Prove the following statement by contradiction. If a and b are rational numbers, b 0, and ris an irrational number, then a + br is irrational. Proof by contradiction: Select an appropriate statement to start the proof. O Suppose not. That is, suppose there exist irrational numbers a and b such that b O Suppose not. That is, suppose there exist rational numbers a and b such that b 0, r is an irrational number, and a + br is irrational. O Suppose not. That is, suppose there exist rational numbers a and b such that b O Suppose not. That is, suppose there exist rational numbers a and b such that b + 0, ris an irrational number, and a + br is rational. O Suppose not. That suppose there exist irrational numbers a and b such that b 0, ris an irrational number, and a + bris rational. Then by definition of rational, a-b-, and a + br where c, d, i, j, m, and n are ---Select--- ✓and-Select-- 0, r is a rational number, and a + br is rational. 0, r is a rational number, and a + br is irrational. ✓. Since b + 0, we also have that / * 0. Solving this equation for r and representing the result as a single quotient in terms of c, d, i, j, m, and n gives that substitution, Note that r-Select--a ratio of two integers because products and differences of integers --Select- integers. Also the denominator of r?0. Therefore, by definition of rational, it follows that r is-Select-, which contradicts the supposition. Hence the supposition is false and the given statement is true.