Question

Prove that √11 is irrational

Answer 1

listen put √11in place of √5

Answer 2

$\huge{\underline{\bf{\red{Solution:-}}}}$

Let √11 be rational number

So,

√11= a/b

where a and b are intigers and b ≠ 0.

Then,

√11= a/b

squaring on both sides

$: \implies \sf11=\frac{a^2}{b^2}$

$: \implies \sf11b^2=a^2..........(i)$

$: \implies \sf11\:\:divides \:\:a^2[\therefore\:11\:divides\:11b^2]$

$: \implies \sf11\:divides \:a$

So,

11 is a prime and divides b²

Then, 11 also divides b.

Let a = 11c for some intiger c.

putting a = 11c in (i)

$: \implies \sf11b^2=121c^2$

$: \implies \sf\:b^2=11c^2$

$: \implies \sf11\: divides\:b^2$so,11 also divides 11c².

$: \implies \sf11\: divides\:b$

So,

11 is a prime and 11 divides b² and b also.

Then, 11 is a common factor of a and b.

but, this contradicts the fact that a and b have no common factor other then 1.

So, this contradiction is arissen because we assume that √11 is rational .

Hence,

√11 is irrational.

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