Theorems on The Order Properties of The Real Numbers

Theorems on The Order Properties of The Real Numbers

We will now look at some various theorems regarding The Order Properties of Real Numbers. Note that each of the following theorems are relatively elementary, and so it is important not to pre-assume prior knowledge in the following proofs. Furthermore, there may be many ways to prove the following theorems so the reader is encouraged to attempt to prove these theorems in alternative ways if feasible.

 Theorem 1: Let $a$, $b$ and $c$ be real numbers. If $a > b$ and $b > c$ then $a > c$.
• Proof: Since $a > b$ then $(a - b) \in P$. Similarly since $b > c$ then $(b - c) \in P$. To prove that $a > c$, we need to show that $(a - c) \in P$. By the Order Axiom 1, it follows that since $(a - b), (b- c) \in P$ then we have that $((a - b) + (b - c)) \in P$ and so:
(1)
\begin{align} \quad \quad ((a - b) + (b - c)) = a + (-b + (b - c)) = a + ((-b + b) + (-c)) = a + (0 - c) = a - c \end{align}
• Therefore $(a - c) \in P$ which implies that $a > c$. $\blacksquare$
 Theorem 2: Let $a$ and $b$ be real numbers. Then either $a > b$, $a = b$, or $b > a$.
• Proof: By the Order Axiom 3, it follows that either $(a - b) \in P$ or $-(a - b) \in P$ or $(a - b) = 0$.
• If $(a -b) \in P$ then $a > b$.
• If $-(a - b) \in P$ then $-(a - b) = (-1) \cdot (a - b) = (-1) \cdot (a) + (-1) \cdot (-b) = -a -(-b) = -a + b = b - a$, and so $(b - a) \in P$, in other words $b > a$.
• If $a = b$, then $(a - b) = 0$.
• Therefore all three cases have been verified. $\blacksquare$.
 Theorem 3: Let $a$ and $b$ be real numbers. If $a ≥ b$ and $b ≥ a$ then $a = b$.
• Proof: $a ≥ b$ implies that $a > b$ or $(a - b) = 0$, while $b ≥ a$ implies that $b > a$ or $(b - a) = 0$. Therefore $(a - b) = (b - a)$, and so $2a = 2b$, that is $a = b$. $\blacksquare$

Before we look at the next theorem we should note that $a^2 := a \cdot a$, and in fact $a^n := \underbrace{a \cdot ... \cdot a}_{\mathrm{n-times}}$

 Theorem 4: Let $a$ be a real number such that $a \neq 0$. Then $a^2 > 0$.
• Proof: By the Order Axiom 3, since $a \in \mathbb{R}$ it follows that $a \in P$, $-a \in P$ or $a = 0$. But $a \neq 0$, so we will consider the cases where $a \in P$ or $-a \in P$.
• If $a \in P$ then by the Order Axiom 2 the product $a \cdot a = a^2 \in P$ and so $a^2 > 0$.
• If $-a \in P$ then by the Order Axiom 2 the product $(-a) \cdot (-a) = [(-1) \cdot a] \cdot [(1) \cdot a] = [(-1) \cdot (-1)] \cdot a^2 = [1] \cdot a^2 = a^2 \in P$ and so $a^2 > 0$. $\blacksquare$
 Theorem 5: $1 > 0$.
• Proof: We note that $1 = 1 \cdot 1 = 1^2$ so by Theorem 4 it follows that $1^2 = 1 > 0$. $\blacksquare$
 Theorem 6: If $a$, $b$, and $c$ are real numbers such that $a > b$ then $(a + c) > (b + c)$.
• Proof: If $a > b$ then $(a - b) \in P$. So then $(a + c - b - c) = (a + c) - (b + c) \in P$ which implies that $(a + c) > (b + c)$. $\blacksquare$
 Theorem 7: If $a$, $b$, and $c$ are real numbers such that $a > b$ and $c > 0$ then $c \cdot a > c \cdot b$.
• Proof: If $a > b$ then $(a - b) \in P$. Furthermore, if $c > 0$ then $c \in P$. Therefore $c \cdot (a - b) = (c \cdot a - c \cdot b) \in P$. Thus $c \cdot a > c \cdot b$. $\blacksquare$
 Theorem 8: If $a$, $b$, and $c$ are real numbers such that $a > b$ and $c < 0$ then $c \cdot a < c \cdot b$.
• Proof: If $a > b$ then $(a - b) \in P$. Furthermore if $c < 0$ then $-c \in P$. Therefore $-c \cdot (a - b) = ((-c) \cdot (a)) + ((-c) \cdot (-b)) = (-c \cdot a) + (c \cdot b) = c \cdot b - c \cdot a$. So $(c \cdot b - c \cdot a) \in P$ which implies that $c \cdot b > c \cdot a$ or equivalently $c \cdot a < c \cdot b$. $\blacksquare$.
 Theorem 9: If $a$ is a real number such that $a > 0$ then $a^{-1} > 0$.
• Proof: If $a > 0$ then $a \in P$. Now since $a \neq 0$, it follows that $a^{-1} \in \mathbb{R}$. By the Order Axiom 3, $a^{-1} \in P$, $-a^{-1} \in P$ or $a^{-1} = 0$ which is equivalent to saying that $a^{-1} > 0$, $a^{-1} < 0$, or $a^{-1} = 0$. We note that $a^{-1} \neq 0$ since then $aa^{-1} = 1$ and by an earlier theorem $a^{-1}$ can't be 0.
• Now suppose $a^{-1} < 0$. Then $-a^{-1} \in P$ and so $a \cdot (-a^{-1}) \overset{M1} = -(a \cdot a^{-1}) \overset{M4} = -1$. But then by Order Axiom 1, $-1 \in P$, which implies $-1 > 0$ which is a contradiction since by Theorem 8 implies that $1 < 0$ which is false by Theorem 5. So by Order Axiom 3 we have that $a^{-1} > 0$. $\blacksquare$
 Theorem 10: If $a$ and $b$ are real numbers such that $a \cdot b > 0$, then either $a > 0$ and $b > 0$ or $a < 0$ and $b < 0$.
• Proof: Suppose that $a \cdot b > 0$ and without loss of generality suppose that that $a > 0$. Now suppose that $b < 0$ (we will show this will arrive at a contradiction).
• We have that $a \in P$ and by the Order Axiom 3, $-b \in P$. So then $(a) \cdot (-b) = -a \cdot b \in P$ which implies from the Order Axiom 3 that $a \cdot b \not \in P$, which is a contradiction. So our assumption that $b < 0$ was false, so $b > 0$.
• The same logic can be applied to show that it is also possible that $-a < 0$ and $-b < 0$. $\blacksquare$.
 Theorem 11: Let $a$ be a real number. If for every $\epsilon > 0$, $0 ≤ a < \epsilon$ then $a = 0$.
• Proof: We note that $a ≥ 0$ implies that either $a > 0$ or $a = 0$. If $a = 0$, we're done. Now suppose instead $a > 0$ and choose $\epsilon_0 = \frac{1}{2} a$. Then $0 ≤ a < \epsilon_{0} = \frac{1}{2}a$ is untrue, so then $a$ cannot be greater than $0$. $\blacksquare$.