a common factor 5.This contradicts the fact that p and q have no common factor (except 1)

Hence,√5 is a not a rational number.It is an irrational number.