prove that
[tex] \sqrt{5} [/tex]
is an irrational number.​

2 thoughts on “prove that <br />[tex] \sqrt{5} [/tex]<br />is an irrational number.​”

1. By using the process of contradiction,

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

   [tex] \longrightarrow \sf { \sqrt{5} = \dfrac{p}{q} } [/tex]

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

   Now, squaring both sides.

   [tex] \longrightarrow \sf { {(\sqrt{5})}^2 = { \bigg ( \dfrac{p}{q} \bigg ) }^{2} } [/tex]

   [tex] \longrightarrow \sf { 5 = \dfrac{ {p}^{2} }{ {q}^{2} } } [/tex]

   [tex] \longrightarrow \sf { 5 \times {q}^{2}= {p}^{2} } [/tex]

   [tex] \longrightarrow \sf { 5{q}^{2}= {p}^{2} } [/tex]. . . . . ( 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 = [tex] \sf {\dfrac{25}{5} } [/tex]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!

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2. Step-by-step explanation:

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