1. Prove \( n = \lceil n/2 \rceil + \lfloor n/2 \rfloor \) where \(n\) is a nonnegative integer.

We note that \(k = \lceil n/2 \rceil\) and \(m = \lfloor n/2 \rfloor\) are equivalent to \(k-1< n/2 \leq k\) and \(m \leq n/2 < m+1 \) respectively. Since if \(a\leq b\) and \(c < d \), then \(a+c < b+d\), the inequalities show that \(k+m-1 < n < k+m+1\). There is only one integer in this interval: \(m+k = \lceil n/2 \rceil + \lfloor n/2 \rfloor = n\) since there are \(\lceil k+m+1 \rceil - \lfloor k+m-1 \rfloor - 1 = k+m\) many integers in the interval.