Use the following building blocks to assemble a proof by contradiction that the

Question

Use the following building blocks to assemble a proof by contradiction that the sum of a rational number and an irrational number is irrational. Not all blocks belong in the proof. By definition of rational number, there must be integers p and q, q notequalto 0, such that x = p/q. Suppose x and y are both rational and their sum is irrational. This is again a quotient of two integers with a nonzero denominator, therefore rational. Therefore, x+y must be irrational. Therefore, we have concluded that y is irrational, a contradiction. Now assume that x+y is irrational. Therefore, x must be rational. Further suppose, to get a contradiction, that x+y is rational. Suppose x and yare both irrational and their sum is rational. By substitution, we find (p/q)+y = a/b, and therefore y = (a/b)-(p/q). Therefore, y must be rational. Suppose x is rational and y is irrational. Likewise, there must be integers a and b, b notequalto 0, such that x+y = a/b. Therefore, we have concluded that y is rational, a contradiction. We now simplify: y = (aq-pb)/bq.

Explanation / Answer

Answer:- 1, 12, 8, 10, 15, 11, 5, 4

1. x = p/q : q is not zero, then x is a rational number.

12. suppose x is rational and y is irrational.

8. further suppose to get a contradiction that x+y is a rational.

10. So we find (p/q) + y = (a/b) => y = (a/b) - (p/q).

11. We now simplify: y = (aq-pb)/bq.

5. Therefore, we have concluded that y is irrational, is a contradiction.

4. Hence x + y must be irrational.

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