21127 Lecture Notes - Lecture 12: Contraposition, Logical Biconditional, Rational Number
16 Dec 2015
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Document Summary
Here are several more examples of good, well-written proofs that properly implement the strategies we went over on friday. There is no integral solution to 3a + 15b = 100. 3a + 15b (cid:54)= 100: proof by contradiction, suppose we have a solution. 3a + 15b = 3(a + 5b) = 100. Since a, b z, we know a + 5b z. We conclude that there is no such integral solution to the equation. 3 / z, so a + 5b / z. This is meant to introduce the concept of uniqueness. We say an object with a certain property is unique if it has the right property but no other object has that property. That is, we would say x is the unique element of s with property p (x) if and only if. X s. p (x) ( y s. y (cid:54)= x p (y)) X s. p (x) ( y s {x}.
