1.4 Properties of the real numbers  (Page 2/2)

We show now that $R$ contains elements other than the rational numbers in $Q.$ Of course this holds for any complete ordered field. The next theorem makes this quite explicit.

If $x$ is a positive real number, then there exists a positive real number $y$ such that $y^2=x.$ That is, every positive real number $x$ has a positive square root in $R.$ Moreover, there is only one positive square root of $x.$

Let $S$ be the set of positive real numbers $t$ for which $t^2\le x.$ Then $S$ is nonempty Indeed, If $x>1,$ then 1 is in $S$ because $1^2=1\times 1<1\times x=x.$ And, if $x\le 1,$ then $x$ itself is in $S$ , because $x^2=x\times x\le 1\times x=x.$

Also, $S$ is bounded above. In fact, the number $1+x/2$ is an upper bound of $S.$ Indeed, arguing by contradiction, suppose there were a $t$ in $S$ such that $t>1+x/2.$ Then

which is a contradiction. Therefore, $1+x/2$ is an upper bound of $S,$ and so $S$ is bounded above.

Now let $y=supS.$ We wish to show that $y^2=x.$ We show first that $y^2\le x,$ and then we will show that $y^2\ge x.$ It will then follow from the tricotomy law that $y^2=x.$ We prove both these inequalities by contradiction.

So, assume first that $y^2>x,$ and write $\alpha$ for the positive number $y^2-x.$ Let $ϵ$ be the positive number $\alpha /\left(2y\right),$ and, using Theorem 1.5, choose a $t\in S$ such that $t>y-ϵ.$ Then $y+t\le \left(2y\right),$ and $y-t<ϵ=\alpha /2y.$ So,

$$\begin{array}{ccc}\hfill \alpha & =& y^2-x\hfill \\ & =& y^2-t^2+t^2-x\hfill \\ & \le & y^2-t^2\hfill \\ & =& (y+t)(y-t)\hfill \\ & \le & 2y(y-t)\hfill \\ & <& 2yϵ\hfill \\ & <& 2y\times \frac{\alpha }{2y}\hfill \\ & =& \alpha ,\hfill \end{array}$$

which is a contradiction. Therefore $y^2$ is not greater than $x.$

Now we show that $y^2$ is not less than $x.$ Again, arguing by contradiction, suppose it is,and let $ϵ$ be the positive number $x-y^2.$ Choose a positive number $\delta$ that is less than $y$ and also less than $ϵ/\left(3y\right).$ Let $s=y+\delta .$ Then $s$ is not in $S,$ whence $s^2>x,$ so that we must have

$$\begin{array}{ccc}\hfill ϵ& =& x-y^2\hfill \\ & =& x-s^2+s^2-y^2\hfill \\ & \le & s^2-y^2\hfill \\ & =& (s+y)(s-y)\hfill \\ & =& (2y+\delta )\delta \hfill \\ & <& 3y\delta \hfill \\ & <& ϵ,\hfill \end{array}$$

which again is a contradiction.

This completes the proof that $y^2=x;$ i.e., that $x$ has a positive square root.

Finally, if $y^{\text{'}}$ were another positive number for which ${y^{\text{'}}}^2=x,$ we show that $y=y^{\text{'}}$ by ruling out the other two cases: $y<y^{\text{'}}$ and $y>y^{\text{'}}.$ For instance, if $y<y^{\text{'}},$ then we would have that $y^2<{y^{\text{'}}}^2,$ giving that

$$x=y^2<{y^{\text{'}}}^2=x,$$

implying that $x<x,$ and this is a contradiction.

If $x$ is a positive real number, then the symbol $\sqrt{x}$ will denote the unique positive number $y$ for which $y^2=x.$ Of course, $\sqrt{0}$ denotes the number 0.

REMARK Part (c) of [link] shows that the field $Q$ contains no number whose square is 2, and [link] shows that the field $R$ does contain a number whose square is 2. We have therefore “proved” that the real numbers is a largerset than the rational numbers. It may come as a surprise to learn that we only now have beenable to prove that. Look back through the chapter to be sure. It follows also that $Q$ itself is not a complete ordered field. If it were, it would be isomorphic to $R,$ by Theorem 1.2, so that it would have to contain a square root of 2, which it does not.

A real number $x$ that is not a rational number, i.e., is not an element of the subset $Q$ of $R,$ is called an irrational number.