Prove without writing all cases that there are more odd numbers from 1 to 199 than even numbers from 1 to 199 Prove without writing all cases that there are more odd numbers from 1 to 199 than even numbers from 1 to 199
How do we do this in the general case from an arbitrary starting point which was 1 in this case and ending point (199 in this case)?
 A: Hint:
There are $100$ even and $100$ odd numbers from $1$ to $200$.
A: Let´s say the starting number is $m$ and the last number is $n$. Then you evaluate if $n-m+1$ is divisible by 2. If this is the case then the number of odd numbers equals the number of even numbers.
If $m-n+1$ is not divisible by 2 then we have 
a) more odd numbers than even numbers if $m$ is an odd number. 
b) more even numbers than odd numbers if $m$ is an even number. 
A: Two adjacent natural numbers will be of different parity.
$\{\color{red}{1,2},\color{blue}{3,4},\color{green}{5,6},\dots\}$
It follows that if there are an even number of consecutive numbers then there are the same number of numbers of each parity type.
Otherwise, it follows that if there are an odd number of consecutive numbers then there is one more number of numbers of the type of the final appearing number.
A: Define functions $O$ and $E$ on the positive integers as follows:
$O(n)$ is the number of odd integers in the set $\{1,2,3,\ldots,n\}$
and $E(n)$ is the number of even integers in the set $\{1,2,3,\ldots,n\}.$
You can show by induction that $O(n)=E(n)$ if $n$ is even and $O(n)=E(n)+1$ if $n$ is odd. 
Then the number of odd integers in the range from $m$ to $n$ inclusive (that is, in the set $\{m,m+1,\ldots,n\}$)
is $O(n)-O(m-1)$
and the number of even integers in the same range is $E(n)-E(m-1).$
The number of odd integers in that range is greater than, equal to, or less than the number of even integers when the difference
$$
(O(n)-O(m-1))-(E(n)-E(m-1)) = (O(n)-E(n))-(O(m-1)-E(m-1))
$$
is greater than, equal to, or less than  zero, respectively.
Now use the fact that $O(k)-E(k)$ is $1$ if $k$ is odd and $0$ if $k$ is even. 
