Question

Prove that 7 + 3√5 is an irrational number​

Answer 1

lat us assume that
$7 + 3 \sqrt{5}$
is a rational no
therefore
$7 + 3 \sqrt{5} = \frac{a}{b} \: \: where \: a \: and \: b \: are \: co \: primes$
so ,

$3 \sqrt{5} = \frac{a}{b} -7$
$3 \sqrt{5 } = \frac{a - 7b}{b}$
$\sqrt{5} = \frac{a -7b}{3b}$

but
$\sqrt{5} \: \: is \: irrational \: no \: \: \\ therefore \: an \: irrational \: no \: cannotbe \: equal \: to \: rational \: no$
$so \: \: 7 + 3 \sqrt{5} \: \: is \: \: irrational$
hope it helps

pls mark it as brainliest.....

Answer 2

Answer:

Assume that 7+3√5 is rational

7+3√5 = p/q where p and q are integers and q ≠ 0

3√5 = p/q - 7/1

3√5 = p-7q/3b, a rational number

√5 is rational which is also a contradiction to our assumption

So our assumption is wrong.