# Real numbers, sequences and series: part V

Real Numbers.

Our scheme of real numbers has so far been extended to rational numbers so that we could not only add and subtract without any hindrances, but could also multiply and divide without any restriction, save division by zero. But, when it comes to extraction of square roots, it seems “incomplete”. For example, there is no rational number whose square gives 2. This question crops up when we try to relate numbers to geometry.

We have seen before that we can always represent a rational number on a line. Now suppose we have a square with sides of unit length. By the theorem of Pythagoras, the square of the length of diagonal is equal to $1^{2}+1^{2}=2$. We can also see that the length of the diagonal can be geometrically represented on the line. Now, we ask here, does there exist a rational number whose square is 2, or equivalently, does the equation $x^{2}=2$ have a solution in rational numbers? This amounts to asking if we can find two non-zero integers p and q such that $p^{2}=2q^{2}$Equation I.

We can demonstrate that there are no such integers. Suppose that there were two such integers p and q satisfying equation I. We may assume without loss of generality, that p and q have no common factors. If there were any, we could cancel them from both the sides of equation I till there were no common factors. Now as there are no common factors between p and q, for equation I to hold, 2 must be a factor of $p^{2}$, and hence, of p. So we may write $p=2m$ Equation 2.

Substituting this in Equation 2, we get $2m^{2}=q^{2}$Equation 3.

As before, we conclude that for equation 3 to hold, 2 must divide q giving $q=2n$ for some integer n. This contradicts our assumption that p and q have no common factors. This only proves that equation I has no solution in integers (except $p=0, q=0$). Thus, the length of the diagonal of the unit square, though it has a point representing it on the line, does not correspond to a rational number. This shows that is not large enough to accommodate a number such as the length of the diagonal of a unit square. So we must extend Q. We could simply include the new number which we write as $\sqrt{2}$ in the scheme along with Q. But, then this ad hoc extension may not stand up to all our demands to include newer and newer numbers which arise out of algebraic equations, e.g., $\sqrt{6}$, etc. Even if we somehow include these numbers in our scheme, we must know how to perform arithmetic operations like addition, subtraction, multiplication and division in it. Also, how does one decide which of a given pair of new numbers is smaller? There are many ways in extending to accommodate all these new numbers. We illustrate one of the simplest ways of doing this.

Let $E=\{ a \in Q: a^{2}<2\}$. It is clear that E has an upper bound, for example, 2. Next, we note that if $y \in E$ is an upper bound of E, then there exists $n \in N$ with $n(2-y^{2})>1+2y$ (we can always choose an n because of the Archimedean property of Q). Then, one has $2-(y+\frac{1}{n})^{2}=(2-y^{2})-\frac{1}{n}(2y+\frac{1}{n}) \geq (2-y^{2})-\frac{2y+1}{n}>0$, showing that $y^{2}<(y+\frac{1}{n})^{2}<2$ and thus no upper bound of E can be in E. Let $x \in Q$ be one such upper bound of E, then so is $\frac{x}{2}+\frac{1}{x}$ since $(\frac{x}{2}+\frac{1}{x})^{2}-2=(\frac{x}{2}-\frac{1}{x})^{2}>0$. On the other hand, $x-(\frac{x}{2}+\frac{1}{x})=\frac{x^{2}-2}{2x}>0$ since $x \not\in E$. Therefore, the E has no least upper bound in Q.

This tells us what to do. We now extend the field of rationals to a larger field containing it which has all properties of along with an additional property, called the least upper bound property and abbreviated as lub property, namely, every set that is bounded above has a least upper bound.

More precisely, we postulate that there is a field $(\Re, +, .)$, $\Re$ containing Q and satisfying all the properties of Q listed above and an additional one:

The least upper bound property (lub): If $A \subseteq \Re$ is bounded above, then A admits a least upper bound, that is, (1) there is a $\gamma \in \Re$ such that $\alpha \leq \gamma$ for every $\alpha \in A$ and (2) for every $\gamma^{'}< \gamma$, there is a $\beta \in A$ such that $\gamma^{'} < \beta$.

Remark:

As we constructed out of Z, there are ways of constructing $\Re$ out of Q. However, an explicit construction of $\Re$ from is beyond the scope of the present blog. Nevertheless, we are going to use the listed properties of $\Re$ in all subsequent discussions and deduce many interesting consequences.

Since $\Re$ has all the properties of including the order property, we write (as for Q), $\Re_{+}=\{ \alpha \in \Re|\alpha >0\}$

and this has the same properties as $Q_{+}$. Now, we use the same notation like $\leq, <, \geq, >$ for elements of $\Re_{+}$, exactly as we did for the elements of Q. As before, we write $\beta > \alpha$ or $\beta \geq \alpha$ according as $\alpha < \beta$ or $\alpha \leq \beta$.

Clearly, 1, $1+1, 1+1+1, \ldots$ belong to $\Re$ and represent the natural numbers. So, one sets up a one to one correspondence between $n \in N$ and $1+1+1+\ldots+1 (n times)$ to conclude that $N \subset \Re$. Writing $-n$ as the additive inverse of $n \in N$, of course $0 \in \Re$, we see that $Z \subseteq \Re$. Since $\Re$ is a field, for $m,n \in Z$, $n \neq 0$, we must have $m.n^{-1} \in \Re$

where $n^{-1}$ is the multiplicative inverse of n. We agree to write $m.n^{-1}=\frac{m}{n}$. This way we have $Q \subseteq \Re$. We have thus effectively extended to $\Re$.

More later,

Nalin Pithwa

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