Algebraic Properties of Real Numbers

Algebraic Properties of Real Numbers

It is important to first review the Axioms of the Field of Real Numbers page before reading forward. We will be proving many (rather simple) theorems using these axioms and subsequent results that we prove. Note that the proofs given below are not necessarily the only way to prove each theorem. Some properties can be proved in a variety of ways just by using the axioms.

We will also note that in the later proofs on this page, we will use earlier proofs to verify these properties.

Theorem 1: If $a$ and $b$ are real numbers such that $a + b = a$, then $b = 0$.
  • Proof:
(1)
\begin{align} a + b = a \\ a + b + (-a) = a + (-a) \\ a + (-a) + b \overset{A2} = a + (-a) \\ (a + (-a)) + b \overset{A2} = (a + (-a)) \\ (0) + b \overset{A4} = (0) \\ b \overset{A3} = 0 \quad \blacksquare \end{align}
Theorem 2: If $a$ and $b$ are real numbers such that $a \cdot b = a$, then $b = 1$.
  • Proof:
(2)
\begin{align} a\cdot b = a \\ a^{-1}(ab) = a^{-1}a \\ (a^{-1}a)b \overset{M2} = a^{-1}a \\ (1)b \overset{M4} = 1 \\ b \overset{M3} = 1 \quad \blacksquare \end{align}
Theorem 3: If $a$ is a real number then $a \cdot 0 = 0$.
  • Proof:
(3)
\begin{align} a \cdot 0 = 0 \\ a + a \cdot 0 = a + 0 \\ a(1 + 0) \overset{D} = a + 0 \\ a(1) = a + 0 \\ a(1) \overset{A3} = a \\ a \overset{M3} = a \quad \blacksquare \end{align}
  • This implies that $a \cdot 0 = 0$ since we have shown that $a + (a \cdot 0) = a + 0$.
Theorem 4: If $a$ and $b$ are real numbers where $b \neq 0$, then if $a \cdot b = 1$, then $b = a^{-1}$.
  • Proof:
(4)
\begin{align} b = b 1 \cdot b \overset{M3} = b \\ (a \cdot a^{-1}) \cdot b \overset{A4} = b \\ (a^{-1}) \cdot a \cdot b \overset{M1} = b \\ a^{-1}\cdot (1) = b \\ a^{-1} \overset{M3} = 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: Without loss of generality, assume that $b \neq 0$. Then we get that:
(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 \overset{M2} = 0 \\ 1 \cdot b \overset{M4} = 0 \\ b \overset{M3} = 0 \quad \blacksquare \end{align}
Theorem 6: If $a$ and $b$ are real numbers such that $a + b = 0$ then $b = -a$.
  • Proof:
(6)
\begin{align} a + b = 0 \\ (-a) + a + b = (-a) + 0 \\ ((-a) + a) + b \overset{A2} = (-a) + 0 \\ 0 + b \overset{A4} = (-a) + 0 \\ b \overset{A3} = (-a) + 0 \\ b \overset{A3} = -a \quad \blacksquare \end{align}
Theorem 7: If $a$ is a real number, then $(-1)a = -a$.
  • Proof:
(7)
\begin{align} (-1) \cdot a = -a \\ a + (-1) \cdot a = a + (-a) \\ 1a + (-1) \cdot a \overset{M3} = a + (-a) \\ (1 + (-1)) \cdot a \overset{D} = a + (-a) \\ (0) \cdot a = a + (-a) \\ 0 \overset{Theorem\:3}= a + (-a) \\ 0 \overset{A4} = 0 \end{align}
  • Therefore $(-1)a = -a$ since $a + (-1)a = a + (-a)$. $\blacksquare$
Theorem 8: If $a$ is a real number then $-(-a) = a$.
  • Proof:
(8)
\begin{align} -(-a) = a \\ (-a) - (-a) = (-a) + a \\ (-a) - (-a) \overset{A4} 0 \\ a + (-a) - (-a) = a + 0 \\ (a + (-a)) - (-a) \overset{A2} = a + 0 \\ 0 - (-a) \overset{A4} = a + 0 \\ -(-a) \overset{A3} = a \quad \blacksquare \end{align}
Theorem 9: $(-1) \cdot (-1) = 1$.
  • Proof:
(9)
\begin{align} (-1) \cdot (-1) = 1 \\ a \cdot (-1) \cdot (-1) = a \cdot 1 \\ (-1) \cdot a \cdot (-1) \overset{M1} = a \cdot 1 \\ (-a)(-1) \overset{Theorem \: 7} = a \cdot 1 \\ (-1)(-a) \overset{M1} = a \cdot 1 \\ -(-a) \overset{Theorem \: 7} = a \cdot 1 \\ -(-a) \overset{M3} = a \end{align}
  • We showed in Theorem 8 that $-(-a) = a$, therefore, $(-1)(-1) = 1$ since $a \cdot (-1) \cdot (-1) = a \cdot 1$. $\blacksquare$
Theorem 10: If $a$ and $b$ are real numbers then $-(a + b) = (-a) + (-b)$.
  • Proof:
(10)
\begin{align} -(a + b) = (-a) + (-b) \\ (-1) \cdot (a + b) \overset{Theorem \: 7} = (-a) + (-b) \\ (-1) \cdot a + (-1) \cdot b \overset{D} = (-a) + (-b) \\ (-a) + (-b) \overset{Theorem \: 7} = (-a) + (-b) \quad \blacksquare \end{align}
Theorem 11: If $a$ and $b$ are real numbers then $(-a) \cdot (-b) = a \cdot b$.
  • Proof:
(11)
\begin{align} (-a) \cdot (-b) = a \cdot b \\ (-1) \cdot (-a) \cdot (-1) \cdot (b) \overset{Theorem \:7} = a \cdot b \\ (-1) \cdot (-1) \cdot (a) \cdot b \overset{M1} = a \cdot b \\ 1 \cdot a \cdot b \overset{Theorem \: 9} = a \cdot b \\ a \cdot b \overset{M3} = a \cdot b \quad \blacksquare \end{align}
Theorem 12: If $a$ and $b$ are real numbers such that $b \neq 0$ then $-(a \cdot b^{-1}) = (-a) \cdot b^{-1}$.
  • Proof:
(12)
\begin{align} -(a \cdot b^{-1}) = (-a) \cdot b^{-1} \\ (-1) \cdot (a \cdot b^{-1} \overset{Theorem \: 7} = (-a) \cdot b^{-1} \\ ((-1) \cdot a) \cdot b^{-1} \overset{M2} = (-a) \cdot b^{-1} \\ (-a) \cdot b^{-1} \overset{Theorem \: 7} = (-a) \cdot b^{-1} \quad \blacksquare \end{align}
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