# completeness property of real numbers

I am studying about how a real number is defined by its properties. The three type of properties that make the real numbers what they are.

1. Algebraic properties i.e, the axioms of addition, subtraction multiplication and division.

2. Order properties i.e., the inequality properties

3. Completeness property

Here is the question : I am not able to understand the completeness property and please explain it to me in detail(it says upper bound lower bound ......)as I am a self learner.

• Don't worry about understanding it - it's controversial: math.stackexchange.com/questions/928366/… – user117644 Sep 15 '14 at 15:30
• It's all about least upper bound property. Its very important because every notion of calculus comes from it. – Henry Sep 15 '14 at 15:31
• @spark That depends on what calculus methodology one subscribes to, eg SIA calculus is wholly algebraic. – user117644 Sep 15 '14 at 15:33
• I know that very important properties as other properties like intermediate value property comes out from it but i am not able to understand the stuff technically in the sense through equations or anything else – Jasser Sep 15 '14 at 15:37
• You obviously don't believe me, your choice... – user117644 Sep 15 '14 at 15:57

The crucial problem with rational numbers is that they are incomplete. It was discovered about 2300 years ago that there is no rational number whose square is 2. But why should there be a square root of 2? One reason is that it's easy to find a sequence of rational numbers that appears to be getting closer and closer to the square root of 2:

$$\frac11, \frac 32, \frac 75, \frac{17}{12}, \frac{41}{29},\ldots, \frac ab, \frac {a+2b}{a+b},\ldots$$

and one can show that although the terms in this sequence get closer and closer together, there isn't anything they get closer to, because if there was, its square would be $2$, and there is no rational number whose square is 2. Or one can consider the sequence of rational numbers $$1, 1.4, 1.41, 1.414, 1.4142, \ldots$$ which is similar: the terms get closer and closer together as you look farther along the sequence, but they do not get close to any rational number, again because if they did that rational number would be a square root of 2, and there is no such rational.

So the rational numbers are literally incomplete; there are “missing” numbers.

The real numbers solve this problem: they are a system of numbers that contains the rationals, but has the property that if $S$ is any sequence whose members eventually get closer and closer together, like the examples above, then there is some real number that the elements of $S$ approach as closely as desired; $S$ converges to some real number. This is the “completeness” property you are looking for.

• In the last line "S converges to real number", you mean irrational number(real number) right. So it is actually infinite series which helps to solve this completeness property i.e., taking the summation of such rational nubers and tending the summation to infinity – Jasser Sep 16 '14 at 7:18
• No, I meant real number. Some sequences converge to rational values. And I said nothing about summations or series. – MJD Sep 16 '14 at 12:06
• sorry for the misinterpretion. now i got that into my head. – Jasser Sep 16 '14 at 12:57

Suppose I have a set $X \subseteq \mathbb{R}$ such that for all $x \in X$ we have $x \le t$ for some number $t$. Then the completeness axiom guarantees the existence of a smallest number, call it $s$, such that $x \le s$.

You write this down more formally by saying that if $X$ is bounded, ie. if all $x$ are such that $x \le t$ for some $t$, then there exists a number $s \in \mathbb{R}$ satisfying the following:

• $x \le s$

• If $x \le t'$ then $s \le t'$

we call $s$ the supremum of $X$ and denote it by $\sup(X)$.

The completeness axiom is declaring the existence of a supremum on a bounded set.