# Let $n$ be an integer, prove that $\lfloor n/2 \rfloor \geq (n-1)/2$

So far, I used the definition of floors to provide an interval. Then I did some algebra in order to get $$(n-1)/2$$. And I am able to deduce that $$\lfloor n/2 \rfloor > (n-1)/2$$, but now I'm stuck on proving them to be equivalent to each other.

$$(n/2) - 1 < \lfloor n/2 \rfloor \leq n/2$$ $$\lfloor n/2 \rfloor > (n-2)/2$$ $$\lfloor n/2 \rfloor -1/2 > (n-1)/2$$

Thus, $$\lfloor n/2 \rfloor > (n-1)/2$$.

But how do I prove that they could also be equivalent to each other?

When $$n$$ is even, $$\lfloor n/2 \rfloor = n/2$$.
When $$n$$ is odd, $$n/2 = (n-1)/2 +1/2$$. Since $$n-1$$ is even, $$(n-1)/2$$ is an integer and $$1/2$$ is a fraction. Thus, $$\lfloor n/2 \rfloor = (n-1)/2$$.