Exercise 4.10

Main Problem.

(a)

Proof. First, we know that $0\le h<1$. If $h=0$ then clearly $h=0=0^2=h^2$ so that $0\le h^2\le h$ is true. If $h\ne 0$ then $0<h<1$ so that $0=0\cdot h<h\cdot h=h^2<1\cdot h=h$ by property (6) since $h>0$ so that again $0\le h^2\le h$ is true.

We then have

$$\begin{array}{llllll}\hfill {(x+h)}^2& =(x+h)(x+h)& \hfill & & \hfill & \\ \hfill & =x^2+2\mathit{xh}+h^2& \hfill & & \hfill & \\ \hfill & \le x^2+2\mathit{xh}+h& \hfill \text{(since }{h}^2\le h\text{)}& & \hfill & & \hfill \\ \hfill & =x^2+h(2x+1).& \hfill & & \hfill & \end{array}$$

Also

$$\begin{array}{llllll}\hfill {(x-h)}^2& =(x-h)(x-h)& \hfill & & \hfill & \\ \hfill & =x^2-2\mathit{xh}+h^2& \hfill & & \hfill & \\ \hfill & \ge x^2-2\mathit{xh}+0& \hfill \text{(since }{h}^2\ge 0\text{)}& & \hfill & & \hfill \\ \hfill & =x^2-h(2x),& \hfill & & \hfill & \end{array}$$

which show the desired results. □

(b)

We modify this result so that the $h$ in the second part is not just positive but also $h<x$. In fact, without this stipulation, the theorem becomes obvious since any arbitrarily large $h$ will suffice. Because then then $x-h$ is arbitrarily large in magnitude (but negative) so that ${(x-h)}^2$ can be made arbitrarily large so that of course ${(x-h)}^2>a$. Adding the stipulation that $0<h<x$ makes the theorem more useful and is necessary for it to be of use in part (c) below.

Proof. Suppose that $x>0$. Then clearly $2x>2\cdot 0=0$ as well. Also it then follows that $2x+1>1>0$.

If $x^2<a$ then clearly $0<a-x^2$. Hence we have that $0<(a-x^2)∕(2x+1)$ by Exercise 4.2 parts (i) and (h) since both $a-x^2$ and $2x+1$ are positive. So let $y=\min <!--\; FUNCTION\; APPLICATION\; -->(1,(a-x^2)∕(2x+1))$ so that clearly both $y\le 1$ and $y\le (a-x^2)∕(2x+1)$. Since $0<1$ and $0<(a-x^2)∕(2x+1)$, we have that $0<y$ so that it follows from Exercise 4.9 part (d) that there is a rational $h$ such that $0<h<y$. Hence $0<h<y\le 1$ so that, by part (a), we have

$$\begin{array}{llllll}\hfill {(x+h)}^2& \le x^2+h(2x+1)& \hfill & & \hfill & \\ \hfill & <x^2+\left(\frac{a-x^2}{2x+1}\right)\left(2x+1\right)& \hfill \text{(since }h<y\le (a-x^2)∕(2x+1)\text{ and }2x+1>0\text{)}& & \hfill & & \hfill \\ \hfill & =x^2+(a-x^2)& \hfill & & \hfill & \\ \hfill & =a.& \hfill & & \hfill & \end{array}$$

If $x^2>a$ then clearly $x^2-a>0$. Then we have again that $(x^2-a)∕(2x)$ is positive since we showed previously that $2x$ is. So let $y=\min <!--\; FUNCTION\; APPLICATION\; -->(1,(x^2-a)∕(2x),x)$ so that clearly $y\le 1$, $y\le (x^2-a)∕(2x)$, and $y\le x$. Since both 1, $(x^2-a)∕(2x)$, and $x$ are all positive it follows that $0<y$ so that there is a rational $h$ such that $0<h<y$ by Exercise 4.9 part (d). Therefore $0>-h>-y$. Since $0<h<y\le 1$ we have by part (a) that

$$\begin{array}{llllll}\hfill {(x-h)}^2& \ge x^2-h(2x)& \hfill & & \hfill & \\ \hfill & >x^2-\left(\frac{x^2-a}{2x}\right)\left(2x\right)& \hfill \text{(since }-h>-y\ge -(x^2-a)∕(2x)\text{ and }2x>0\text{)}& & \hfill & & \hfill \\ \hfill & =x^2-(x^2-a)& \hfill & & \hfill & \\ \hfill & =a,& \hfill & & \hfill & \end{array}$$

which show the desired results since clearly $0<h<y\le x$. □

(c)

Proof. Suppose that $a>0$ and let $B=\left\{x\in ℝ\mid x^2<a\right\}$.

If $a<1$ then $0<a<1$ so that $a^2=a\cdot a<1\cdot a=a$ so that $a$ itself is in $B$ (and of course $a$ is positive). Now consider any $x\in B$ so that $x^2<a$. Then $x^2<a<1$ so that also $x<1$ by Lemma 1 . Since $x\in B$ was arbitrary, this shows that $1$ is an upper bound of $B$.

If $a\ge 1$ then ${(1∕2)}^2=1∕2^2=1∕4<1\le a$ so that $1∕2\in B$ (and of course $1∕2$ is positive). Now consider any $x\in B$ so that $x^2<a$. If $x\le 1$ then $x\le 1\le a$. On the other hand, if $x>1$ then $x^2=x\cdot x>1\cdot x=x$ since $x>1>0$ so that $x<x^2<a$. Thus in both cases $x\le a$ so that $a$ is an upper bound of $B$ since $x$ was arbitrary.

Therefore in each case $B$ contains a positive element (so that $b\ne \varnothing$) and $B$ is bounded above. It then follows that $B$ has a least upper bound $b$ (so that $b=\sup <!--\; FUNCTION\; APPLICATION\; -->B$). Clearly since $B$ has a positive element $x$, it follows that $0<x\le b$ so that $b$ is positive.

Now suppose that $b^2<a$. Then by definition $b\in B$ so that $b$ has to be the largest element of $b$ since it is the least upper bound. Since we know that $b$ is positive and $b^2<a$, it follows from part (b) that there is an $h>0$ where ${(b+h)}^2<a$ and hence $b+h\in B$. However, since $h>0$, it follows that $b<b+h$, which contradicts the fact that $b$ is the greatest element of $B$. Hence it cannot be that $b^2<a$.

So suppose that $b^2>a$. Then again by part (b) there is an $h$ where $0<h<b$ such that ${(b-h)}^2>a$. Now, since $h>0$, it follows that $b-h<b$ so that $n-h$ is not an upper bound of $B$ (since then $b$ would not be the least upper bound). Hence there is an $x\in B$ such that $b-h<x$, noting that $x^2<a$ by the definition of $B$. Since $h<b$, we have that $0<b-h<x$ so that ${(b-h)}^2<x^2<a$ by Lemma 2 . But this contradicts the established fact that ${(b-h)}^2>a$ so that it cannot be that $b^2>a$.

Thus the only possibility remaining is that $b^2=a$ as desired. □

(d)

Proof. Suppose that $b$ and $c$ are positive and that $b^2=c^2$. If it were the case that $b<c$ then $0<b<c$ so that $0<b^2<c^2$ by Lemma 2 so that clearly $b^2\ne c^2$. As this is a contradiction, it has to be that $b\ge c$. An analogous argument shows that $b>c$ also leads to a contradiction so that $b\le c$. Hence it must be that $b=c$ as desired. □