Question

prove that
$3 + 2 \sqrt{7 =}$
us an irrational

​

Answer 1

Answer:

16807

Hope it helps........

Answer 2

$\rm\huge\blue{\underline{\underline{ Question : }}}$

Prove that 3 + 2√7 is an irrational number.

$\rm\huge\blue{\underline{\underline{ Solution : }}}$

Let us assume that 3 + 2√7 is a rational number.

$\sf\:\implies 3 + 2\sqrt{7} = \frac{a}{b}$

• [ a & b are co - primes. ]

$\sf\:\implies 2\sqrt{7} = \frac{a}{b} - 3$

$\sf\:\implies 2\sqrt{7} = \frac{a - 3b}{b}$

$\sf\:\implies \sqrt{7} = \frac{a - 3b}{2b}$

↪ Now, (a - 3b)/2b is a rational number. And √7 is an irrational number.

↪ So our assumption is wrong.

↪ This assumption has arisen because we assumed that 3 + 2√7 is a rational number.

$\underline{\boxed{\bf{\purple{ \therefore 3 + 2\sqrt{7}\: is \:an\: irrational \:number.}}}}\:\orange{\bigstar}$