Math, asked by MansiWriterBlyton, 10 months ago

Given that √5 is irrational, prove that 2√5-3 is an irrational number​

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Answered by harshagrawal1029
6

may it help u

change the value of question

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Answered by Anonymous
2

 \maltese \: \: {\underline{\bold{Question :}}}

Given that √5 is irrational , prove that 2 √5 – 3 is an irrational number.

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 \maltese \: \: {\underline{\bold{Solution :}}}

We have to prove that 2√5 - 3 is an irrational number.

Let us assume the contrary the 2√5 - 3 is a rational number.

So we can represent 2√5 - 3 in the form of \frac{p}{q} where p and q are co-prime integers and q ≠ 0.

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2√5 - 3 = \frac{p}{q}

⇒ 2√5 = 3 + \frac{p}{q}

⇒ 2√5 = \frac{3q \: + \:  p}{q}

⇒ √5 = \frac{1}{2} \times \frac{3q \: + \:  p}{q}

⇒ √5 =\frac{3q \: + \: q}{2q}

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Here 3 , 2 , p and q are integers so \frac{3q \: + \: q}{2q} is a rational number.

But we are given that √5 is irrational number so our assumption that 2√5 - 3 is an rational number was wrong.

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So we conclude that 2√5 - 3 is an irrational number.

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