Question

prove that 5-3/7√3 is a irrational number ​

Answer 1

Answer:

Answer: let 5-2/7√3 be rational no. here 7,5,2,q,and p are integers. therefore 5-2/7√3 is irrational no.

Answer 2

Answer:

plz mark as brainliest

Step-by-step explanation:

Let it be a rational number

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b}

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} = > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} = > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5 = > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b}

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} = > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5 = > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b} \sqrt{3} = \frac{7(5b - a)}{3b}

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} = > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5 = > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b} \sqrt{3} = \frac{7(5b - a)}{3b} it seems that √3 is rational

Let it be a rational number5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} = > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5 = > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b} \sqrt{3} = \frac{7(5b - a)}{3b} it seems that √3 is rationalso our assumption get wrong because √3 is irrational number so 5-3/7√3 is irrational