Question

Prove that 3+2✓11 is irrational

Answer 1

Answer:

The answer is

Step-by-step explanation:

3+2√11 is an irrational number

Let

3+2√11 is not a irrational number

therefore 3+2√11 is a rational number

3+2√11=a/b [ a,b are co - primes & + ve integers ]

2√11=a/b -3

2√11=a-3b/b

√11=a-3b/2b

but it is impossible [√11is an irrational number]

our assumption is wrong

therefore 3+2√11 is an irrational number

This is the answer

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Answer 2

$\underline\mathfrak{To \: \: Prove:-}$

• $\: \: \: \: \: {3} \: + \: {2}\sqrt{11} \: \: is, \: \: irrational$

$\huge\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: Let \: \: us \: \: assume \: \: that \: \: {3} \: + \: {2}\sqrt{11} \: \: is \: \: a \: \: irrational$

$\: \: \: \: \: So, \: \: the \: \: number \: \: is \: \: rational \: \: in \: \: the \: \: fraction \: \: of \: \: \frac{p}{q}$

• $\: \: \: \: \: where \: \: \: \: \: \: {(q \: = \: {0} \: \: \: \: and \: \: \: \: p, \: \: q \: \: are \: \: integer)}$

$\underline{Now,}$

$\: \: \: \: \: \leadsto {3} \: + \: {2}\sqrt{11} \: \: = \: \: \frac{p}{q}$

$\: \: \: \: \: \leadsto {2}\sqrt{11} \: \: = \: \: \frac{p}{q} \: - \: 3$

$\: \: \: \: \: \leadsto {2}\sqrt{11} \: \: = \: \: \frac{p \: - \: {3q}}{q}$

$\: \: \: \: \: \leadsto \sqrt{11} \: \: = \: \: \frac{p \: - \: {3q}}{2q}$

• $\: \: \: \: \: p \: \: and \: \: q \: \: both \: \: integer \: \: so \: \: = \: \: \frac{p \: - \: {3q}}{2q}\: \: is \: \: a \: \: rational \: \: number$

• $\: \: \: \: \: \sqrt{11} \: \: is \: \: a \: \: irrational \: \: number$

• $\: \: \: \: \: {3} \: + \: \sqrt{11} \: \: is \: \: a \: \: irrational \: \: number$

• $\: \: \: \: \: \underline{A \: \: rational \: \: number \: \: is \: \: never \: \: be \: \: equal \: \: to \: \: irrational \: \: number.}$

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