Mathematical-Reasoning · Unit 8 · Video 2 · Interactive Practice

Writing a Proof by Contradiction: x/y + y/x and Negations

IKey Formulas

Formula	Name	Use in a proof by contradiction
$\neg(P \to Q) \equiv P \wedge \neg Q$	Negating a conditional	Keep the hypothesis, negate the conclusion: the negation has no "if"
$\neg(P \vee Q) \equiv \neg P \wedge \neg Q$	De Morgan's laws	An "or" becomes an "and" (and $\neg(P \wedge Q) \equiv \neg P \vee \neg Q$ )
$\neg(\forall x \in U)\,P(x) \equiv (\exists x \in U)\,\neg P(x)$	Negating a quantifier	"For each" becomes "there exists", with the $\neg$ moved inside
$t^2 \geq 0$ for every $t \in \mathbb{R}$ , and $t^2 > 0$ if $t \neq 0$	Squares of reals	The impossible statement in both proofs: $(x - y)^2 \leq 0$ with $x \neq y$ , and $(2x - 1)^2 < 0$

Key Insight: A proof by contradiction has no conclusion to work backward from, so only the know column is used: announce the method, assume the negation, and work forward until a statement that cannot be true appears.

IIVisualization 1 — Can Any Point Satisfy Row P?

Row P assumes $x \neq y$ , $x > 0$ , $y > 0$ and $\frac{x}{y} + \frac{y}{x} \leq 2$ ; no point $(x, y)$ meets all four.

x 1.5 y 2

IIIVisualization 2 — Writing the Negation, One Layer at a Time

Each rule removes the $\neg$ from one layer of the statement and moves it one level inward.

Statement

IVVisualization 3 — Working Forward From the Negation

Every row of the know column restates the negation, so every row asks the graph for a point that does not exist.

Row P x 0.3

VQuiz Questions

Problem 1 · Assume the Negation

Given: Proposition: for each real number $x$ , if $x > 0$ , then $x + \frac{1}{x} \geq 2$ . Which assumption begins a proof by contradiction of this proposition?

✅ Correct! The quantifier flips to "there exists", and the negated conditional keeps the hypothesis

x > 0

and negates the conclusion: not

\geq 2

means

< 2

❌ The "if" must go. The negation of a conditional is an "and" statement,

\neg(P \to Q) \equiv P \wedge \neg Q

. With the "if" kept,

x = -1

makes this statement true only because its hypothesis

x > 0

is false.

❌ The quantifier did not change. The negation of "for each

x

\ldots

" is "there exists

x

such that not

\ldots

": one

x

that breaks the claim is enough.

❌ The hypothesis was negated.

\neg(P \to Q) \equiv P \wedge \neg Q

keeps

P

, so

x > 0

stays; only the conclusion

x + \frac{1}{x} \geq 2

is negated.

Show solution

The proposition has the form $(\forall x \in \mathbb{R})\big(P(x) \to Q(x)\big)$ with $P(x)$ : $x > 0$ and $Q(x)$ : $x + \frac{1}{x} \geq 2$ .

Quantifier: $\neg(\forall x \in \mathbb{R})\big(P(x) \to Q(x)\big) \equiv (\exists x \in \mathbb{R})\,\neg\big(P(x) \to Q(x)\big)$ .

Conditional: $\neg\big(P(x) \to Q(x)\big) \equiv P(x) \wedge \neg Q(x)$ .

Reword: not $x + \frac{1}{x} \geq 2$ means $x + \frac{1}{x} < 2$ .

So the proof begins: We will use a proof by contradiction. Assume there exists a real number $x$ such that $x > 0$ and $x + \frac{1}{x} < 2$ .

Problem 2 · Negating an "Or" Statement

Given: Proposition: for each integer $n$ , $n$ is odd or $4 \mid n^2$ . Which statement is its negation?

✅ Correct! "For each" becomes "there exists", and by De Morgan's law the negated "or" becomes an "and": both parts must fail for the same

n

❌ The "or" was kept. This statement needs only one part to fail:

n = 3

makes it true through

4 \nmid 9

, although

3

is odd. By

\neg(P \vee Q) \equiv \neg P \wedge \neg Q

the negation needs both parts to fail.

❌ The quantifier did not change. The negation of "for each

n

" is "there exists

n

": a single counterexample is what the proposition rules out.

❌ The parts were not negated. This joins the two original parts with "and"; the negation needs "not odd" and "

4 \nmid n^2

Show solution

The form is $(\forall n \in \mathbb{Z})\big(P(n) \vee Q(n)\big)$ with $P(n)$ : $n$ is odd and $Q(n)$ : $4 \mid n^2$ . There is no conditional.

\neg(\forall n \in \mathbb{Z})\big(P(n) \vee Q(n)\big) \equiv (\exists n \in \mathbb{Z})\,\neg\big(P(n) \vee Q(n)\big) \equiv (\exists n \in \mathbb{Z})\big(\neg P(n) \wedge \neg Q(n)\big).

In words: there exists an integer $n$ such that $n$ is not odd and $4 \nmid n^2$ . These are the assumptions a proof by contradiction of the proposition would begin with.

Problem 3 · Work the Know Column

Given: Continuing Problem 1, the proof assumes $x > 0$ and $x + \frac{1}{x} < 2$ . Multiply both sides by $x$ , move every term to the left side, and factor.

Which row do you reach?

Why is that row impossible?

✅ Correct!

(x - 1)^2 < 0

is impossible for every real

x

, so the assumption

x > 0

and

x + \frac{1}{x} < 2

cannot hold, and the proposition is true.

❌ The direction changed. Multiplying by

x

keeps

<

because

x > 0

; only a negative factor reverses an inequality.

❌ Check the factoring. The row before factoring is

x^2 - 2x + 1 < 0

, and

x^2 - 2x + 1 = (x - 1)^2

, while

(x + 1)^2 = x^2 + 2x + 1

❌ Both sides must be multiplied.

x\left(x + \frac{1}{x}\right) = x^2 + 1

and

2 \cdot x = 2x

, so the row is

x^2 + 1 < 2x

❌ No such assumption was made. The assumptions are

x > 0

and

x + \frac{1}{x} < 2

; unlike the

\frac{x}{y} + \frac{y}{x}

proof, the contradiction here needs no "

\neq

" condition.

❌ Not a contradiction.

x < 1

and

x > 0

hold together, for example at

x = \frac{1}{2}

; the row clashes with a fact about squares instead.

❌ The row came from that assumption. A row derived from an assumption agrees with it; the clash is with a fact true for every real number.

❌ Check that row. Redo the algebra from

x + \frac{1}{x} < 2

❌ Not quite. Look for a fact that holds for every real number.

Show solution

Step	Know	Reason
$P$	$x > 0$ and $x + \frac{1}{x} < 2$	Assumption (negation)
$P1$	$x^2 + 1 < 2x$	Multiply by $x > 0$
$P2$	$x^2 - 2x + 1 < 0$	Subtract $2x$
$P3$	$(x - 1)^2 < 0$	Factor

But $x - 1$ is a real number, and the square of every real number is $\geq 0$ , so $(x - 1)^2 \geq 0$ . No number is both $< 0$ and $\geq 0$ . This is a contradiction, so the proposition is true: for each $x > 0$ , $x + \frac{1}{x} \geq 2$ .

Equality is allowed in the proposition: at $x = 1$ , $x + \frac{1}{x} = 2$ and $(x - 1)^2 = 0$ , which matches $\geq 2$ , not $> 2$ .

Problem 4 · A Missing Hypothesis

Given: A student "proves" that for all nonzero real numbers $x$ and $y$ , if $x \neq y$ , then $\frac{x}{y} + \frac{y}{x} > 2$ . The proof: "We will use a proof by contradiction. Assume $x \neq y$ and $\frac{x}{y} + \frac{y}{x} \leq 2$ . Multiplying by $xy$ gives $x^2 + y^2 \leq 2xy$ , so $(x - y)^2 \leq 0$ . Since $x \neq y$ , $(x - y)^2 > 0$ , a contradiction." What is wrong?

✅ Correct! Without

x > 0

and

y > 0

xy

may be negative and the multiplication reverses the inequality. The statement itself fails at

x = 1

y = -1

, where

\frac{x}{y} + \frac{y}{x} = -2

❌ Test a point. At

x = 1

y = -1

x \neq y

, both are nonzero, and

\frac{x}{y} + \frac{y}{x} = -1 + (-1) = -2

, which is not

> 2

. A valid proof cannot prove a false statement.

❌ The negation was right.

\neg(P \to Q) \equiv P \wedge \neg Q

keeps the hypothesis

x \neq y

; only the conclusion is negated.

❌ That step is fine. If

x \neq y

then

x - y \neq 0

, and the square of a nonzero real number is

> 0

; the error comes earlier.

Show solution

Multiplying both sides of an inequality by a number keeps its direction only when the number is positive. In the video's proof, $x > 0$ and $y > 0$ made $xy > 0$ . Here only $x \neq 0$ and $y \neq 0$ are given, so $xy$ can be negative.

The statement is in fact false. With $x = 1$ and $y = -1$ :

\frac{x}{y} + \frac{y}{x} = \frac{1}{-1} + \frac{-1}{1} = -2 \leq 2.

Here $xy = -1 < 0$ , and multiplying $-2 \leq 2$ by $-1$ gives $2 \geq -2$ , so $x^2 + y^2 \leq 2xy$ (that is, $2 \leq -2$ ) does not follow. The proof breaks at the multiplication step, which is exactly where the video's proof used $x > 0$ and $y > 0$ .

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