Finding $x$ such that $\lfloor x + 1/2 \rfloor = \lfloor x \rfloor$

Question

Find all real numbers $x$ satisfying:

a. $\lfloor x + 1/2 \rfloor = \lfloor x \rfloor$.

b. $\lfloor x + 1/2 \rfloor = \lceil x \rceil$.

Answer 1

This is homework, so you should really post what you have tried so far.

Nevertheless, I will give you the following hint: the floor function returns the integer part of any real number $x$, or in other word, it rounds every number down to the nearest integer.

So $\lfloor 5.9\rfloor = 5$ and $\lfloor 5.1\rfloor = 5$. If you added $1/2 = 0.5$ to $5.9$, would you still get $5$? What about if you added it to $5.1$?

Edit to provide further hints:

Let $5 \le x < 5.5$. Or, in other words, let $x \in [5,5.5)$. Does any value of $x$ in this set work for part (a)? Does every value of $x$ work?

Answer 2

Given a real number $x$, write $x=n+s$, where $n\in\mathbb Z$ and $0\leq s<1$. That is, we have $\lfloor x\rfloor=n$. Consider the following two situations:

1. $s\in[0,1/2)$;
2. $s\in[1/2,1)$.

What can you say about $\lfloor x+1/2\rfloor$ in these two separate situations?

Answer 3

You're just fine up to "$\lfloor s+\frac{1}{2}\rfloor=0\quad$ I don't know what's next". Now you should look at the $s$ values that will keep $s+\frac{1}{2}$ from going above $1$, since then $\lfloor s+\frac{1}{2} \rfloor$ would no longer be $0$. You'll find that there is a limited range of possibilities for $s$, which when you add the integer part $n$ back in, will give you the form of the $x$ answers you need. Once you see this, part (b) should be easy.