Beginning Activity Beginning Activity 2: Constructing a Proof by Contradiction
Consider the following proposition: For all real numbers \(x\) and \(y\text{,}\) if \(x \ne y\text{,}\) \(x > 0,\text{ and } y > 0\text{,}\) then \(\dfrac{x}{y} + \dfrac{y}{x} > 2\text{.}\)
Proposition 3.17.
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
until we obtain a contradiction.
1.
Try the following algebraic operations on the inequality in (4). First, multiply both sides of the inequality by \(xy\text{,}\) which is a positive real number since \(x > 0\) and \(y > 0\text{.}\) Then, subtract \(2xy\) 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.