Theorem 2.17

If \(P, Q\) aare statements (or open sentences), then

\[ \begin{gather} P \implies Q\\ \iff\\ (\neg{P}) \vee Q \end{gather} \]

Theorem 2.18 - Fundamental Logical Equivalences

1. • \((P \vee Q) \iff (Q \vee P)\)
   • \((P \wedge Q) \iff (Q \wedge P)\)
2. • \(P \vee (Q \vee R) \iff (P \vee Q) \vee R\)
   • \(P \wedge (Q \wedge R) \iff (P \wedge Q) \wedge R\)
3. • \(P \vee (Q \wedge R) \iff (P \vee Q) \wedge (P \vee R)\)
   • \(P \wedge (Q \vee R) \iff (P \wedge Q) \vee (P \wedge R)\)
4. • \(\neg{(P \vee Q)} \iff (\neg{P}) \wedge (\neg{Q})\)
   • \(\neg{(P \wedge Q)} \iff (\neg{P}) \vee (\neg{Q})\)

Example: Write negations of the following open sentences:

• internet was unstable during lecture, couldn't get these copied down. should be in recording from September 1st!

Quantified Statements

missed intro because of bad internet

• For all/every/each \(x \in S, P(x)\) is written \(\forall x \in S, P(x)\)
  • universal quantifier
• There exists/is/for some/at least one/ \(x \in S\) such that \(P(x)\)
  • \(\exists x \in S, P(x)\)
  • existential quantifier

\[ ~[\exists x \in \mathbb{R}: x^2=3] \iff \forall x \in \mathbb{R}: x^2 \neq 3 \]

Characterizations

Chapter 3

Trivial and Vacuous Proofs

We usually consider

\[ P(x) \implies Q(x) \]

• where the variable(s) \(x \in \mathbb{S}\), universal set

Observations:

• \(P(x) \implies Q(x)\) is True if it is true for all \(x \in \mathbb{S}\)
• \(P(x) \implies Q(x)\) is False if it is false for at least one \(x \in \mathbb{S}\).

Next, recall:

\(P(x)\)	\(Q(x)\)	\(P(x) \implies Q(x)\)
T	T	T
T	F	F
F	T	T
F	F	T

1. Observe: If \(Q(x)\) (the conclusion) is always True, then \(P(x) \implies Q(x)\) (the implication) is True.
2. Called: A trivial proof (the hypothesis is irrelevant)

Example:

1. Let \(x \in \mathbb{R}\).
2. If \(x < 0\), then \(x^2+1 > 0\).
   • Observe: \(\forall x \in \mathbb{R}\), we have \(x^2 \geq 0\).
   • So \(x^2+1 \geq 0 +1 > 0\)
   • Meaning the conclusion is always true.
   • So, this result is trivially true.

Example:

1. Observe: If \(P(x)\) (the hypothesis) is always false, then \(P(x) \implies Q(x)\) is true.
2. Called: A vacuous proof (the conclusion is irrelevant)

Example:

1. \(x \in \mathbb{R}\). If \(x^2 - 2x + 2 \leq 0\) then \(x^3 \geq 8\).
2. Proof:
   • Observe \(\forall x \in \mathbb{R}\), we have
     • \(x^2 - 2x + 2 = (x^2 - 2x + (-1)^2) + 2\)
     • \(= (x-1)^2 + 1 \geq 0 + 1\)
     • \(> 0\)
   • Meaning the hypothesis is always false.
   • So this result is vacuously true.

Recall the classic fact: For all sets \(A, \emptyset \subseteq A\). We can now prove this!

• If \(x \in \emptyset\), then \(x \in A\).
• Note that \(x \in \emptyset\) is always false.
• Therefore, this statement is vacuously true!

Direct Proofs

Familiar Integer Facts We Assume

• the negation of an integer is an integer
• the sum, difference, or product of integers is an integer.

Result 3.4: If \(n\) is an odd integer, then \(3n+7\) is an even integer.

Proof:

• Assume \(n\) is an odd integer.
• So \(n = 2k+1, k \in \mathbb{Z}\).
• We want to show that \(3n+1 = 2a, where a \in \mathbb{Z}\)
• Observe (look!): \[ \begin{align} 3n+7 &= 3(2k+1)+7\\ &= 6k+10\\ &= 2(3k+5)\\ \end{align} \]
• Noting that \(3k+5 \in \mathbb{Z}\), 3n+7 is even. \(\blacksquare\)

General Steps

1. Begin with the hypotehsis
2. Deduce what you can from the hypothesis
3. State your goal/conclusion
4. PROVE the goal!
   • Be guided by the upcoming conclusion
5. End with the conclusion.

Result 3.6: If \(n\) is an odd integer, then \(4n^3+2n-1\) is odd.

Proof:

• Assume \(n\) is an odd integer.
• So \(n = 2y+1, y \in \mathbb{Z}\).
• We want to show that \(4n^3+2n-1 = 2a + 1, a \in \mathbb{Z}\).
• Observe:

\[ \begin{align} 4n^3 + 2n - 1 &= 4(2y+1)^3+2(2y+1)-1\\ &= 2\left[2(2y+1)^3+(2y+1) - 1\right] + 1 \end{align} \]

• Thus, \(4n^3+2n-1\) is odd. \(\blacksquare\)

Proof by Contrapositive

Fact: \(P(x) \implies Q(x) \iff \left[ \neg Q(x) \implies \neg P(x)\right]\)

Proof:

\(P(x)\)	\(Q(x)\)	\(\neg P(x)\)	\(\neg Q(x)\)	\(P(x) \implies Q(x)\)	\(\neg Q(x) \implies \neg P(x)\)
T	T	F	F	T	T
T	F	F	T	F	F
F	T	T	F	T	T
F	F	T	T	T	T

This results in a proof technique called proof by contrapositive.

Result 3.10: let \(x \in \mathbb{Z}\). If \(5x-7\) is even, then \(x\) is odd.

Proof (Contrapositive):

• Assume \(x\) is even, so \(x = 2k, k \in \mathbb{Z}\).
• We want to show that given \(x\) is even, then \(5x-7\) is odd (\(5x-7 = 2n+1, n \in \mathbb{Z}\)).
• Observe: \[ \begin{align} 5x-7 &= 5(2k)-7\\ &= 2(5k)-8 + 1\\ &= 2(5k-4) + 1 \end{align} \]
• So, \(5x-7\) is odd. \(\blacksquare\)

The contrapositive is often used when the hypothesis is difficult, but the conclusion is relatively simple.

Words of Conclusion

So, thus, therefore, hence, whence, thence, then, it follows that, we have, ...

Result: let \(x \in \mathbb{Z}\). \[ x -7 \text{ is even} \iff x \text{ is odd.} \]

Proof:

• Assume \(x\) is odd. Then \(x = 2k+1, k \in \mathbb{Z}\).
• We want to show that \(11x-7 = 2a, a \in \mathbb{Z}\).
• Observe: \[ \begin{align} 11x - 7 &= 11(2k+1) - 7\\ &= 2(11k+11) + 11 - 7\\ &= 2(11k) + 4\\ &= 2(11k+2)\\ \end{align} \]
• So \(11x-7\) is even.

For the forward direction, we'll prove by contrapositive.

• Assume \(x\) is even, so \(x = 2k, k \in \mathbb{Z}\).
• We want to show that \(11x-7 = 2a+1, a \in \mathbb{Z}\).
• Observe: \[ \begin{align} 11x-7 &= 11(2k)-7\\ &= 2(11k) - 8 + 1\\ &= 2(11k-4)+1 \end{align} \]
• So, \(11x-7\) is odd. \(\blacksquare\)

Theorem 3.12

• Let \(x \in \mathbb{Z}\). \(x^2\) is even \(\iff x\) is even.

Proof

• Assume \(x\) is even. In other words, \(x = 2k, k \in \mathbb{Z}\).
• We want to show that \(x^2 = 2a, a \in \mathbb{Z}\).
• Observe: \[ x^2 = (2k)^2 = 2(2k^2) \]
• We'll prove the forward direction by contrapositive.
• Assume \(x\) is odd. In other words, \(x = 2k+1, k \in \mathbb{Z}\).
• We want to show that \(x^2 = 2a + 1, a \in \mathbb{Z}\).
• Observe: \[ \begin{align} x^2 &= (2k+1)^2\\ &= 4k^2+4k+1\\ &= 2(2k^2+2k)+1 \end{align} \]
• So, \(x^2\) is odd. \(\blacksquare\)

What about \(x^2\) is odd \(\iff x\) is odd? We just proved that too!

Result to prove: let \(x \in \mathbb{Z}\). If \(5x-7\) is odd, then \(9x+2\) is even.

• Neither a direct proof nor a contrapositive proof seems useful, or at least neither seem like a clean approach.
• Technique: Introduce a lemma.
  • A lemma helps bridge the gap between hypothesis and conclusion.
  • A lemma is a separate result on its own!

Lemma: let \(x \in \mathbb{Z}\).

• if \(5x-7\) is odd, then \(x\) is even. (invented hypothesis to simplify)
• We'll prove by contrapositive.
• Assume \(x\) is odd, so \(x = 2k+1, k \in \mathbb{Z}\).
• We want to show that \(5x-7 = 2a, a \in \mathbb{Z}\).
• Observe: \[ \begin{align} 5x-7 &= 5(2k+1) - 7\\ &= 2(5k)+5 - 7\\ &= 2(5k) - 2\\ &= 2(5k-1) \end{align} \]
• So, \(5x-7\) is even. \(\blacksquare\)

Result 3.14: let \(x \in \mathbb{Z}\).

• If \(5x-7\) is odd, then \(9x+2\) is even.

Proof: Assume \(5x-7\) is odd. By the above lemma, \(x\) is even.

• So \(x = 2k, k \in \mathbb{Z}\).
• We want to show that \(9x+2 = 2a, a \in \mathbb{Z}\).
• Observe:

\[ \begin{align} 9x + 2 &= 9(2k)+2\\ &= 2(9k+1) \end{align} \]

• So, \(9x+2\) is even. \(\blacksquare\)

Proof by Cases

Examples:

• For \(n \in \mathbb{Z}\), \(n\) is even or \(n\) is odd.
• For \(x, y \in \mathbb{R}, x \neq 0, y \neq 0\), \(xy > 0\) or \(xy < 0\).
  • alternatively..
  • \((x > 0 \text{ and } y > 0)\) or \((x < 0 \text{ and } y < 0)\) or \((x > 0 \text{ and } y < 0)\) or \((x < 0 \text{ and } y > 0)\)

Result 3.15: If \(n \in \mathbb{Z}\), then \(n^2 + 3n+5\) is odd.

Proof: Assume \(n \in \mathbb{Z}\).

• Notice that \(n\) can only be even or odd. If we prove the theorem for both cases, we've proven the theorem generally. Let the conclusion guide you!
• The parity of \(n\) is relevant only because the parity of an expression depending on \(n\) is relevant.
• Case 1: \(n\) is even, so \(n = 2k, k \in \mathbb{Z}\). \[ \begin{align} n^2 + 3n + 5 &= (2k)^2+3(2k)+4+1\\ &= 2(2k^2+3k+2)+1 \end{align} \]
  • So \(n^2+3n+5\) is odd.
• Case 2: \(n\) is odd, so \(n = 2k+1, k \in \mathbb{Z}\). \[ \begin{align} n^2 + 3n + 5 &= (2k+1)^2+3(2k+1)+5\\ &= 4k^2 + 4k + 1 + 6k + 3 + 5\\ &= 4k^2 + 10k + 9\\ &= 2(2k^2 + 5k + 4) + 1 \end{align} \]
  • So \(n^2+3n+5\) is odd. \(\blacksquare\)

An even-ness - oddness tale:

Theorem 3.16: let \(x, y \in \mathbb{Z}\). \(x\) and \(y\) are of the same parity \(\iff x+y\) is even.

• If you're trying to prove a theorem based on \(x\) and \(y\) and it doesn't matter which is which, this is referred to as the theorem being symmetric in \(x\) and \(y\). This means we can swap \(x\) and \(y\) and the conclusion doesn't change. We refer to this as "Without Loss of Generality", or "WLOG".

Theorem 3.17: let \(a, b \in \mathbb{Z}\). \(ab\) is even \(\iff a\) even or \(b\) even.

• Backwards: Assume \(a\) is even or \(b\) is even.
• We want to show that \(ab = 2a, a \in \mathbb{Z}\).
• WLOG, assume \(a\) is even.
  • So \(a = 2k, k \in \mathbb{Z}\).
  • Observe: \(ab = (2k)b = 2(kb)\).
  • So \(ab\) is even.
• Forwards (by contrapositive): Assume \(a\) is odd and \(b\) is odd.
• \(a = 2k+1, b=2j+1, k, j \in \mathbb{Z}\).
• Observe: \[ \begin{align} ab &= (2k+1)(2j+1)\\ &= 4kj + 2k + 2j + 1\\ &= 2(2kj+k+j)+1 \end{align} \]
• So \(ab\) is odd. \(\blacksquare\)

Proof Evaluations

Chapter 4

Proofs Involving Divisibility of Integers

Divides: if \(a, b \in \mathbb{Z}\) with \(a \neq 0\), we say "\(a\) divides \(b\)" if \(\exists c \in \mathbb{Z}: b = ac\).

• We write \(a \mid b\).

• We also say:

  • \(b\) is a multiple of \(a\).
  • \(a\) is a divisor of \(b\).

• If \(a\) does not divide \(b\), we write \(a \nmid b\).

• Old: \(ab\) even \(\iff a\) is even or \(b\) is even.

• New: \(2 \mid ab \iff 2 \mid a \vee 2 \mid b\)

Result 4.1: let \(a, b, c \in \mathbb{Z}\) with \(a \neq 0\) and \(b \neq 0\).

• If \(a \mid b\) and \(b \mid c\), then \(a \mid c\).
• Proof:
  • Assume \(a \mid b\) and \(b \mid c\).
  • Then \(b = ax\) and \(c = by\), \(x, y \in \mathbb{Z}\).
  • We want to show that \(a \mid c\), meaning \(c = a \cdot \square, \square \in \mathbb{Z}\).
  • Observe:

\[ \begin{align} c = by = (ax)y = a(xy) \end{align} \]

• So \(a \mid c\). \(\blacksquare\)

Result 4.3: let \(a, b, c, x, y \in \mathbb{Z}\), where \(a \neq 0\).

• If \(a \mid b\) and \(a \mid c\), then \(a \mid (bx + cy)\).
• Proof:
  • Assume \(a \mid b\) and \(a \mid c\), \(a \neq 0\).
  • Wrong: Then \(b = ax\) and \(c = ay, x, y \in \mathbb{Z}\).
    • We've already used \(x\) and \(y!\)
  • Then \(b = ar\) and \(c = as, r, s \in \mathbb{Z}\).
  • We want to show that \(a \mid (bx+cy\) meaning \(bx+cy = a \cdot \square, \square \in \mathbb{Z}\).
  • Observe:

\[ \begin{align} bx + cy &= (ar)x+(as)y\\ &= a(rx+sy) \end{align} \]

• So \(a \mid (bx+cy)\). \(\blacksquare\)

Result 4.4: let \(x \in \mathbb{Z}\). If \(2 \mid (x-1)\), then \(4 \mid (x-1)\).

• Proof:
• Assume \(2 \mid (x^2-1)\).
  • So \(x^2 - 1 = 2y, y \in \mathbb{Z}\).
  • We want to show that \(4 \mid (x^2-1)\), meaning \(x^2-1 = 4 \cdot square, \square \in \mathbb{Z}\).
    • It's not clear what to do here!
    • That indicates that we probably need to try another path.
  • Try isolating the x term: \(x^2 = 2y+1\).
    • Notice: \(2y+1\) is odd, and we have a theorem saying that \(r^2\) odd \(\iff r\) odd!
    • This means \(x\) is odd.
  • So \(x = 2u+1, u \in \mathbb{Z}\).
  • Observe:

\[ \begin{align} x^2 - 1 &= (2u+1)^2-1\\ &= 4u^2+4u+1-1\\ &= 4(u^2+u) \end{align} \]

• So \(4 \mid (x^2-1)\). \(\blacksquare\)

Result 4.5: Let \(x, y \in \mathbb{Z}\). If \(3 \nmid xy\), then \(3 \nmid x\) and \(3 \nmid y\).

• Proof by Contrapositive
  • Assume \(~(3 \nmid x)\) and \(\neg 3\nmid y\)
    • \(3 \mid x\) or \(3 \mid y\).
  • So \(x = 3s\) or \(y = 3t\), \(s, t \in \mathbb{Z}\).
  • We want to show that \(3 \mid xy\), meaning \(xy = 3\square, \square \in \mathbb{Z}\).
  • WLOG, assume \(3\mid x\), meaning \(x = 3s\).
  • Observe: \(xy = (3s)y = 3(sy)\).
  • So \(3 \mid xy\).

Handy Observations: let \(n \in \mathbb{Z}\).

1. \(n = 2q\) or \(n = 2q+1\), \(q \in \mathbb{Z}\).
2. \(n = 3q\) or \(n = 3q+1\) or \(n = 3q+2\), \(q \in \mathbb{Z}\).
3. For multiples of 2, we have 2 sets. Multiples of 3, 3 sets. Multiples of 4?
   • Notice the pattern :)

Result 4.6 let \(x \in \mathbb{Z}\).

• If \(3 \nmid (x^2-1)\), then \(3 \mid x\).
• Proof by Contrapositive:
  • Assume \(3 \nmid x\).
  • So \(x\) is not a multiple of 3.
  • So by the handy observation, \(x = 3q+1\) or \(x = 3q+2\), \(q \in \mathbb{Z}\).
  • We want to show that \(3 \mid (x^2-1)\), meaning \(x^2-1 = 3\square, \square \in \mathbb{Z}\).
    • Case 1: \(x = 3q+1\) \[ \begin{align} x^2-1 &= (3q+1)^2 - 1\\ &= 9q^2+6q+1-1\\ &= 3(3q^2+2q) \end{align} \]
      • So \(3 \mid (x^2 - 1)\).
    • Case 2: \(x = 3q+2\). \[ \begin{align} x^2 - 1 &= (3q+2)^2\\ &= 9q^2 + 12q + 4 - 1\\ &= 9q^2 + 12q + 3\\ = 3(3q^2+4q+1) \end{align} \]
      • So \(3 \mid (x^2-1)\).

Proofs Involving Congruence of Integers

Motivation

Let \(x, y \in \mathbb{Z}\).

• \(n = 2\): \(x, y\) must have the form \(2q\) or \(2q+1, q \in \mathbb{Z}\).
  • Observe: \(x - y = \{2a-2b, (2a+1) - (2b+1)\} = 2(a-b)\).
• \(n = 3\): \(x, y\) must have the form \(3q, 3q+1, 3q+2, q \in \mathbb{Z}\).
  • Observe: \(x - y = \{3a-3b, (3a+1)-(3b+1), (3a+2)-(3b+2)\} = 3(a-b)\).

For \(a, b, n \in \mathbb{Z}, n \geq 2\), we say \(a\) is congruent to \(b \pmod n\) written \(a \equiv b \pmod n\)

• "\(a\) equals \(b\) \(\pm\) a multiple of \(n\)"
  • means \(n \mid (a-b)\).

New Interpretation: \(x \in \mathbb{Z}\).

• \(x = 2q\) or \(x = 2q+1, q \in \mathbb{Z}\).
• \(x \equiv 0 \pmod 2\) or \(x \equiv 1 \pmod 2\).

Result 4.9: let \(a, b, k, n \in \mathbb{Z}, n \geq 2\).

• If \(a \equiv b \pmod n\), then \(ka \equiv kb \pmod n\).
• Proof:
  • Assume \(a \equiv b \pmod n\).
  • This means \(n \mid (a - b)\).
  • Hence \(a - b = nx, x \in \mathbb{Z}\).
  • We want to show that \(ka \equiv kb \pmod n\), meaning \(n \mid (ka - kb)\)
    • hence \(ka - kb = n\square, \square \in \mathbb{Z}\).
  • Observe:

\[ \begin{align} ka - kb &= k(a - b)\\ &= k(nx)\\ &= n(kx) \end{align} \]

• Thus \(ka \equiv kb \pmod n\). \(\blacksquare\)

Result 4.10: let \(a, b, c, d, n \in \mathbb{Z}, n \geq 2\).

• If \(a \equiv b \pmod n\) and \(c \equiv d \pmod n\)
• Then \(a + c \equiv b + d \pmod n\)
• Proof:
  • Assume \(a \equiv b \pmod n\) and \(c \equiv d \pmod n\).
    • meaning \(n \mid (a-b)\) and \(n \mid (c-d)\)
  • Hence \(a - b = nx\) and \(c - d = ny, x, y \in \mathbb{Z}\)
  • We want to show that \(a + c \equiv b+d \pmod n\), meaning \(n \mid \left[(a+c)-(b+d)\right]\)
  • Observe:

\[ \begin{align} (a + c) - (b + d) &= (a - b) + (c - d)\\ &= nx + ny\\ &= n(x + y) \end{align} \]

• So \(a + c \equiv b + d \pmod n\)

Result 4.11: let \(a, b, c, d, n \in \mathbb{Z}, n \geq 2\).

• If \(a \equiv b \pmod n\) and \(c \equiv d \pmod n\)
• Then \(ac \equiv bd \pmod n\).
• Proof: Assume \(a \equiv b \pmod n\) and \(c \equiv d \pmod n\).
• So, \(a - b = nx\) and \(c - d = ny, x, y \in \mathbb{Z}\).
• We want to show that \(ac \equiv bd \pmod n\), so \(ac-bd = n\square, \square \in \mathbb{Z}\).
  • we have \(a = b+nx\) and \(c = d+ny\)
• Observe:

\[ \begin{align} ac-bd &= (b+nx)(d+ny)-bd\\ &= bd+bny + nxd + nxny - bd\\ &= n(by+xd+xny) \end{align} \]

• So \(ac \equiv bd \pmod n\). \(\blacksquare\)

Result 4.12: let \(n \in \mathbb{Z}\).

• If \(n^2 \not\equiv n \pmod 3\)
• then \(n \not\equiv 0 \pmod 3\) and \(n \not\equiv 1 \pmod 3\)
  • For equals:
    • \(n^2 \neq n \implies [n \neq 0 \wedge n \neq 1]\) means \([n = 0\) or \(n = 1] \implies n^2=n\)
• Proof by Contrapositive
  • Assume \(\neg[n \not\equiv 0 \pmod 3]\) and \(n \not\equiv 1 \pmod 3\).
    • \(n \equiv 0 \pmod 3\) OR \(n \equiv 1 \pmod 3\).
  • We want to show that \(n^2 \equiv n \pmod 3\), so \(n^2 - n = 3\square, \square \in \mathbb{Z}\).
  • Case 1: \(n \equiv 0 \pmod 3\)
    • So \(n - 0 = 3k, k \in \mathbb{Z}\).
    • Observe: \(n^2 - n = (3k)^2-(3k) = 3(3k^2-k)\).
    • So \(n^2 \equiv n \pmod 3\).
  • Case 2: \(n \equiv 1 \pmod 3\).
    • So \(n - 1 = 3j, j \in \mathbb{Z}\).
      • \(n = 1+3j\).
    • Observe: \[ \begin{align} n^2-n &= (1+3j)^2 - (1+3j)\\ &= 1 + 6j + 9j^2 - 1 -3j\\ &= 3(2j+3j^2-j) \end{align} \]
    • So \(n^2 \equiv n \pmod 3\).

Proofs Involving Real Numbers

Facts

• \(a^2 \geq 0, \forall a \in \mathbb{R}\)
• \(a^n \geq 0, \forall a \in \mathbb{R}\) if \(n \in \mathbb{Z}\) is even.
• \(a^n < 0, \forall a < 0\) if \(n \in \mathbb{Z}\) is odd.
• \([ab > 0] \iff a > 0\) INCOMPLETE

Let \(a, b, c \in \mathbb{R}\).

• \(a \geq b\), divide by \(c \geq 0\)
  • \(ac \geq bc\)
• \(a \geq b\), divide by \(c > 0\)
  • \(\tfrac{a}{c} \geq \tfrac{b}{c}\).
• \(a > b\), multiply or divide by \(c > 0\)
  • \(ac > bc\)
  • \(\tfrac{a}{c} > \tfrac{b}{c}\)
• \(a > b\), multiply or divide by \(c < 0\)
  • \(ac < bc\)
  • \(\tfrac{a}{c} < \tfrac{b}{c}\)

Theorem 4.13 (Zero Product Rule)

• Let \(x, y \in \mathbb{R}\).

• If \(xy = 0\), then \(x = 0\) or \(y = 0\).

  • Note: this provides a "proof technique"
    • If a product of reals is 0, then one of those reals must be 0.

• Proof: Assume \(xy = 0\).

  • WLOG, consider \(x\).
  • Case 1: \(x = 0\): Done :)
  • Case 2: \(x \neq 0\).
    • So \(\tfrac{1}{x}\) is defined.
    • Observe: \[ \begin{align} xy &= 0\\ \tfrac{1}{x}(xy) &= \tfrac{1}{x}(0)\\ y &= 0 \quad \blacksquare \end{align} \]

• Alternate Proof (Contrapositive)

  • Assume \(\neg(x = 0 \vee y = 0)\)
    • So \(x \neq 0 \wedge y \neq 0\)
  • From here, solve by cases (\(x > 0, y > 0\), etc)

Result 4.14: Let \(x \in \mathbb{R}\).

• If \(x^3-5x^2+3x = 15\), then \(x = 5\).
• Proof: Assume \(x^3 - 5x^2 + 3x = 15\).

\[ \begin{align} x^3 - 5x^2 + 3x - 15 &= 0\\ x^2(x-5) + 3(x-5) &= 0\\ (x^2+3)(x-5) &= 0\\ \end{align} \]

• By the Zero Product Rule, either \(x^2+3 = 0\) or \(x-5 = 0\).
  • \(x^2+3 = 0 \implies x^2=-3\) (contradiction)
  • \(x - 5 = 0 \implies x=5\) \(\blacksquare\)