Question

Prove that 2 + 3√7 is irrational number.
Prove that 3 + 4√5 is irrational number.​


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Answer 1

Answer:

Heya Here's the required Answer mate

Given:

• 3 + 4√5

To prove:

• 3 + 4√5 is an irrational number.

Proof:

• Let us assume that 3 + 4√5 is a rational number.

• So, it can be written in the form a/b
• 3 + 4√5 = a/b

• Here a and b are coprime numbers and b ≠ 0

• Solving 3 + 4√5 = a/b we get,
• =>4√5 = a/b – 3
• =>4√5 = (a-3b)/b
• =>√5 = (a-3b)/4b

• This shows (a-3b)/4b is a rational number. But we know that √5 is an irrational number.

• So, it contradicts our assumption. Our assumption of 3 + 4√5 is a rational number is incorrect.

• 3 + 4√5 is an irrational number

Hope it helps bete

Answer 2

$\begin{gathered} \bf \huge \underbrace{Answer} \\ \\ \sf \: given : - \\ \\ \sf \: \sqrt{2} \: \: is \: \: \: an \: \: irrational \: \: \: number \\ \\ now \\ \\ let \: \: us \: \: assume \: \: that \: \: 7 - 3 \sqrt{2} \: \: rational \: number \: \: and \: \: also \: \: co \: \: \: prime \: \: number \: \: \: a \: \: and \: \: b \: \: (b ≠0) \\ \\ \sf \implies \:7 - 3 \sqrt{2} = \frac{a}{b} \\ \\ \sf \implies \: - 3 \sqrt{2} = \frac{a - 7b}{b} \\ \\ \sf \implies \: \sqrt{2} = \frac{a - 7b}{ - 3b} \\ \\ \: \frac{a - 7b}{ - 3b} \: \: is \: \: a \: \: rational \: \: number. \\ \\ but \: \: in \: \: given \: \: \sqrt{2} \: \: is \: \: a \: \: irrational \: \: number \\ \\ so \: \: our \: \: assumption \: \: \: is \: \: wrong \: \: \\ \\ therefore \\ \\ 7 - 3 \sqrt{2} \: \: is \: \: a \: \: irrational \: \: number \\ \\ \bf \huge \blue{hence \: \: proved}\end{gathered}$