Checking cases
Cases
Proposition: If \(n\) is a natural number, then \(1+(-1)^n(2n-1)\) is a multiple of \(4\).
Cases
Proof: Let \(K(n)=1+(-1)^n(2n-1)\). We consider the cases where \(n\) is odd and even separately. When \(n\) is even, \(K(n)=1+(2n-1)=2n\). Since \(n\) is even, \(n=2m\) for some \(m\), and therefore \(K(n)=2(2m)=4m\). Therefore \(K(n)\) is a multiple of \(4\).
When \(n\) is odd, \(K(n)=1-(2n-1)=2-2n\). Since \(n\) is odd, \(n=2m+1\) for some \(m\), and therefore \[K(n)=2-2(2m+1)=2-4m-2=4m\] and again \(K(n)\) is a multiple of \(4\).
Cases
Proposition: (The Triangle Inequality) For any real numbers \(x\) and \(y\), we have \[ |x+y|\le |x|+|y| \]
Note: \(|x|=x\) if \(x\ge 0\), otherwise \(|x|=-x\).
Cases
Proof: There are four cases to consider, depending on the signs of \(x\) and \(y\), and we take them in turn.
\(x\ge 0\) and \(y\ge 0\). Then \(x+y\ge 0.\) Therefore, in this case, \(|x|=x\), \(|y|=y\), and \(|x+y|=x+y\) and so \(|x+y|=|x|+|y|\).
If \(x<0\) and \(y<0\) then \(|x+y|=-x-y=|x|+|y|\).
Cases
- \(x\ge 0\) and \(y<0\).
Cases
- \(x\ge 0\) and \(y<0\).
Note that in this case \(|x|=x\) and \(|y|=-y\).
We also see that \(y<x+y<x\). If \(x+y\ge 0\), then \(|x+y|=x+y=|x|-|y|\le |x|+|y|\). If \(x+y<0\), then \(|x+y|=-x-y=-|x|+|y|\le |x|+|y|\).
The 4th case, \(x<0\) and \(y\ge 0\), follows by the same argument.