Question

prove that
$\sqrt{5}$
is an irrational number.​

Answer 1

By using the process of contradiction,

Let us assume that √5 is a rational number. Then,

$\longrightarrow \sf { \sqrt{5} = \dfrac{p}{q} }$

Here, p and q are integers that haven't common factor and q ≠ 0.

Now, squaring both sides.

$\longrightarrow \sf { {(\sqrt{5})}^2 = { \bigg ( \dfrac{p}{q} \bigg ) }^{2} }$

$\longrightarrow \sf { 5 = \dfrac{ {p}^{2} }{ {q}^{2} } }$

$\longrightarrow \sf { 5 \times {q}^{2}= {p}^{2} }$

$\longrightarrow \sf { 5{q}^{2}= {p}^{2} }$. . . . . ( equation 1 )

→ 5 is a factor of p². So,

→ 5 is a factor of p.

Now, let p = 5m for some natural number m. So,

→ p = 5m

→ p² = (5m)²⠀⠀⠀⠀(Squaring both sides)

→ p² = 25m²

Substituting the value of p² from the equation (i).

→ 5q² = 25m²

→ q² = $\sf {\dfrac{25}{5} }$m²

→ q² = 5m²

We get that 5 is a factor of q². So,

→ 5 is a factor of q.

5 is a factor of both p and q. This contradicts the assumption that p and q have no common factor. We get that our assumption is wrong. So, √5 can't be a rational number. Hence, √5 is an irrational number.

Hence, proved!

Answer 2

Step-by-step explanation:

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