Tag: Reals

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 $. \qquad \blacksquare$

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

by RoRi

July 1, 2015

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.

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 $. \qquad \blacksquare$

There is exactly one integer between any real number x and x+1.

by RoRi

June 30, 2015

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

Proof. By I.3.12, Exercise #4 we know that for $x \in \mathbb{R}$ there is exactly one $n \in \mathbb{Z}$ such that

$n \leq x < n+1.$

If $n = x$ , then we already done since $x \leq n < x+1$ .
If $n \neq x$ , then we have $n < x$ and so $n+1 < x+1$ . But we already have $x < n+1$ ; hence,

$x < n+1 < x+1 \quad \implies \quad x \leq n+1 < x+1.$

So, $n+1$ is an integer with the desired properties. Uniqueness follows from the uniqueness of our choice of $n. \qquad \blacksquare$

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

by RoRi

June 30, 2015

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 $. \qquad \blacksquare$

For a positive real x, there is always a positive integer whose reciprocal is smaller than x.

by RoRi

June 30, 2015

Prove that if $0 < x$ , then there is an $n \in \mathbb{Z}^+$ such that $1/n < x$ .

Proof. Since $x > 0$ , we know that for any $y \in \mathbb{R}$ there exists an $n \in \mathbb{Z}^+$ such that $nx > y$ (Theorem I.30, p. 26 of Apostol). Let $y = 1$ , then we have $nx > 1$ and so $\frac{1}{n} < x. \qquad \blacksquare$

Prove that every fixed real number is between two integers.

by RoRi

June 30, 2015

Prove that if $x \in \mathbb{R}$ is fixed, then there exist $m, n \in \mathbb{Z}$ such that $m < x < n$ .

Proof. Since the set of positive integers is unbounded above (Example #1, p. 24 of Apostol) we know there exists an $n \in \mathbb{Z}^+$ such that $x < n$ (otherwise $x$ would be an upper bound on $\mathbb{Z}^+$ ).
Then, by the same logic there is some $(-m) \in \mathbb{Z}^+$ such that $-x < -m$ ; hence, $m < x$ . Since $(-m) \in \mathbb{Z}^+$ , we know $-m \in \mathbb{Z}$ . Thus, we have found $m,n \in \mathbb{Z}$ such that $m < x < n. \qquad \blacksquare$

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

by RoRi

June 30, 2015

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 $. \qquad \blacksquare$

Prove a squeeze-like property for reals.

by RoRi

June 30, 2015

If $x \in \mathbb{R}$ and $0 \leq x < h$ for all $h \in \mathbb{R}^+$ , then $x = 0$ .

Proof. Suppose otherwise, that $0 \leq x < h$ for all $h \in \mathbb{R}^+$ , but $x \neq 0$ . Then, we have $0 \leq x$ and $0 \neq x$ implies $0 < x$ , so $x$ is a positive real number. Then let $h = x$ (since $x \in \mathbb{R}^+$ this is a valid choice of $h$ ). By assumption we then have $x < h = x$ (since $x < h$ for all $h \in \mathbb{R}^+$ ), but this is a contradiction since $x = x$ (equality is symmetric) implies $x \not< x$ . Therefore, we must have $x = 0. \qquad \blacksquare$