Consider the following statement. 3√2-7 is irrational. This statement is true, but the following proposed proof by contradiction is incorrect. Proposed proof: Suppose not. That is, suppose 3√√2-7 is rational. By definition of rational, there exist real numbers a and b with a 3√2-7= and b + 0. b Using algebra to solve for √2 gives a + 7b 3b But a + 7b and 3b are real numbers and 3b + 0. 3b√/2 - 7b = a; and so √2 = Therefore, by definition of rational, √/2 is rational. This contradicts Theorem 4.7.1, which states that √2 is irrational. Hence the supposition is false. Identify the errors in the proposed proof. (Select all that apply.) ✔ There is an error in the algebra solving for √2. O For a proof by contradiction you should suppose that 3√/2 - 7 is irrational. To apply the definition of rational, a and b must be integers. ✔ The square root of 2 is rational, so there is no contradiction. ✔The statement to be proved is assumed. X