Cut: A cut in (The set of rational numbers) is a set such that

.

.

If , and , then .

does not have a least element.

Real numbers: The set of real numbers is defined as the set of all cuts in .

If , are cuts, then we say that iff .

Theorem 1 is complete.

Proof: Let be bounded below.

Now consider .

We first show that is a cut.

Since as it is a cut and there exists atleast one , .

, such that .

and , and hence, .

If had a least element, say x, then since , for some . But for every , and hence such that since does not have a least element. Hence, does not have a least element. Therefore, is a cut.

Now, for every , . Hence, . Hence, . Therefore, is a lower bound for B.

Consider any other lower bound, say for B. Now, and hence .

Hence, . This implies that and . Hence, is the greatest lower bound for B.

We have shown that for any arbitrary set , there exists a greatest lower bound. Hence, for every , there exists a greatest lower bound. Hence, satisfies the greatest lower bound property. Therefore, satisfies the least upper bound property. This implies that is complete.