Question

So that (2+√5) are irrational number

Answer 1

$\red{\underline{\underline{\tt{Question:}}}}$

Show that (2+√5) is an irrational number.

$\blue{\underline{\underline{\tt{Answer:}}}}$

Let us assume that (2+√5) is a rational number.

Now,

$\bf 2+\sqrt{5}=\frac{p}{q}$

∵ In which (p), (q) are integer and q ≠ 0. HCF(p,q) = 1

$\bf \rightarrow \sqrt{5}=\frac{p}{q}-2$

∵ Taking 2 to RHS.

$\bf \rightarrow \sqrt{5} =\frac{p-2q}{q}$

Here,

(p), (q), (-2q) are integers and q ≠ 0. So,

$\bf \boxed{\frac{p-2q}{q} } \:\:\underline{is\: a\: rational\: number}$

But $\red{\boxed{\sqrt{5}}}$ is not a rational number.

$\bf \sqrt{5} \neq \frac{p-2q}{q}$

→ Hence Proved,

(2+√5) is an irrational number.

$\rule{200}2$

Here,

$\star \red{\underline{\underline{\tt{Proving\:\:\sqrt{5}\:\: an \:\:irrational\:\:number-}}}}$

Let us assume that $\red{\sf{\sqrt{5}}}$ is a rational number.

So,

$\bf \sqrt{5}=\frac{p}{q}$

∵ Where (p) and (q) are integers and q ≠ 0. HCF(p,q) = 1

$\rightarrow \bf (\sqrt{5})^{2} = (\frac{p}{q})^{2}\:\: \underline{[Sqaring \:\:both \:\:sides]}$

$\rightarrow \bf 5 = \frac{(p)^{2}}{(q)^{2}}$

$\rightarrow \bf 5q^{2}= p^{2}------(i)$

Here,

$\blue{\sf{5\:\: is\:\: a \:\:factor \:\:of\:\: p^{2}}}$

$\blue{\sf{5\:\: is\:\: also\:\: a \:\:factor \:\:of\:\: p}}$

Now,

Let p = 5k

→ (p)² = (5k)²

→ (p)² = 25k² $\bf ------(ii)$

Now, comparing eq.(i) and (ii), we get

5q² = 25k²

→ q² = $\bf \frac{25k^{2}}{5}$

→ q² = 5k²

Here,

5 is a factor of q²

5 is also a factor of q

From the above discussion we found that 5 is a factor (common factor) of (p) and (q) i.e HCF(p,q) = 5.

But the HCF(p,q) should be 1 ans we assume √5 as rational number.

So,

This arise due to our wrong assumption.

Hence

√5 is an irrational number

Proved.

Answer 2

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$\huge\tt{TO~PROOF:}$

• (2+√5) is a irrational number

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$\huge\tt{SOLUTION:}$

If possible, let us assume 2+ √5 is a rational number.

⇒2+ √5 = q/p (where p,q ∈ z,q ≠ 0)

⇒2− p/q = √5

⇒2p−p/q = -√5

⇒ 5 is a rational number

∵ 2q−p/q is a rational number

But, -√5 is not a rational number

∴ Our assumption 2+ √5 is a rational number is wrong.

Then,

⇒2+ √5 is an irrational number.

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