2. Prove that 3 + 2√5 is irrational.

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2 thoughts on “2. Prove that 3 + 2√5 is irrational. <br /><br />❌❌spammers keep away❌❌<br /><br />❣spam = 20 answers reported.❣<br />​”

1. To Prove:

   3+2√5 is irrational.

   → let take that 3+2√5 is rational number

   → so, we can write this answer as

   ⇒3+2√5 = a/b

   Here a & b use two co-prime number and b ≠ 0.

   [tex]⇒ \tt2 \sqrt{5} = \frac{a}{b} – 3[/tex]

   [tex]⇒ \tt2 \sqrt{5} = \frac{a – 3b}{b} [/tex]

   [tex] \therefore\tt \sqrt{5} = \frac{a – 3b}{2b} [/tex]

   [tex] \small \sf\: Here \: \tt \: a \sf \: and \: \tt \: b \: \sf \: are \: integer \: so \: \tt \: \frac{a – 3b}{2b} \sf \: is \: a \: rational \: number \: but \: \tt \sqrt{5} \sfis \: a \:irrational \: number \: so \: it \: is \: contradict[/tex]

   [tex] \sf – Hence \tt \: 3 + 2 \sqrt{5} \sf\: is \: irrational[/tex]

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2. Answer:

   Step-by-step explanation:

   Let us assume tat 3+2[tex]\sqrt{5}[/tex] is a rational number of the form p/q, where p & q are integers & q not equals to zero and HCF (p ,q) = 1

   Now,

   3+2[tex]\sqrt{5}[/tex] = p/q

   = 2[tex]\sqrt{5}[/tex] = p/q – 3

   =[tex]\sqrt{5}[/tex] = (p/q – 3)

   We know that, p/q = Rational

   =p/q – 3 = Rational [Rational – Rational = Rational]

   It follows [tex]\sqrt{5}[/tex] is Rational

   And, It contradicts the fact that [tex]\sqrt{5}[/tex] is irrational

   Hence, 3+2[tex]\sqrt{5}[/tex] is an irrational number.

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