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COM-1 Solutions: Formal Notation and Logic

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Section 5 — Guided Practice Solutions

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Problem 1 — The Equals Sign (C1)

Variants 0 and 1: The answer is B — the equals signs create a false chain. In Variant 0, "3x = 12 = x = 4" claims that 12 = x = 4, which is false. In Variant 1, "2 + 8 = 10 ÷ 2 = 5" claims that 2 + 8 = 5, which is false.

Why the other options are wrong (Variants 0 & 1):

Variant 2: The answer is B — this is actually not a run-on equation. "4 × 7 = 28 = 28 cm²" connects statements that are genuinely equal to each other. This is valid (though two lines would be cleaner). The key lesson: not every chain of equals signs is a run-on. A run-on occurs when the things being connected are not actually equal.

Variant 3: The answer is C — the chain "2x + 6 = 14 = 2x = 8 = x = 4" claims 14 = 2x and 14 = 8, both of which are false. The arithmetic (subtract 6, then divide by 2) is perfectly valid; it's the way the steps are joined that's wrong.

Variant 4: The answer is B — the chain "f(3) = 9 = 9 + 1 = 10" claims \( f(3) = 9 \). But \( f(3) = 3^2 + 1 = 10 \), not 9. The intermediate step "= 9" is a false link in the chain. A corrected version would write \( f(3) = 3^2 + 1 = 9 + 1 = 10 \), where every link is true.

What to check if you got this wrong: Ask yourself — is every = sign in the chain actually asserting equality? If "A = B" but A and B have different values, the chain is broken.

Problem 2 — Interval Notation (C5)

Variant 0: \( -2 \leq x < 5 \) → Correct answer: [-2, 5)

Variant 1: \( x > 3 \) → Correct answer: (3, ∞)

Variant 2: \( x \leq -1 \) → Correct answer: (-∞, -1]

Variant 3: \( -3 < x \leq 7 \) → Correct answer: (-3, 7]

Variant 4: \( x \leq 4 \) or \( x > 9 \) → Correct answer: (-∞, 4] ∪ (9, ∞)

Common trap — infinity with brackets: Never write \( [\infty) \) or \( (-\infty] \). Infinity is not a number you can reach, so it is never included. Always use parentheses for infinity: \( (-\infty, \ldots) \) or \( (\ldots, \infty) \).

Problem 3 — Set Membership (C4) — Correct answer: II and III only

  • I. \( \sqrt{2} \in \mathbb{Q} \) — FALSE. \( \sqrt{2} \) is irrational. Proof sketch: assume \( \sqrt{2} = p/q \) in lowest terms; squaring gives \( 2q^2 = p^2 \), so \( p \) is even, say \( p = 2k \); then \( 2q^2 = 4k^2 \Rightarrow q^2 = 2k^2 \), so \( q \) is also even — contradicting "lowest terms." \( \sqrt{2} \in \mathbb{R} \) but \( \sqrt{2} \notin \mathbb{Q} \).
  • II. \( -5 \in \mathbb{Z} \) — TRUE. The integers include all negative whole numbers.
  • III. \( \frac{3}{4} \in \mathbb{R} \) — TRUE. Every rational number is real: \( \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \).
  • IV. \( \pi \in \mathbb{Q} \) — FALSE. \( \pi \) is transcendental (not even a root of any polynomial with rational coefficients), so certainly not rational.

Problem 4 — Exact Values (C6) — Correct answer: √3 (exact form)

\( \sqrt{3} \) is the exact, lossless representation. The decimal approximations 1.73, 1.732, ≈ 1.7 all introduce rounding error. In calculus, if a later step involves \( (\sqrt{3})^2 \), you'd get exactly \( 3 \) — not 2.9929 or 2.999424.

Rule of thumb: The only time you should write a decimal is when the problem specifically asks for a decimal answer, or when you're making a final numerical comparison (e.g., "is this length greater than 2?").

Section 6 — Independent Practice Solutions

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Problem 1 — Fix the Run-On Equation (C1 + C3)

All five variants follow the same pattern: Each step of the original horizontal "chain" becomes its own line, with exactly one operation applied.

Variant 0: \( 5x - 3 = 22 \)

\[ 5x - 3 = 22 \] \[ 5x = 25 \quad (\text{add } 3 \text{ to both sides}) \] \[ x = 5 \quad (\text{divide both sides by } 5) \]

Variant 1: \( 2(x-4) = 10 \)

\[ 2(x - 4) = 10 \] \[ 2x - 8 = 10 \quad (\text{distribute}) \] \[ 2x = 18 \quad (\text{add } 8) \] \[ x = 9 \quad (\text{divide by } 2) \]

Variant 2: \( \dfrac{x}{3} + 1 = 5 \)

\[ \frac{x}{3} + 1 = 5 \] \[ \frac{x}{3} = 4 \quad (\text{subtract } 1) \] \[ x = 12 \quad (\text{multiply by } 3) \]

Variant 3: \( 3(x + 2) = x + 10 \)

\[ 3(x + 2) = x + 10 \] \[ 3x + 6 = x + 10 \quad (\text{distribute}) \] \[ 2x = 4 \quad (\text{subtract } x + 6 \text{ from both sides}) \] \[ x = 2 \quad (\text{divide by } 2) \]

Variant 4: \( x^2 - 5x + 6 = 0 \)

\[ x^2 - 5x + 6 = 0 \] \[ (x - 2)(x - 3) = 0 \quad (\text{factor}) \] \[ \Rightarrow x = 2 \quad \text{or} \quad x = 3 \]

Note the implication arrow \( \Rightarrow \): the factored equation tells us the solutions — we write \( \Rightarrow x = 2 \text{ or } x = 3 \), not "= x = 2 or x = 3".

What to check: Is every line a complete, true equation? Does each line follow from the previous by one clear operation? If you can't say "I applied [operation] to both sides," the step needs its own line.

Problems 2 & 3 — Generator Problems

These problems are randomly generated, so specific solutions depend on the problem you received. Use the pattern below to verify your work.

For interval conversion (Problem 2):

For set-builder to roster (Problem 3):

Problem 4 — Rewrite in Vertical Form (C3)

All five variants follow the same logic: Break the horizontal chain into a vertical stack, one algebraic manipulation per line.

Variant 0: Simplify \( 3(2x+1) - 4x \)

\[ 3(2x + 1) - 4x \] \[ = 6x + 3 - 4x \quad (\text{distribute the 3}) \] \[ = 2x + 3 \quad (\text{collect like terms}) \]

Variant 1: Simplify \( (x+2)^2 - (x-1)^2 \)

\[ (x+2)^2 - (x-1)^2 \] \[ = (x^2 + 4x + 4) - (x^2 - 2x + 1) \quad (\text{expand both squares}) \] \[ = x^2 + 4x + 4 - x^2 + 2x - 1 \quad (\text{distribute the minus sign}) \] \[ = 6x + 3 \quad (\text{collect like terms}) \]

Variant 2: Simplify \( \dfrac{x^2-4}{x-2} \)

\[ \frac{x^2 - 4}{x - 2} \] \[ = \frac{(x-2)(x+2)}{x-2} \quad (\text{factor the numerator}) \] \[ = x + 2, \quad x \neq 2 \quad (\text{cancel, with domain restriction}) \]

The domain restriction \( x \neq 2 \) is mandatory — we divided by \( x-2 \), which requires \( x \neq 2 \). Omitting it is a notation error.

Variant 3: Simplify \( \dfrac{6x^2+9x}{3x} \)

\[ \frac{6x^2 + 9x}{3x} \] \[ = \frac{3x(2x + 3)}{3x} \quad (\text{factor the numerator}) \] \[ = 2x + 3, \quad x \neq 0 \quad (\text{cancel, with domain restriction}) \]

Domain restriction \( x \neq 0 \) is required — dividing by \( 3x \) is only valid when \( x \neq 0 \).

Variant 4: Factor \( (2x-1)^2 + 3(2x-1) \)

\[ (2x - 1)^2 + 3(2x - 1) \] \[ = (2x - 1)\bigl[(2x - 1) + 3\bigr] \quad (\text{factor out the common } (2x-1)) \] \[ = (2x - 1)(2x + 2) \quad (\text{simplify inside the brackets}) \] \[ = 2(2x - 1)(x + 1) \quad (\text{factor out 2 from } (2x+2)) \]

The key insight: treat \( (2x-1) \) as a single unit and factor it out before expanding — much cleaner than expanding first.

Problem 5 — All Six Concepts (C1–C6)

The student's work contained violations of five different concepts:

Original (error-filled):
"x² - 4 = 0 = x² = 4 = x = 2, and the domain is about (-∞, 1.99999…) ∪ (2.0001, ∞)."

  1. C1: Run-on equation "x² - 4 = 0 = x² = 4 = x = 2" — these are not all equal.
  2. C2: The step from \( x^2 = 4 \) to \( x = 2 \) is an implication, not an equality. Also: the student missed \( x = -2 \).
  3. C3: All work should be vertical, one step per line.
  4. C5: Domain should be expressed as \( (-\infty, 2) \cup (2, \infty) \), not in words or approximations.
  5. C6: The boundary point is exactly \( 2 \), not approximately \( 1.99999 \) or \( 2.0001 \).

Corrected work:

\[ x^2 - 4 = 0 \] \[ (x-2)(x+2) = 0 \] \[ \Rightarrow x = 2 \text{ or } x = -2 \]

If the problem specifies domain restriction \( x \neq 2 \), then \( x = -2 \) is the only valid solution.
Domain: \( (-\infty, 2) \cup (2, \infty) \)

Section 7 — Mastery Check Solutions

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Apply Question — Domain of \( h(x) = \sqrt{2x - 6} \)

Step 1: The correct condition is \( 2x - 6 \geq 0 \) (non-negative, not strictly positive, because \( \sqrt{0} = 0 \) is defined).

\[ 2x - 6 \geq 0 \] \[ 2x \geq 6 \] \[ x \geq 3 \]

Domain: \( [3, \infty) \)

In set notation: \( \{ x \in \mathbb{R} \mid x \geq 3 \} \)

What to check if you got \( (3, \infty) \): You may have used strict \( > 0 \) instead of \( \geq 0 \). The value \( x = 3 \) gives \( h(3) = \sqrt0 = 0 \) — perfectly valid. The domain includes 3, so use a square bracket.

Analyze Question — Error in the Domain Reasoning

The student's error was applying only the "denominator ≠ 0" restriction and missing the "radicand ≥ 0" restriction.

Why the student got lucky: The condition \( x + 1 \neq 0 \) gives \( x \neq -1 \), which happens to be a subset of the correct restriction \( x > -1 \). But the reasoning path was wrong — the student didn't consider that negative values of \( x + 1 \) also make the function undefined.

Correct reasoning:

For \( \dfrac{1}{\sqrt{x+1}} \) to be defined, we need both:

(1) Square root: \( x + 1 \geq 0 \Rightarrow x \geq -1 \)

(2) Denominator: \( \sqrt{x+1} \neq 0 \Rightarrow x + 1 \neq 0 \Rightarrow x \neq -1 \)

Combining: \( x \geq -1 \) AND \( x \neq -1 \), so \( x > -1 \).

Domain: \( (-1, \infty) \)

Section 8 — Boss Fight Solutions

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Path A — The Editor: \( k(x) = \sqrt{x-2} + \dfrac{1}{x-5} \)

Complete list of errors in the student's work:

  1. C6: "x > 2.0" — the .0 implies measurement/approximation. The boundary is exactly \( 2 \).
  2. Mathematical error: "x - 2 must be positive" should be "non-negative" (\( \geq 0 \), not \( > 0 \)). The value \( x = 2 \) is valid: \( \sqrt{0} = 0 \).
  3. C1: "domain = x > 2 and x ≠ 5 = (2, ∞) minus 5" — equals signs connecting logical statements.
  4. C5: "(2, ∞) minus 5" is not interval notation. Correct: \( [2, 5) \cup (5, \infty) \).
  5. C6: "(2, 4.9999) ∪ (5.0001, ∞)" — the excluded point is exactly \( 5 \), not approximate.

Corrected solution:

For \( k(x) = \sqrt{x-2} + \dfrac{1}{x-5} \):

\[ x - 2 \geq 0 \Rightarrow x \geq 2 \] \[ x - 5 \neq 0 \Rightarrow x \neq 5 \]

Combining: \( x \geq 2 \) and \( x \neq 5 \).

Domain: \( [2, 5) \cup (5, \infty) \)

Path B — The Translator: Cell Phone Plan

Correct fully-notated solution:

Let \( n \) = number of text messages sent in one month, \( n \in \{ x \in \mathbb{Z} \mid 0 \leq x \leq 200 \} \).

Cost function:

\[ C(n) = 15 + 0.05n \quad (\text{dollars, exact}) \]

Range of costs:

\[ C(0) = 15 + 0.05(0) = 15 \] \[ C(200) = 15 + 0.05(200) = 15 + 10 = 25 \]

Since \( C(n) \) is increasing (slope \( 0.05 > 0 \)), the minimum cost is \( \$15 \) and the maximum is \( \$25 \).

Set of possible monthly costs: \( C \in [15, 25] \)

Key checks:

  • Did you define \( n \) with set notation before using it?
  • Did you write \( 0.05 \) (exact) rather than "5 cents" (imprecise)?
  • Did you use interval notation for the final answer?
  • Did you work vertically to find the min and max?

Section 9 — Challenge Problem Solutions

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Challenge 1 — Set Operations

\( A = \{0, 1, 2, 3, 4, 5\} \) and \( B = \{3, 4, 5, 6, 7, 8\} \)

(a) \( A \cap B = \{3, 4, 5\} \) — elements in both sets. In set-builder: \( \{ x \in \mathbb{Z} \mid 3 \leq x \leq 5 \} \).

(b) \( A \cup B = \{0, 1, 2, 3, 4, 5, 6, 7, 8\} \) — elements in either set. In set-builder: \( \{ x \in \mathbb{Z} \mid 0 \leq x \leq 8 \} \).

(c) \( 4 \in A \cap B \): True — 4 is in both \( A \) and \( B \). \( 2 \in A \cap B \): False — 2 is in \( A \) but not in \( B \).

Challenge 2 — Implication Chain and Biconditional

Forward: \( x^2 < 9 \Rightarrow -3 < x < 3 \)

The converse \( -3 < x < 3 \Rightarrow x^2 < 9 \) is also true: if \( |x| < 3 \), then \( x^2 < 9 \).

Since both directions hold: \( x^2 < 9 \Leftrightarrow -3 < x < 3 \).

Equivalently: \( x^2 < 9 \Leftrightarrow x \in (-3, 3) \).

This is a good example of when \( \Leftrightarrow \) is appropriate: the two conditions describe exactly the same set of numbers, just in different forms.

Challenge 3 — Preview: Slope with Exact Values

(a) \( m = \dfrac{\sqrt{3} - 0}{1 - 0} = \sqrt{3} \)

(b) \( y = \sqrt{3} \cdot 4 = 4\sqrt{3} \)

Note: \( 4\sqrt{3} \approx 6.928 \), but writing \( 4\sqrt{3} \) is exact and far more useful. In the next lesson (COM-2), you'll use exact slope values to write equations of lines, find midpoints, and analyze geometric relationships.

If you made it through all three challenges, you're well beyond the minimum for this lesson. The notation habits built here will save you time and marks throughout this course — and in every math course you take.