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

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.

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.

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.

  1. 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.

  2. 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.
  3. 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.

  4. 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.
  5. 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.

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