Question

prove that √5 is irrational​

Answer 1

${\huge{\bold{\underline{\sf{\mathfrak{AnSweR:-}}}}}}$

To prove :-

$\large\sf\green{\sqrt{5} \: is \: irrational}$

Proof:-

Let us assume that $\large\sf\sqrt{5}$ is rational

So,

$\large\tt\sqrt{5} = \frac{a}{b}$ ( a and b are co-prime number , b≠0 )

Squaring both sides

$\large\implies\sf\sqrt{5}^{2}= \frac{a^2}{b^2}$

$\large\implies\sf\ 5b^2 = a^2 \\ \\ \large\implies\sf\ a^2 =5b^2$

Therefore , $\large\bf\ a^2$ divides 5 so “a” also divides 5

Now,

$\large\sf\implies\ a = 5c$

Squaring both sides

$\large\sf\implies\ (a)^2 = (5c)^2 \\ \\ \large\sf\implies\ a^2 = 25c^2$

Therefore, $\large\bf\ a^2$ divides 25 and a also divides 25

•°• Here, a has more than one factor so it is not co-prime.

So,

$\large\sf\implies\ 5b^2 = 25c^2 \\ \\ \large\sf\implies\ b^2 = 5c^2$

•°•This contradicts the fact that b is co- prime

Hence , Our assumption is Wrong

$\large\sf\therefore\pink{\sqrt{5} \: is \: irrational}$

Answer 2

$\bold{\underline{\underline{\huge{\sf{ANSWER\::}}}}}}$

Given:

√5 is irrational number.

To find:

Prove that.

Explanation:

Let if possible √5 is not irrational, so √5 is a rational number.

∴ by definition of rational number, we can write as;

→ $\sqrt{5} =\frac{p}{q}$

where p and q are integers such that p & q have no common factor except 1.

• squaring both the sides, we get;

⇒ $(\sqrt{5} )^{2} =\:\frac{p^{2} }{q^{2} }$

⇒ $5=\frac{p^{2} }{q^{2} }$

⇒ 5q² = p²                

⇒ $\frac{5}{p^{2} }$             [∵2/2q² & 2q² =p²]

⇒ $\frac{5}{p}$.......................(1)

⇒ p= 5r for some integer r

Therefore,

⇒ p² = 25r²

⇒ 5q² = 25r²

⇒ q² = 5r²

⇒ 5/q²

⇒ 5/q...............................(2)

From equation (1) & (2) we conclude that 5 is a common factor of p and q which is a contradiction, because we have assumed that p & q have no common factor except 1.

∴ Our assumption is wrong.

Hence,

√5 is an irrational number.