Theorems on The Properties of The Real Numbers

We are now going to look at a bunch of theorems we can now prove using The Axioms of the Field of Real Numbers. All of these theorems are elementary in that they should be relatively obvious to the reader. However, it is important to exercise a bit of caution with the proofs of these theorems (and similar theorems), as we don't want to be hasty in accidentally assuming prior knowledge in our proofs.

Also note that the proofs given are just one of potentially many ways for which these theorems can be proven. Furthermore, in subsequent proofs, we will use earlier proven theorems as justification.

Note also that we are ONLY using the ten axioms from the page above and previously proven theorems. Be sure to be able to identify which axioms/theorems are used when not explicitly stated.

Theorem 1: If $a$ and $b$ are real numbers such that $a + b = a$, then $b = 0$.

• Proof: Suppose that $a + b = a$. We will manipulate both sides of this equation to arrive at the conclusion that $b = 0$.

(1)

\begin{align} a + b = a \\ (a + b) + (-a) = a + (-a) \\ (b + a) + (-a) = a + (-a) \\ b + (a + (-a)) = a + (-a) \\ b + 0 = 0 \\ b = 0 \quad \blacksquare \end{align}

Theorem 2: If $a$ and $b$ are real numbers such that $a \cdot b = a$, then $b = 1$.

• Proof: Suppose that $a \cdot b = a$. We will manipulate both sides of this equation to arrive at the conclusion that $b = 1$.

(2)

\begin{align} a \cdot b = a \\ a^{-1} \cdot (a \cdot b) = a^{-1} \cdot a \\ (a^{-1} \cdot a) \cdot b \overset{M2} = a^{-1} \cdot a \\ 1 \cdot b = 1 \\ b = 1 \quad \blacksquare \end{align}

Theorem 3: If $a$ is a real number then $a \cdot 0 = 0$.

• Proof: Let $a \in \mathbb{R}$.

(3)

\begin{align} a \cdot 0 = 0 \\ a + a \cdot 0 = a + 0 \\ a \cdot 1 + a \cdot 0 = a + 0 \\ a \cdot (1 + 0) = a + 0 \\ a \cdot 1 = a + 0 \\ a = a \\ \end{align}

• Since we have shown that $a + (a \cdot 0) = a + 0$ implies that $a = a$, then we must have that $a \cdot 0 = 0$. $\blacksquare$

Theorem 4: If $a$ and $b$ are real numbers where $b \neq 0$, then if $a \cdot b = 1$, then $b = a^{-1}$.

• Proof: Let $a, b \in \mathbb{R}$ be such that $b \neq 0$ and suppose that $a \cdot b = 1$.

(4)

\begin{align} b = b 1 \cdot b = b \\ (a \cdot a^{-1}) \cdot b = b \\ a^{-1} \cdot (a \cdot b )= b \\ a^{-1}\cdot 1 = b \\ a^{-1} = b \quad \blacksquare \end{align}

Theorem 5: If $a$ and $b$ are real numbers and $a \cdot b = 0$ then $a = 0$ or $b = 0$ or both $a, b = 0$.

• Proof: If both $a, b = 0$ then $0 \cdot 0 = 0$. Now without loss of generality, assume that $a \neq 0$.

(5)

\begin{align} a \cdot b = 0 \\ a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0 \\ a^{-1} \cdot (a \cdot b) \overset{Theorem 3} = 0 \\ (a^{-1} \cdot a) \cdot b = 0 \\ 1 \cdot b = 0 \\ b = 0 \end{align}

• So if $a \neq 0$ then $b = 0$. $\blacksquare$

Theorem 6: If $a$ and $b$ are real numbers such that $a + b = 0$ then $b = -a$.

• Proof: Let $a, b \in \mathbb{R}$ be such that $a + b = 0$.

(6)

\begin{align} a + b = 0 \\ (-a) + (a + b) = (-a) + 0 \\ ((-a) + a) + b = -a \\ 0 + b = -a \\ b = -a \quad \blacksquare \end{align}

Theorem 7: If $a$ is a real number, then $-1 \cdot a = -a$.

• Proof: Let $a \in \mathbb{R}$. To show that $-1 \cdot a = -a$ we will show that both values are additive inverses of $a$, and then by theorem 6 (which says that additive inverses are unique), we can conclude that $-1 \cdot a = -a$.

• First $a + -1 \cdot a = 1 \cdot a + -1 \cdot a = (1 - 1) \cdot a = 0 \cdot a \overset{Theorem 3} = 0$. So $a$ is the additive inverse of $-1 \cdot a$.

• Furthermore $a + (-a) = 0$. Since additive inverses are unique we have that $-1 \cdot a = -a$. $\blacksquare$

Theorem 8: If $a$ is a real number then $-(-a) = a$.

• Proof: Let $a \in \mathbb{R}$. Notice that $(-a) - (-a) = [(-1) \cdot a] - [(-1) \cdot a] = (-1) \cdot (a - a) = -1 \cdot 0 = 0$. So the additive inverse to $-a$ is $-(-a)$. But the additive inverse to $-a$ is also $a$ which implies that $-(-a) = a$. $\blacksquare$

Theorem 9: $(-1) \cdot (-1) = 1$.

• Proof: Notice that $(-1) \cdot (-1) = - (-1) \overset{Theorem 8} = 1$. $\blacksquare$

Theorem 10: If $a$ and $b$ are real numbers then $-(a + b) = (-a) + (-b)$.

• Proof: Let $a, b \in \mathbb{R}$. Then $-(a + b) = (-1) \cdot (a + b) = (-1) \cdot a + (-1) \cdot b = (-a) + (-b)$. $\blacksquare$

Theorem 11: If $a$ and $b$ are real numbers then $(-a) \cdot (-b) = a \cdot b$.

• Proof: Let $a, b \in \mathbb{R}$. Then $(-a) \cdot (-b) = [(-1) \cdot a] \cdot [(-1) \cdot b] = [(-1) \cdot (-1)] \cdot [a \cdot b] \overset{Theorem 9} = 1 \cdot [a \cdot b] = a \cdot b$. $\blacksquare$

Theorem 12: If $a$ and $b$ are real numbers such that $b \neq 0$ then $-(a \cdot b^{-1}) = -a \cdot b^{-1}$.

• Proof: Let $a, b \in \mathbb{R}$ be such that $b \neq 0$. Then $-(a \cdot b^{-1}) = (-1) \cdot (a \cdot b^{-1} ) = [(-1) \cdot a] \cdot b^{-1} = [-a] \cdot b^{-1} = -a \cdot b^{-1}$. $\blacksquare$