20 1. The Real Numbers

Now it is easy to see that 2 is not the square of a rational number. If it

were, that number would have to be an integer, by the above theorem. The only

possibilities are −1, 0,1 since all other integers have squares that are too large. Of

course, none of the numbers −1, 0,1 has its square equal to 2.

Other geometric objects also lead to the conclusion that the system of rational

numbers is not suﬃcient for the measurement of objects that occur in the natural

world. The area π of a circle of radius 1 is not a rational number, for example.

In fact, the rational number system is riddled with holes where there ought to be

numbers. This problem is fixed by the introduction of the system of real numbers

which is the topic of the next section.

Exercise Set 1.3

1. Given that N has an operation of addition which is commutative and associa-

tive, how would you define such an addition operation in Z?

2. Referring to the previous exercise, answer the same question for the operation

of multiplication.

3. Prove that if Z satisfies the axioms for a commutative ring, then Q satisfies A1

and M1.

4. Prove that if Z satisfies the axioms for a commutative ring, then Q satisfies A2

and M2.

In the next three exercises you are to prove the given statement assuming x, y, z

are elements of a field. You may use the results of examples and theorems from

this section.

5. (−x)(−y) = xy.

6. xz = yz implies x = y, provided z = 0.

7. xy = 0 implies x = 0 or y = 0.

In the next three exercises you are to prove the given statement assuming x, y, z

are elements of an ordered field. Again, you may use the results of examples and

theorems from this section.

8. x 0 and y 0 imply xy 0.

9. x 0 implies

x−1

0.

10. 0 x y implies

y−1 x−1.

11. Prove that the equation x2 = 5 has no rational solution.

12. Generalize Theorem 1.3.9 by proving that every rational solution of a polyno-

mial equation

xn

+

an−1xn−1

+ · · · + a1x + a0 = 0,

with integer coeﬃcients ak, is an integer solution.

13. Prove that if m and n are positive integers with no common factors other than

1 (i.e. m and n are relatively prime), then there are integers a and b such that

1 = am+bn. Hint: Let S be the set of all positive integers of the form am+bn,