Question

prove that 7+3 root 5 is irrational number​

Answer 1

♪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.

$\rule{200}{8}$

Answer 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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