Question

prove that √5 is irrational number​

Answer 1

★$Question:-$

prove that $\sqrt{5}$ is irrational number

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★$Answer:-$

Let us assume, to the contrary, that $\sqrt{5}$ is a rational number and its simplest form is $\frac{a}{b}$, where a and b are integers having no common factor other than 1 and b≠0.

Now,

$\huge\sqrt{5}$ = $\huge\frac{a}{b}$

$\huge5$=$\huge\frac{a²}{b²}$

${5b²}$= ${a²}$ ___(1)

⇒{a² is divisible by 5

⇒a is divisible by 5

Let a=5c for some integer 'c'

On substituting a=5c in equation (1) we get,

$5b²$= $(5c²)$

$5b²$=$25c²$

$b²$= $5c²$

⇒b² is divisible by 5

⇒b is divisible by 5

Since a and b are both divisible by 5, 5 is common factor of a and b.

But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is a rational number.

Hence, $\sqrt{5}$ is irrational.

Answer 2

Concept :

• We have to prove √5 is a irrational number.

• An irrational number is a number that cannot be expressed as a fraction for any integers .

• Irrational numbers have decimal expansions that neither terminate nor become periodic.

Let's start :

$\underline{ \boxed{ \sf{ \purple{ \sqrt{5} = \frac{a}{b} }}}}$

$\underline{ \boxed{ \sf{ \purple{ \sqrt{5b} = a}}}}$

Squaring both sides.

$\underline{ \boxed{ \sf{ \purple{( \sqrt{5 {b}^{2} }) = {a}^{2} }}}}$

$\: \: \: \: \: \: \: \: \: \sf \rightarrow5 {b}^{2} = {a}^{2}$

$\underline{ \boxed{ \sf{ \purple{ \frac{ {a}^{2} }{5} = {b}^{2} }}}}$

Hence, 5 divides a²

So, 5 shall divided a also ...( 1 )

∴ we can say.

$\underline{ \boxed{ \sf{ \purple{ \frac{a}{5} = c \: where \: c \: is \: some \: integer}}}}$

So, a = 5c

So we know that

$\: \: \: \: \: \: \: \: \: \: \: \sf \rightarrow5 {b}^{2} = {a}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \sf \: \rightarrow \: putting \: a \: = 5c$

$\: \: \: \: \: \: \: \: \: \: \: \sf \implies5 {b}^{2} = (5c) {}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \rightarrow \sf5 {b}^{2} = 25 {c}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \rightarrow \sf {b}^{2} = 5 {c}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \sf{ \purple{ \frac{ {b}^{2} }{5} = {c}^{2} }}}}$

Hence 5 divides b²

So, 5 divides By ( 1 ) and ( 2 )

∴ By contradiction

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \sf{ \purple{ \sqrt{5} \: is \: irrational \: }}}}$

Other information :

• irrational number is a real number and not a complex number, because it is possible to represent these numbers in the number line.

• Irrational numbers are a set of real numbers that cannot be expressed in the form of fractions or ratios.

• Ex: π, √2, e, √5