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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

Prove there is an irrational number between any two real numbers.

Let x, y \in \mathbb{R} be given with x < y. Prove that there exists an irrational number z such that x < z < y.


Note: To do this problem, I think we need to assume the existence of an irrational number. We will prove the existence of such a number (the \sqrt{2}) in I.3.12, Exercise #12.

Proof. Since the rationals are dense in the reals I.3.12, Exercise #6, we know that for x, y \in \mathbb{R} with x<y there exist r,s  \in \mathbb{Q} such that

    \[ x < r < s < y. \]

Now, assume the existence of an irrational number, say w (see note preceding the proof about this). Since w \in \mathbb{R} we know -w \in \mathbb{R} and from the order axioms exactly one of w or -w is positive (w is nonzero since 0 \in \mathbb{Q}). Without loss of generality, let w > 0. Then, since s-r > 0, we know there exists an integer n such that

    \[ n (s-r) > w \implies s > \frac{w}{n} + r \]

Also, since n,w > 0, we have \frac{w}{n} > 0; thus, r < r + \frac{w}{n}.
Then, by I.3.12, Exercise #7 we have \frac{w}{n} irrational and hence r+\frac{w}{n} irrational.
Thus, letting z = r + \frac{w}{n}, we have x < z < y with z irrational. \qquad \blacksquare

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

Prove that any real number lies between exactly one pair of consecutive integers.

Prove that for fixed x \in \mathbb{R}, there is exactly one n \in \mathbb{Z} such that n \leq x < n+1.


First, we prove a lemma.

Lemma: 1 is the smallest element of \mathbb{Z}^+.
Proof. Consider the set S = \{ x \in \mathbb{R} \mid x \geq 1 \}. Then, S is an inductive set since 1 \in S and for all x \in S, x+1 \in S. Thus, \mathbb{Z}^+ \subseteq S (since, by definition, \mathbb{Z}^+ is the set of elements common to every inductive set). But, for any n \in \mathbb{Z}, if n < 1, then n \notin S and hence n \notin \mathbb{Z}^+. Thus, 1 is the least element of \mathbb{Z}^+.

Now, for the theorem of the exercise.
Proof. First, we prove existence.
Let S = \{ n \in \mathbb{Z} \mid n \leq x \}. We know S is non-empty since by I.3.12, Exercise #2 we know there exist integers m and n such that m < x < n. This m must be in our set S. We also know S is bounded above since x is an upper bound by our definition of S. Therefore, by the least upper bound axiom (Axiom 10, p. 25), we know \sup S exists.
Then, by Theorem I.32, we know that for any positive real number h, there is some n \in S such that

    \[ n > \sup S - h. \]

Letting h = 1, we have

    \[ n > \sup S - 1 \implies \sup S < n+1. \]

Therefore, since the supremum of S is an upper bound, we know n+1 \notin S; therefore, x < n+1 by definition of S. But we already know n \in S, and so by definition of S we must have n \leq x. Therefore, we have found an integer n such that

    \[ n \leq x < n+1. \]

Now, for uniqueness. Suppose there are integers n, n' \in \mathbb{Z} with this property. Then we have

    \begin{alignat*}{3}   &n &\leq x &< n+1  \qquad &\text{and} \qquad &n' &\leq x &< n'+1.\\   \implies& -n &\geq -x &> -n-1 \qquad &\text{and} \qquad &n'+1 &> x &\geq n. \end{alignat*}\end{align*}

Then, adding the terms of these inequalities we have,

    \[ n' + 1 -n > 0 > n'-n-1 \quad \implies \quad 1 > n'-n > -1. \]

Without loss of generality, assume n' \geq n. Then this inequality implies, 0 \leq n' - n < 1. But since 1 is the smallest element of \mathbb{Z}^+ (from the lemma above) and \mathbb{Z} is closed under subtraction, we must have n'-n = 0, i.e., n = n'. Thus, any such n is unique. \qquad \blacksquare

There is a real number between any two given real numbers.

Prove that if x,y \in \mathbb{R} with x < y, then there exists z \in \mathbb{R} such that x < z < y.


Proof. (This is about the point in the blog at which I’m just going to use the basic properties of \mathbb{R} without much comment. As usual, if something is unclear, leave a comment, and I’ll clarify.)

    \begin{align*}   x < y & \implies (y-x) > 0 \\   &\implies 2(y-x) > (y-x) \\   &\implies (y-x) > \frac{y-x}{2}. \end{align*}

Then, \frac{y-x}{2} > 0 since y-x > 0 and \frac{1}{2} > 0 (since 2 > 0 and using I.3.5, Exercise #4) and so their product is also greater than 0.
Then, by adding x,

    \[ 0 < \frac{y-x}{2} < y-x \implies x < \frac{y+x}{2} < y.  \]

So, letting z = \frac{y+x}{2}, we have z \in \mathbb{R} with

    \[ x < z < y, \]

as requested. \qquad \blacksquare