The Rules of the Game
The real numbers \( \mathbb{R} \) form a field. This means they satisfy standard algebraic properties like commutativity, associativity, and distributivity. But \( \mathbb{R} \) is also an ordered field, meaning there is a strict ordering \( \lt \) that behaves nicely with addition and multiplication.
One of the most important consequences of these properties is the Triangle Inequality, which states that for any \( a, b \in \mathbb{R} \): \[ |a + b| \le |a| + |b| \]
Visualizing the Triangle Inequality
Drag the handles for \( a \) (Blue) and \( b \) (Red) along the number line. Observe how the absolute value of their sum \( |a+b| \) compares to the sum of their absolute values \( |a| + |b| \).
Equality holds when \( a \) and \( b \) have the same sign!
The Archimedean Property
If \( x \gt 0 \) and \( \epsilon \gt 0 \), then for some natural number \( n \in \mathbb{N} \), we have \( n\epsilon \gt x \).
"No matter how far the destination \( x \), and how small your step \( \epsilon \), you will eventually get there."
Practice Problems
True/False
1. For all real numbers \( a \) and \( b \), \( |a + b| = |a| + |b| \).
Show Solution
FALSE - Counterexample: \( a = 1, b = -1 \). Then \( |1 + (-1)| = 0 \) but \( |1| + |-1| = 2 \). The correct statement is \( |a + b| \le |a| + |b| \) (triangle inequality).
2. If \( a \lt b \) and \( c \lt d \), then \( ac \lt bd \).
Show Solution
FALSE - Counterexample: \( a = -2, b = -1, c = 1, d = 2 \). Then \( -2 \lt -1 \) and \( 1 \lt 2 \), but \( ac = -2 \) and \( bd = -2 \), so \( ac = bd \). Need additional conditions (e.g., all positive).
3. If \( |a| \lt |b| \), then \( a \lt b \).
Show Solution
FALSE - Counterexample: \( a = -5, b = 1 \). Then \( |-5| = 5 \gt 1 = |1| \), so \( |a| \gt |b| \) but \( a \lt b \).
4. The inequality \( |a - b| \le |a| + |b| \) holds for all real numbers \( a \) and \( b \).
Show Solution
TRUE - By the triangle inequality: \( |a + (-b)| \le |a| + |-b| = |a| + |b| \), and \( |a - b| = |a + (-b)| \).
Fill in the Blank
5. Prove that if \( a \le b \) and \( b \le c \), then \( a \le c \).
By property _____, we conclude that \( a \le c \).
Show Solution
Property used: O3 (Transitivity)
6. Prove \( |(-5)(3)| = |-5| \cdot |3| \).
\( |(-5)(3)| = |-15| = \) _____
\( |-5| \cdot |3| = \) _____ \( \cdot \) _____ \( = \) _____
Show Solution
\( |(-5)(3)| = 15 \)
\( |-5| \cdot |3| = 5 \cdot 3 = 15 \)
7. Solve the inequality \( |x - 2| \lt 5 \).
\( \iff \) _____ \( \lt x - 2 \lt \) _____ (by \( |a| \lt c \iff -c \lt a \lt c \))
\( \iff \) _____ \( \lt x \lt \) _____
Answer: \( x \in \) (_____, _____)
Show Solution
\( -5 \lt x - 2 \lt 5 \)
\( -3 \lt x \lt 7 \)
Answer: \( (-3, 7) \)
Full Problems
8. Prove: If \( ac = bc \) and \( c \ne 0 \), then \( a = b \).
Show Solution
Given \( ac = bc \) with \( c \ne 0 \). By M4, \( c^{-1} \) exists. Multiply both sides by \( c^{-1} \):
\( (ac)c^{-1} = (bc)c^{-1} \)
\( a(cc^{-1}) = b(cc^{-1}) \) (by M1)
\( a \cdot 1 = b \cdot 1 \) (by M4)
\( a = b \) (by M3) \( \blacksquare \)
9. Prove: If \( 0 \lt a \lt b \), then \( a^2 \lt b^2 \).
Show Solution
Given \( 0 \lt a \lt b \).
By O5 with \( a \lt b \) and \( 0 \lt a \): \( a \cdot a \lt b \cdot a \), i.e., \( a^2 \lt ab \)
By O5 with \( a \lt b \) and \( 0 \lt b \): \( ab \lt b \cdot b \), i.e., \( ab \lt b^2 \)
By O3 (transitivity): \( a^2 \lt ab \) and \( ab \lt b^2 \) implies \( a^2 \lt b^2 \) \( \blacksquare \)
10. Solve: \( |2x + 3| \le 7 \)
Show Solution
\( -7 \le 2x + 3 \le 7 \)
\( -10 \le 2x \le 4 \)
\( -5 \le x \le 2 \)
Answer: \( x \in [-5, 2] \)
11. Prove: \( ||a| - |b|| \le |a - b| \) for all \( a, b \in \mathbb{R} \).
Show Solution
Apply triangle inequality to \( |a| = |(a - b) + b| \): \( |a| \le |a - b| + |b| \), so \( |a| - |b| \le |a - b| \)
Apply triangle inequality to \( |b| = |(b - a) + a| \): \( |b| \le |b - a| + |a| = |a - b| + |a| \), so \( |b| - |a| \le |a - b| \)
This means \( -(|a| - |b|) \le |a - b| \) and \( |a| - |b| \le |a - b| \)
Therefore: \( ||a| - |b|| \le |a - b| \) \( \blacksquare \)
12. Prove: If \( |a - 5| \lt 2 \) and \( |b - 3| \lt 1 \), then \( |a + b - 8| \lt 3 \).
Show Solution
\( |a + b - 8| = |(a - 5) + (b - 3)| \)
\( \le |a - 5| + |b - 3| \) (triangle inequality)
\( \lt 2 + 1 = 3 \) (given conditions) \( \blacksquare \)