Question

$\huge\underline\mathfrak\pink{Question}$

:- Prove that √5 is an irrational number.
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Answer 1

AnswEr:

Let us Consider that $\sf\sqrt{5}$ is an rational number.

So, it can be written in the form of $\sf\dfrac{p}{q}.$ Where q is not equal to zero, p & q are co - primes number.

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Such That,

$:\implies\sf\; \sqrt{5} \; =\; \dfrac{p}{q}$

$:\implies\sf\; \sqrt{5}\; q\; = \; p$

$\dag\bold{\underline{\sf{\red{Squaring\; Both\; Sides}}}}$

$:\implies\sf\; \sqrt{5} q^2 \; = \; p^2$

$:\implies\sf\; 5q^2\; = \; p^2$

$\bold{\underline{\sf{Here,\; p^2\; is \; divisible\; by\; 5}}}$

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Let us consider that p = 5k where k is an integer.

$\dag\bold{\underline{\sf{Again\; squaring\; both\; Sides}}}$

$:\implies\sf\;p^2 = 25k^2$

Substituting 5q² = p²

$:\implies\sf\;5q^2\; =\; 25k^2$

$:\implies\sf\;q^2\; =\; 5k^2$

Here, we can see that q² is also divisible by 5.

Now we can say that the both p & q have atleast 5 as it's common factor.

So, it arises contradiction because of our wrong assumption.

$\bold{\underline{\sf{\red{Hence,\; \sqrt{5} \; is \; an \; irrational\; number.}}}}$

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Answer 2

$\huge\underline\mathfrak\pink{AnswEr:-}$

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Let us assume the contrary, that √5 is rational no.

Now, √5 = $</strong><strong>\</strong><strong>h</strong><strong>u</strong><strong>g</strong><strong>e</strong><strong>\frac{a}{b}$, where a and b are coprimes and b ≠ 0.

→ √5b = a

$\huge\underline\mathfrak\blue{Squaring \ on \ both \ sides, \ we \ get:-}$

a² = 5b² ....(1)

→ a² is divisible by 5.

→ a is also divisible by 5. ....(2)

.°. a = 5m, m is a natural number.

Putting the value of a in equation (1), we have

(5m)² = 5b²

→ 25m² = 5b²

→ b² = 5m²

→ b² is divisible by 5.

→ b is also divisible by 5. ....(3)

From eq. (2) and (3), we find that a and b have 5 as a common factor. This contradiction the fact a and b are coprime. Thus √5 is not a rational number.

Hence, √5 is an irrational number.

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