Question

prove that 8-2 root 3 is irrational number​

Answer 1

♦Question:-

Prove that, 8-2√3 is irrational no.

$\rule{200}{1}$

★Proof:-

→Let 8-2√3 be rational.

→i.e,8-2√3=$\frac{a}{b}$ [where a & b are co-prime integers and a&b≠0]

•Shift 8 from LHS to RHS•

→2√3=$\frac{a}{b}$+8

•Cross multiply RHS•

→2√3=$\frac{a+8b}{b}$

•Shift 2 from LHS to RHS•

→√3=$\frac{a+8b}{2b}$

★This creates a contradiction as √3 is irrational and $\frac{a+8b}{2b}$ is rational★

→8-2√3 is irrational.

$\rule{200}{8}$

Answer 2

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

Let , 8 - 2√3 is an rational number

$\sf \hookrightarrow 8 - 2 \sqrt{3} = \frac{a}{n} \\ \\ \sf \hookrightarrow </p><p>-2 \sqrt{3} = \frac{a}{b} - 8 \\ \\ \sf \hookrightarrow</p><p>-2 \sqrt{3} = \frac{a - 8b}{b} \\ \\ \sf \hookrightarrow</p><p> \sqrt{3} =\frac{ 8b - a}{b}$

Here , √3 is an irrational number but (8b - a)/8 is an rational number

Since , irrational ≠ rational

Thus , our assumptions is wrong

Hence , 8 - 2√3 is an irrational number

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