Question

prove that √5 is irrational
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Answer 1

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

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

Now, √5 = $\huge\frac{a}{b}$ , where a and b are coprimes and b ≠ 0.

→ √5b = a

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

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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