Question

pove that 5+3 √7 is irrational​

Answer 1

5 + 3√7

We know that any rational number can be written as p/q form, where p & q are integers.

Let 5 + 3√7 as a rational number

$= > 5 + 3 \sqrt{7} = \frac{p}{q} \\ \\ = > 3 \sqrt{7} = \frac{ p}{q} - 5 \\ \\ = > 3 \sqrt{7} = \frac{p - 5q}{q} \\ \\ = > \sqrt{7} = \frac{p - 5q}{q} \div 3 \\ \\ = > \sqrt{7} = \frac{p - 5q}{q} \times \frac{1}{3} \\ \\ = > \sqrt{7} = \frac{p - 5q}{3q}$

We can see that p - 5q/3q is a rational number, but √7 is a irrational number.

So,

our consideration was wrong and

$= > 5 + 3 \sqrt{7} ≠ \frac{p}{q}$

5 + 3√7 is a irrational number.

hope it helps.

Answer 2

Step-by-step explanation:

Let us assume that

$5 +3 \sqrt{7}$

is rational

So,

$5 + 3 \sqrt{7 } = x \div y \:$

Where x and y are positive integers and are co-prime

$x \div y - 5 = 3 \sqrt{7}$

$(x - 5y) \div y = 3 \sqrt{7}$

$(x - 5y) \div 3y = \sqrt{7}$

Here,

$\sqrt{7 \ \: } is \: irrational \: but \: (x - 5y) \div 3y \: is \: rational$

But, Rational≠Irrational

So, our assumption is wrong,

5+3√7 is irrational