Question

prove that √7 is an irrationsl​

Answer 1

Answer:

Here in the above figure is your answer

Answer 2

Answer:

Let's prove √7 an irrational number

Steps for explanation:-

Let √7 be a rational number,

$\sqrt{7} = \frac{p}{q} \: \: where \: \: p \: \: and \: \: q \: \: are \: \: integers \: \: q \: \: no \: equals \: \: to \: \: 0 \: \: and \: \: p \: \: and \: \: q \: \: have \: \: no \: \: common factors\: \: (except \: \: 1)$

Squaring both the sides:-

$7 = \frac{ {p}^{2} }{ {q}^{2} }$

${p}^{2} = 7 {q}^{2} \: \: (1)$

$As \: \: 7 \: \: divides \: \: 7 q^2 \: \: so \: \: 7 \: \: divides \: \: p^2 \: \: but \: \: 7 \: \: is \: \: prime$

• 7 divides p

Let p=7k ,where k is an integer

Substituting the p in (1)

${(7k)}^{2} = 7 {q}^{2}$

$49 {k}^{2} = 7 {q}^{2}$

${q}^{2} = 7 {k}^{2}$

$As \: \: 7 \: \: divides \: \: 7 k^2 \: \: so \: \: 7 \: \: divides \: \: q^2 \: \: but \: \: 7 \: \: is \: \: prime$

• 7 divides q

Thus p and q have a common factor 7.

But,This contradicts the fact that p and q have no common factor(except 1)

Hence,

√7 is not a rational number.It is an irrational number.