Beginning Activity 2: Constructing a Proof by Contradiction

Beginning Activity Beginning Activity 2: Constructing a Proof by Contradiction
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Consider the following proposition:

Proposition 3.17.

For all real numbers $x$ and $y,$ if $x \neq y,$ $x > 0, and y > 0,$ then $\frac{x}{y} + \frac{y}{x} > 2 .$

To start a proof by contradiction, we assume that this statement is false; that is, we assume the negation is true. Because this is a statement with a universal quantifier, we assume that there exist real numbers

x

and

y

such that

x \neq y,

x > 0, y > 0

and that

\frac{x}{y} + \frac{y}{x} \leq 2 .

(Notice that the negation of the conditional sentence is a conjunction.)

For this proof by contradiction, we will only work with the know column of a know-show table. This is because we do not have a specific goal. The goal is to obtain some contradiction, but we do not know ahead of time what that contradiction will be. Using our assumptions, we can perform algebraic operations on the inequality

\begin{matrix} (4) & \frac{x}{y} + \frac{y}{x} \leq 2 \end{matrix}

until we obtain a contradiction.

1.

Try the following algebraic operations on the inequality in (4). First, multiply both sides of the inequality by $x y,$ which is a positive real number since $x > 0$ and $y > 0 .$ Then, subtract $2 x y$ from both sides of this inequality and finally, factor the left side of the resulting inequality.

2.

Explain why the last inequality you obtained leads to a contradiction.

By obtaining a contradiction, we have proved that the proposition cannot be false, and hence, must be true.

Mathematical Reasoning: Writing and Proof (PreTeXt Edition)

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Beginning Activity Beginning Activity 2: Constructing a Proof by Contradiction
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Proposition 3.17.

1.

2.

Beginning Activity Beginning Activity 2: Constructing a Proof by Contradiction A4US

Proposition 3.17.

1.

2.

Beginning Activity Beginning Activity 2: Constructing a Proof by Contradiction
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