# Real Numbers, Sequences and Series: Part 6

Theorem:

Given $x \in \Re_{+}$ and $n \in N$, we can find $y \in \Re$ such that $x=y^{n}$.

Proof:

Let $A={u \in \Re_{+}: u^{n}. If $x<1$, then $x^{n} and hence $x \in A$. On the other hand, if $x \geq 1$, then $1/2 \in A$. Thus, A is non-empty. Next, observe that if $v^{n}>x$, then v is an upper bound of A. In particular, $\{ 1+x\}$ is an upper bound of A.

By the least upper bound property, A admits a least upper bound. Let us denote it by y. We will rule out the possibilities $y^{n}>x$ and $y^{n} implying that $x=y^{n}$.

If $y^{n}, let $a=\frac{x-y^{n}}{nx}$ and $z=y(1+a)$. It can be checked that $z^{n} so that $z \in A$. But, $y , contradicting the fact that y is the least upper bound of A. (we have used the inequalities 7 and 8 in the previous blog).

On the other hand, if $y^{n}>x$, let $\frac{y^{n}-x}{ny^{n}}$ and $w=y(1-b)$. Again, it can be verified that $w^{n}>x$ and hence w is an upper bound of A. But, $w, contradicting the fact that y is the least upper bound of A. Hence, $y^{n}=x$QED.

In particular, we see that there is an element $\alpha \in \Re$ such that $\alpha^{2}=2$ and hence also $(-\alpha)^{2}=2$ which means that the equation $x^{2}=2$ has two solutions. The positive one of those two solutions is $\sqrt{2}$. In fact, the above theorem has guaranteed its extraction of the square root, cube root, even nth root of any positive number. You could ask at this stage, if this guarantees us extraction of square root of a negative number. The answer is clearly no. Indeed, we have

$x^{2} \geq 0$ for $x \in \Re$.

Remark.

We can further extend $\Re$ to include numbers whose squares are negative, which you know leads to the idea of complex numbers.

We have shown that Q is a subset of $\Re$. One can also show that between any two distinct real numbers there is a rational number. This naturally leads to the decimal representation of real number: Given any real number x and any $q \in N$, we can get a unique $a_{0} \in Z$ and unique $a_{1},a_{2}, \ldots a_{q} \in N$ such that $0 \leq a_{1} \leq a_{2} \ldots a_{q} \leq 9$ and

$|x-(a_{0}+a_{1}/10+a_{2}/100+\ldots + \frac{a_{q}}{10^{q}})|<\frac{1}{10^{q}}$

You are invited to try to  prove this familiar decimal representation.

If we have a terminating decimal representation of a real number, then surely it is rational.But, we know that rationals like 1/3, 1/7, 1/11, do not have a terminating decimal expansion.

It is clear that the decimal representation of $\sqrt{2}$ cannot terminate as it is not rational. There are many elements of $\Re$ which are not in Q. Any element of $\Re$ which is not in is called an irrational number, and irrational numbers cannot have terminating decimal representation.

More later,

Nalin Pithwa