The real numbers, denoted ℝ, form the foundation of all calculus and analysis. They consist of every point on the number line — no gaps.
| Set | Symbol | Description | Examples |
|---|---|---|---|
| Natural Numbers | ℕ | Counting numbers | 1, 2, 3, 4, ... |
| Whole Numbers | ℕ₀ | Naturals plus zero | 0, 1, 2, 3, ... |
| Integers | ℤ | Whole numbers and negatives | ..., -2, -1, 0, 1, 2, ... |
| Rational Numbers | ℚ | Ratios of integers (p/q, q ≠ 0) | 1/2, -3/4, 7, 0.333... |
| Irrational Numbers | ℝ \ ℚ | Cannot be expressed as ratio | √2, π, e |
| Real Numbers | ℝ | All rationals and irrationals | Everything above |
Key relationship: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ
A number is rational if and only if its decimal expansion either:
A number is irrational if its decimal expansion neither terminates nor repeats.
Example: Is 0.101001000100001... rational or irrational?
The pattern grows but never repeats — it's irrational.
For all real numbers a, b, c:
| Property | Addition | Multiplication |
|---|---|---|
| Closure | \(a + b \in \mathbb{R}\) | \(a \cdot b \in \mathbb{R}\) |
| Commutative | \(a + b = b + a\) | \(a \cdot b = b \cdot a\) |
| Associative | \((a + b) + c = a + (b + c)\) | \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) |
| Identity | \(a + 0 = a\) | \(a \cdot 1 = a\) |
| Inverse | \(a + (-a) = 0\) | \(a \cdot (1/a) = 1\), for \(a \neq 0\) |
| Distributive | \(a \cdot (b + c) = a \cdot b + a \cdot c\) | |
These aren't just abstract rules — they justify every algebraic manipulation you'll ever do.
Example: Why can we "factor out" in the expression \(3x + 3y\)?
By the distributive property: \(3x + 3y = 3(x + y)\)
Example: Why does \((-1)(-1) = 1\)?
We need: \((-1) + (-1)(-1) = (-1)(1 + (-1)) = (-1)(0) = 0\)
So \((-1)(-1)\) must be the additive inverse of \((-1)\), which is 1.
The real numbers are ordered: for any two distinct reals, one is larger.
For any real numbers a and b, exactly one holds:
| Property | Statement |
|---|---|
| Transitive | If \(a < b\) and \(b < c\), then \(a < c\) |
| Addition | If \(a < b\), then \(a + c < b + c\) |
| Multiplication (positive) | If \(a < b\) and \(c > 0\), then \(ac < bc\) |
| Multiplication (negative) | If \(a < b\) and \(c < 0\), then \(ac > bc\) |
Example: Solve \(-2x < 6\)
Dividing by -2 (negative), we reverse: \(x > -3\)
The absolute value of a real number a is:
$$|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}$$
Geometrically: \(|a|\) is the distance from a to 0 on the number line.
For all real a, b:
| Type | Solution |
|---|---|
| Type 1: \(|x| = k\) (where \(k \geq 0\)) | \(x = k\) or \(x = -k\) |
| Type 2: \(|x| < k\) (where \(k > 0\)) | \(-k < x < k\) |
| Type 3: \(|x| > k\) (where \(k \geq 0\)) | \(x < -k\) or \(x > k\) |
Example: Solve \(|2x - 3| \leq 5\)
This means: \(-5 \leq 2x - 3 \leq 5\)
Add 3: \(-2 \leq 2x \leq 8\)
Divide by 2: \(-1 \leq x \leq 4\)
Intervals describe connected subsets of ℝ.
| Notation | Set-Builder | Description |
|---|---|---|
| \((a, b)\) | \(\{x : a < x < b\}\) | Open interval |
| \([a, b]\) | \(\{x : a \leq x \leq b\}\) | Closed interval |
| \([a, b)\) | \(\{x : a \leq x < b\}\) | Half-open (closed left) |
| \((a, b]\) | \(\{x : a < x \leq b\}\) | Half-open (closed right) |
| \((a, \infty)\) | \(\{x : x > a\}\) | Unbounded above |
| \((-\infty, b]\) | \(\{x : x \leq b\}\) | Unbounded below |
| \((-\infty, \infty)\) | ℝ | All real numbers |
Here's the property that separates ℝ from ℚ:
Why ℚ fails this: Consider \(S = \{x \in \mathbb{Q} : x^2 < 2\}\). This set is bounded above (by 2, for instance), but its least upper bound would be \(\sqrt{2}\), which is not in ℚ.
This axiom is the foundation of real analysis — we'll return to it rigorously in Module M7.
Classify each number as natural, integer, rational, or irrational (choose the most specific):
State which property justifies each step:
$$3 + (x + 2) = 3 + (2 + x) = (3 + 2) + x = 5 + x$$
Solve: \(-3x + 7 > 1\)
Solve and express in interval notation: \(|4 - 2x| < 6\)
Prove that the product of two rational numbers is rational.
Solve: \(|x - 3| + |x + 1| = 6\)
Prove: For all real a, b: \(|a - b| \geq ||a| - |b||\)
(This is the reverse triangle inequality)
Let \(S = \{1 - 1/n : n \in \mathbb{N}\}\).
Let's check each category from most specific to least:
Most specific classification: Integer (ℤ)
First, simplify: 4/2 = 2
Now classify 2:
Most specific classification: Natural Number (ℕ)
First, simplify: √9 = 3 (taking the principal/positive root)
Most specific classification: Natural Number (ℕ)
Can √5 be simplified to a nice number? Let's check if 5 is a perfect square: 1, 4, 9, 16, ... — no, 5 is not a perfect square.
Is √5 rational? Suppose √5 = p/q where p, q are integers with no common factors.
Then: \(5 = p^2/q^2\), so \(p^2 = 5q^2\)
This means \(p^2\) is divisible by 5, so p is divisible by 5 (since 5 is prime).
Write p = 5k for some integer k.
Then: \((5k)^2 = 5q^2 \Rightarrow 25k^2 = 5q^2 \Rightarrow 5k^2 = q^2\)
So \(q^2\) is divisible by 5, meaning q is divisible by 5.
But if both p and q are divisible by 5, they share a common factor — contradiction!
Therefore √5 is irrational.
Most specific classification: Irrational (ℝ \ ℚ)
This decimal repeats with period 2 (the block "12" repeats forever).
Any repeating decimal is rational. Let's prove it by finding the fraction:
Let x = 0.121212...
Multiply by 100 (since the repeating block has 2 digits):
100x = 12.121212...
Subtract the original:
100x - x = 12.121212... - 0.121212...
99x = 12
x = 12/99 = 4/33
Verification: 4 ÷ 33 = 0.121212... ✓
Most specific classification: Rational (ℚ)
Step 1: \(3 + (x + 2) = 3 + (2 + x)\)
Inside the parentheses, we swapped x + 2 to 2 + x.
Property: Commutative Property of Addition (a + b = b + a)
Step 2: \(3 + (2 + x) = (3 + 2) + x\)
We regrouped: instead of adding 3 to the quantity (2 + x), we add (3 + 2) to x.
Property: Associative Property of Addition ((a + b) + c = a + (b + c))
Step 3: \((3 + 2) + x = 5 + x\)
We computed 3 + 2 = 5.
Property: Substitution / Arithmetic
Step 1: Isolate the term with x
Subtract 7 from both sides:
\(-3x + 7 - 7 > 1 - 7\)
\(-3x > -6\)
Step 2: Solve for x
Divide both sides by -3.
\(-3x / (-3) < -6 / (-3)\)
\(x < 2\)
Solution: \(x < 2\), or in interval notation: \((-\infty, 2)\)
Verification: Let's test x = 0 (which should work) and x = 3 (which shouldn't):
Step 1: Apply the absolute value inequality rule
For |expression| < k where k > 0, we have: -k < expression < k
So: \(-6 < 4 - 2x < 6\)
Step 2: Solve the compound inequality
Subtract 4 from all parts:
\(-6 - 4 < 4 - 2x - 4 < 6 - 4\)
\(-10 < -2x < 2\)
Step 3: Divide by -2
\(-10 / (-2) > -2x / (-2) > 2 / (-2)\)
\(5 > x > -1\)
Step 4: Rewrite in standard order
\(5 > x > -1\) is the same as \(-1 < x < 5\)
Solution: \(-1 < x < 5\), or in interval notation: \((-1, 5)\)
Let a and b be rational numbers.
By definition of rational, there exist integers p, q, r, s with \(q \neq 0\) and \(s \neq 0\) such that:
Consider their product:
\(a \cdot b = (p/q) \cdot (r/s) = (p \cdot r) / (q \cdot s)\)
Now we verify this is rational:
Therefore \(a \cdot b = (pr)/(qs)\) is a ratio of two integers with nonzero denominator.
By definition, \(a \cdot b\) is rational. ∎
This problem has two absolute values with different expressions inside. We need to consider cases based on where each expression changes sign.
Step 1: Find critical points
|x - 3| changes sign at x = 3
|x + 1| changes sign at x = -1
These divide the number line into three regions: x < -1, -1 ≤ x < 3, and x ≥ 3
Case 1: x < -1
In this region:
The equation becomes:
\((-x + 3) + (-x - 1) = 6\)
\(-2x + 2 = 6\)
\(-2x = 4\)
\(x = -2\)
Check: Is x = -2 in our region x < -1? Yes, -2 < -1 ✓
Verify: |(-2) - 3| + |(-2) + 1| = |-5| + |-1| = 5 + 1 = 6 ✓
x = -2 is a solution.
Case 2: -1 ≤ x < 3
The equation becomes:
\((-x + 3) + (x + 1) = 6\)
\(4 = 6\)
This is a contradiction! No solution exists in this region.
Case 3: x ≥ 3
The equation becomes:
\((x - 3) + (x + 1) = 6\)
\(2x - 2 = 6\)
\(2x = 8\)
\(x = 4\)
Check: Is x = 4 in our region x ≥ 3? Yes, 4 ≥ 3 ✓
Verify: |4 - 3| + |4 + 1| = |1| + |5| = 1 + 5 = 6 ✓
x = 4 is a solution.
Final Solution: x = -2 or x = 4
We'll use the standard triangle inequality: \(|x + y| \leq |x| + |y|\)
Part 1: Show \(|a| - |b| \leq |a - b|\)
Write \(a = (a - b) + b\)
Apply the triangle inequality:
\(|a| = |(a - b) + b| \leq |a - b| + |b|\)
Subtract |b| from both sides:
\(|a| - |b| \leq |a - b|\) ... (Inequality 1)
Part 2: Show \(|b| - |a| \leq |a - b|\)
By symmetry (swap a and b in the argument above):
\(|b| - |a| \leq |b - a|\)
But \(|b - a| = |-(a - b)| = |a - b|\), so:
\(|b| - |a| \leq |a - b|\)
Multiply both sides by -1 (this reverses the inequality):
\(|a| - |b| \geq -|a - b|\) ... (Inequality 2)
Part 3: Combine the inequalities
From Inequality 1: \(|a| - |b| \leq |a - b|\)
From Inequality 2: \(|a| - |b| \geq -|a - b|\)
Together: \(-|a - b| \leq |a| - |b| \leq |a - b|\)
By the definition of absolute value, this means:
\(||a| - |b|| \leq |a - b|\)
Equivalently: \(|a - b| \geq ||a| - |b||\) ∎
Substitute n = 1, 2, 3, 4, 5:
First five elements: {0, 1/2, 2/3, 3/4, 4/5}
Observe the pattern: as n increases, 1/n decreases, so 1 - 1/n increases.
For all \(n \in \mathbb{N}\):
So every element of S is less than 1. S is bounded above by 1.
We claim sup(S) = 1. To prove this, we need to show:
Part 1: We showed above that 1 - 1/n < 1 for all \(n \in \mathbb{N}\), so 1 is an upper bound. ✓
Part 2: Let M < 1. We need to find an element of S that exceeds M.
We want: \(1 - 1/n > M\)
Rearranging: \(1 - M > 1/n\)
So: \(n > 1/(1 - M)\)
Since 1 - M > 0 (because M < 1), the value 1/(1 - M) is a positive real number.
By the Archimedean property, there exists a natural number \(n > 1/(1 - M)\).
For this n: \(1 - 1/n > M\)
So M is not an upper bound of S. ✓
Therefore, sup(S) = 1.
Is 1 ∈ S?
For 1 to be in S, we'd need: 1 - 1/n = 1 for some \(n \in \mathbb{N}\)
This would require: 1/n = 0
But 1/n > 0 for all \(n \in \mathbb{N}\), so there is no natural number n that gives 1 - 1/n = 1.
No, sup(S) = 1 is NOT an element of S.
This concept is fundamental to real analysis — limits, continuity, and the completeness axiom all build on understanding suprema.
Lesson M1.2 covers Algebraic Manipulation and Factoring — rebuilding fluency with polynomials, fractions, and expressions.