Tag: Rationals

Prove that the rationals satisfy the Archimedean property, but not the least-upper-bound axiom.

by RoRi

July 2, 2015

Recall, the Archimedean property states that if $x > 0$ and $y \in \mathbb{R}$ is arbitrary, then there exists an integer $n > 0$ such that $nx > y$ .
Further, recall that the least upper bound axiom states that every nonempty set $S$ of real numbers which is bounded above has a supremum.
Now, prove that $\mathbb{Q}$ satisfies the Archimedean property, but not the least-upper-bound axiom.

First, we prove that $\mathbb{Q}$ satisfies the Archimedean property.
Proof. This is immediate since if $x,y$ are arbitrary elements in $\mathbb{Q}$ with $x>0$ , then they are also in $\mathbb{R}$ since $\mathbb{Q} \subseteq \mathbb{R}$ . Thus, by the Archimedean property in $\mathbb{R}$ we know there is an $n \in \mathbb{Z}_{>0}$ such that $nx > y$ . Hence, the Archimedean property is satisfied in $\mathbb{Q}. \qquad \blacksquare$

Next, we prove that $\mathbb{Q}$ does not satisfy the least-upper-bound axiom.
Proof. Let

$S = \{ r \in \mathbb{Q} \mid r^2 \leq 2 \}.$

Then $S$ is non-empty since $1 \in S$ . Further, $S$ is bounded above by 4 since $r^2 \leq 2 \ \implies \ r^2 \leq 16 \ \implies \ r \leq 4$ . (Of course, there are better upper bounds available, but we just need any upper bound.)
Now, we must show that $S$ has no supremum in $\mathbb{Q}$ to show that the least-upper-bound property fails in $\mathbb{Q}$ (since this will mean $S$ is a nonempty set which is bounded above, but fails to have a least upper bound in $\mathbb{Q}$ ).
Suppose otherwise, say $b = \sup S$ with $b \in \mathbb{Q}$ . We know $b^2 \neq 2$ (I.3.12, Exercise #11). Thus, by the trichotomy law we must have either $b^2 < 2$ or $b^2 > 2$ .
Case 1: If $b^2 < 2$ , then there exists $r \in \mathbb{Q}$ such that $b < r < \sqrt{2}$ (since the rationals are dense in the reals, see I.3.12, Exercise #6). But then, $r^2 < 2$ (since $r < \sqrt{2} \ \implies \ r \cdot r < \sqrt{2} \cdot \sqrt{2} \ \implies \ r^2 < 2$ ) and $r \in \mathbb{Q}$ implies $r \in S$ with $r > b$ , contradicting that $b$ is an upper bound for $S$ . Hence, we cannot have $b^2 < 2$ .
Case 2: If $b^2 > 2$ , then there exists $r \in \mathbb{Q}$ such that $b > r > \sqrt{2}$ . But then, $r^2 > 2$ , so if $s \in S$ we have $s < r$ ; hence, $r$ is an upper bound for $S$ which is less than $b$ . This contradicts that $b$ is the least upper bound of $S$ . Hence, we also cannot have $b^2 > 2$ .
Thus, there can be no such $b^2 \in \mathbb{Q}$ (since by the trichotomy exactly one of $b^2> 2, \ b^2 < 2, \ b^2 = 2$ must hold, but we have shown these all lead to contradictions).
Hence, $\mathbb{Q}$ does not have the least-upper-bound property $. \qquad \blacksquare$

Prove there is no rational number which squares to 2.

by RoRi

July 1, 2015

Prove that there is no $r \in \mathbb{Q}$ such that $r^2 = 2$ .

Proof. Suppose otherwise, that there is some rational number $r \in \mathbb{Q}$ such that $r^2 = 2$ . Then, since $r \in \mathbb{Q}$ , we know there exist integers $a,b \in \mathbb{Z}$ not both even (I.3.12, Exercise #10 (e) such that $r = \frac{a}{b}$ . Then, since $r^2 = 2$ we have

$2 = \frac{a^2}{b^2} \quad \implies \quad 2b^2 = a^2.$

But, by I.3.12, Exercise #10 (d), we know $2b^2 = a^2$ implies both $a$ and $b$ are even, contradicting our choice of $a$ and $b$ not both even. Hence, there can be no such rational number $. \qquad \blacksquare$

Sums, differences, products and quotients of an irrational and a rational are irrational.

by RoRi

June 30, 2015

Prove that if $x \in \mathbb{Q}$ with $x \neq 0$ , and $y \in \mathbb{R} \smallsetminus \mathbb{Q}$ , then

$x+y,\quad x-y, \quad xy, \quad \frac{x}{y}, \quad \frac{y}{x}$

are all irrational, i.e., are all in $\mathbb{R} \smallsetminus \mathbb{Q}$ .

Proof. Since $x \in \mathbb{Q}$ , we know there are integers $m$ and $n$ such that $x = \frac{m}{n}$ . Now we consider each of the elements we wish to show are irrational.

Suppose otherwise, that $x+y \in \mathbb{Q}$ . Then, there exist $r,s \in \mathbb{Z}$ such that
$x+y = \frac{r}{s} \ \implies \ \frac{m}{n} +y = \frac{r}{s} \ \implies \ y = \frac{r}{s} - \frac{m}{n} \implies \ y = \frac{rn - ms}{sn} \ \implies \ y \in \mathbb{Q}.$

This contradicts our assumption that $y$ is irrational.
Since $y$ is irrational, so is $-y$ (by part (a), the sum of a rational and an irrational cannot be rational and since $y+(-y) = 0$ is rational, cannot have $-y$ rational). But then, $x-y = x+(-y)$ and by part (a) this sum must be irrational.
Suppose otherwise, that $xy \in \mathbb{Q}$ . Then there exist $r,s \in \mathbb{Z}$ such that $xy = \frac{r}{s}$ . Further, since $x \neq 0$ , we know $x^{-1} = \frac{n/m}$ exists.
$xy = \frac{r}{s} \ \implies \ \frac{m}{n} y = \frac{r}{s} \ \implies \ y = \frac{r}{s} \frac{n}{m} \ \implies \ y = \frac{rn}{sm} \in \mathbb{Q}.$

Contradicting our assumption that $y$ is irrational.
First, we since $y$ is irrational, we have $y \neq 0$ , and thus $y^{-1}$ exists. Further, since $yy^{-1} = 1$ is rational, and $y$ is irrational, by (c) we must have $y^{-1}$ irrational as well. Then by (c), since $\frac{x}{y} = xy^{-1}$ , we must have $xy^{-1}$ irrational.
By (d) we know $\frac{x}{y}$ is irrational, and since $\frac{x}{y} \frac{y}{x} = 1$ is rational, by (c), we must have $\frac{y}{x}$ irrational.

The rationals are dense in the reals.

by RoRi

June 30, 2015

Prove that the rational numbers are dense in the reals. I.e., if $x, y \in \mathbb{R}$ with $x < y$ , then there exists an $r \in \mathbb{Q}$ such that $x < r < y$ . It follows that there are then infinitely many such.

Proof. Since $x < y$ , we know $(y-x)> 0$ . Therefore, there exists an $n \in \mathbb{Z}^+$ such that

$n(y-x) > 1 \quad \implies \quad ny > nx+1.$

We also know (I.3.12, Exercise #4) that there exists $m \in \mathbb{Z}$ such that $m \leq nx < m+1$ . Putting these together we have,

$\begin{align*} nx < m+1 \leq nx+1 < ny &\implies \ nx < m+1 < ny \\ &\implies \ x < \frac{m+1}{n} < y. \end{align*}$

Since $m,n \in \mathbb{Z}$ we have $\frac{m+1}{n} \in \mathbb{Q}$ . Hence, letting $r = \frac{m+1}{n}$ we have found $r \in \mathbb{Q}$ such that

$x < r < y.$

This then guarantees infinitely many such rationals since we can just replace $y$ by $r$ (and note that $\mathbb{Q} \subseteq \mathbb{R}$ ) and apply the theorem again to find $r_1 \in \mathbb{Q}$ such that $x < r_1 < r$ . Repeating this process we obtain infinitely many such rationals $. \qquad \blacksquare$

Prove equivalence of different forms for additive inverses of fractions

by RoRi

June 29, 2015

Prove that if $b \neq 0$ then

$- \left( \frac{a}{b} \right) = \frac{-a}{b} = \frac{a}{-b}.$

Proof. We can use the first exercise of this section (Section I.3.3, Exercise #1) and the previous exercise (Section I.3.3, Exercise #8) to compute

$\begin{align*} - \left(\frac{a}{b} \right) &= -(ab^{-1}) &(\text{Def of } b^{-1}) \\ &= -(b^{-1}a) \\ &= \left( (-b)^{-1} a \right) &(\text{Thm I.12}) \\ &= a(-b^{-1}) &(\text{Thm I.12}) \\ &= \frac{a}{-b} &(\text{Def of } (-b)^{-1}). \end{align*}$

Then for the other equality, similarly, we have

$\begin{align*} -\left( \frac{a}{b} \right) &= (-a)b^{-1} \\ &= \frac{-a}{b}. \qquad \blacksquare \end{align*}$