Question

Prove that 3 + 2√5 is irrational.​

Answer 1

⠀⠀ıllıllı uoᴉʇnloS ıllıllı

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Let us say 3 + 2√5 is rational.

Then the co-prime x and y of the given rational number where (y ≠ 0) is such that:

3 + 2√5 = x/y

Rearranging, we get,

2√5 = x/y - 3

√5 =1/2(x/y-3)

Since x and y are integers, thus, 1/2(x/y-3) is a rational number.

Therefore, √5 is also a rational number. But this confronts the fact that √5 is irrational.

• Hence, we get that 3 + 2√5 is irrational.

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Answer 2

Answer:

$let \: be \: assume \: that \: 3 + 2 \sqrt{5} \: is \: a \: rational \: number \\ so \: it \: can \: be \: written \: in \: the \: form \: of \: \frac{p}{q } \\ 3 + 2 \sqrt{5} = \frac{p}{q } \\ \sqrt{5} = \frac{p - 3q}{2q} \\ \frac{p - 3q}{2q} \: is \: a \: rational \: number \: therefore \\ \sqrt{5} \: is \: also \: a \: rational \: number \: but \: it \: is \: contradict \\ that \: \sqrt{5 \: } \: isan \: irrational \: number \: therefore \: our \: assumption \: is \: wrong \: and \\ 3 + 2 \sqrt{5} \: is \: an \: irrational \: number.$

hope it will help you