Direct Proofs
If P, then Q
Proposition: If P is true, then Q is true.
We are asserting the truth of the implication \(P\implies Q\). If \(P\) is false, then the implication is automatically true. So from a practical point of view, we must show that whenver \(P\) is true, \(Q\) is also true.
Proof: Suppose P.
(bunch of stuff)
Therefore Q.
Simple example
Proposition: The sum of two odd integers is even.
Proposition: The sum of two odd integers is even.
Proof:
Suppose \(a\) and \(b\) are odd integers.
Therefore \(a+b\) is even.
Proposition: The sum of two odd integers is even.
Proof:
Suppose \(a\) and \(b\) are odd integers.
Then there are integers \(k\) and \(u\) so that \(a=2k+1\) and \(b=2u+1\).
Therefore \(a+b\) is even.
Proposition: The sum of two odd integers is even.
Proof:
Suppose \(a\) and \(b\) are odd integers.
Then there are integers \(k\) and \(u\) so that \(a=2k+1\) and \(b=2u+1\).
Therefore \[a+b=(2k+1)+(2u+1)=2(k+u)+2=2(k+u+1)\]
Therefore \(a+b\) is even.
Proposition: The sum of two odd integers is even.
Proof:
Suppose \(a\) and \(b\) are odd integers.
Then there are integers \(k\) and \(u\) so that \(a=2k+1\) and \(b=2u+1\).
Therefore \(a+b=(2k+1)+(2u+1)=2(k+u)+2=2(k+u+1)\)
The definition of even integer says that an integer \(x\) is even if there is an integer \(y\) so that \(x=2y\). We found that \(a+b = 2(k+u+1)\) where \(k+u+1\) is an integer.
Therefore \(a+b\) is even.