Question

Prove that |~5 is irrational number.

Answer 1

$QUESTION$

Prove that √5 is an irrational number.

$ANSWER$

Assume √5 as a rational number.

Therefore,

there exists co-prime integer a and b which is ≠0,such that

$\\ \sqrt{5} = \frac{a}{b} \\ \\ \sqrt{5} b = a \\ \\ squaring \: both \: sides \\ \\ 5 {b}^{2} = {a}^{2}....1 \\ \\ therefore \\ 5 \: divides \: a^{2} .....2\\ \\ let \: a = 5c \: where \: c \: is \: some \: integer \: \\ \\ substituting \: this \: value \: of \: a \: in \: eq.1 \\ \\ 5 {b}^{2} = 5{c}^{2} \\ \\ 5 {b}^{2} = 25 {c}^{2} \\ \\ {b}^{2} = \frac{25 {c}^{2} }{5} \\ \\ {b}^{2} = 5 {c}^{2} \\ \\ now \: 5 \: divides \: { b}^{2} .........3 \\ \\ from(2) \: and \: (3) \: we \: get \\ \\ a \: and \: b \: both \: have \: common \: factor. \: \\ \\ This \: contradicts \: the \: fact \: that \: a \: and \: b \: are \: co - prime. \\ \\ Therefore \: the \: assumption \: that \: \sqrt{5} is \: a \: rational \: number \: is \: wrong. \\ \\ Therefore \: \sqrt{5} is \: an \: irrational \: number. \\ \\ Hence \:proved.$

Answer 2

$\huge\mathfrak{A{\underline{\underline{\textbf{\red{nswer :}}}}}}$

Let.. √5 is rational number

√5 = $\dfrac{a}{b}$

[Here a and b are co-prime numbers]

b√5 = a

Squaring on both sides we get;

5b² = a² ....(1)

b² = $\dfrac{{a}{^2}}{5}$

Here 5 divide a² and 5 divide a also.

Now....

a = 5c

[Here c is integer]

Squaring on both sides we get;

a² = 25c²

5b² = 25c² [From (1)]

b² = 5c²

c² = $\dfrac{{b}^{2}}{5}$

Here 5 divide b² and 5 divide b also.

Both a and b are co-prime numbers. And 5 divides both of them.

So, our assumption is wrong.

√5 is irrational number.

Hence proof.

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