$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)} \\$

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

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) } \\$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

$\\$

Hence, 3 + √5 is an irrational number.

7 0

3 years ago

16/___ = -8<br><br> I will mark brainlyst for the first person!

Paraphin [41]

$\frac{16}{ - 2} = - 8$

3 0

3 years ago

Read 2 more answers

Jacy has a collection of 140 coins worth $19.00. She has only nickels, dimes, and quarters. The difference in the

Colt1911 [192]

Answer:50(b)

Step-by-step explanation:

I just took the test

7 0

3 years ago