Answer:sum of rational and irrational number is always an irrational number. Step-by-step explanation:Here we are asked to prove that the sum of an irrational number and a rational number is always an irrational number. We will do it with an assumption that the sum of rational and irrational number is rational. Let us say the the irrational number is x and rational number is [tex]\frac{a}{b}[/tex] where a and b are integers and b ≠ 0 . Hence we are saying that c is rational hence the result can be represented in the form of fraction . Say it will be \frac{m}{n}[/tex] which is rational Hence[tex]x + \frac{a}{b} = \frac{m}{n}[/tex]Let us solve it for x , subtract \frac{a}{b} \frac{a}{b}[/tex] from both sides we get\frac{a}{b} x = \frac{m}{n} - \frac{a}{b}[/tex] \frac{a}{b} x = \frac{mb-an}{nb}[/tex]as a, b , m and n all are integers nb , mb , an and (mb-an) will be integers as the product and sum of two integers are always integers. Hence x comes out to be ratio of two integers , which is a rational number . Hence we have an contradiction here. Our x was an irrational number but the result turned out to say that x be a rational number . Hence our assumption was wrong and the sum of irrational and rational number can not be a rational number. x will be an number which can not be represented in the form of [tex]\frac{a}{b}[/tex] where a and b are integers and b ≠ 0