Math, asked by sukhwinderaulakh82, 9 months ago

prove that 7+3 root 5 is irrational number​

Answers

Answered by Aloi99
5

Question:-

Prove that 7+3√5 is irrational

\rule{200}{1}

Proof:-

→Let 7+3√5 be rational

→i.e, 7+3√5= \frac{a}{b} [where a&b are co-prime integers and a&b≠0]

★Shifting 7 to RHS★

→3√5= \frac{a}{b} -7

★Cross multiply RHS★

→3√5= \frac{a-7b}{b}

★Shift 3 to RHS★

→√5= \frac{a-7b}{3b}

•Since √5 is irrational and  \frac{a-7b}{3bB} is rational, it creates a contradiction. Hence, 7+3√5 is irrational.

\rule{200}{8}

Answered by Anonymous
2

 \mathtt{ \huge{ \fbox{Solution :)}}}

Let , 7 + 3√5 is an rational number

 \sf \mapsto 7 + 3 \sqrt{5}  =  \frac{a}{b} \\  \\\sf \mapsto 3 \sqrt{5}   =  \frac{a}{b} - 7 \\  \\\sf \mapsto 3 \sqrt{5}   =  \frac{a - 7b}{b}  \\  \\ \sf \mapsto  \sqrt{5}  =  \frac{a - 7b}{3b}

Here , √5 is an irrational number but (a - 7b)/3b is an rational number

Since , Irrational number ≠ rational number

Thus , our assumptions is wrong

Hence , 7 + 3√5 is an irrational number

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