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let x and y be rational numbers.

A. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = \left \lfloor x+y \right \rfloor$

B. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \le \left \lfloor x+y \right \rfloor$

C. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \ge \left \lfloor x+y \right \rfloor$

D. None of the above.

Can anyone please explain why the answer is B? Thank You!

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    Hint: think about $x=0.5$ and $y=1.5$. – 2017-02-24

4 Answers 4

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The fact that $\left\lfloor x\right \rfloor\leq x$ should be clear (let me know if it isn't). Now add this to the same inequality with $y$ to get:

$$\left\lfloor x\right \rfloor+\left\lfloor y\right \rfloor\leq x+y$$

Now, taking the floor of both sides does not change the left-hand side, and gives your answer.

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    This is the most elegant answer – 2017-02-24
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    Yes, please explain why ⌊x⌋≤x ? – 2017-02-25
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    @pkim the floor function of $x$ gives the greatest integer less than or equal to $x$. the key part there is *less than or equal to*. does that make sense? – 2017-02-26
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    Yes it is . Thank you! – 2017-02-26
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    @pkim, glad to hear it! when you get answers that are helpful, be sure to accept the one that helped the most and upvote all the ones you found helpful, too! – 2017-02-27
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Let $\left \lfloor x \right \rfloor = x_{0} + x_{1}, x_{0} \in \mathbb{Z}, 0\leq x_{1}<1$

Similarily we define $\left \lfloor y\right \rfloor$

Then $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$

Now there are two cases to consider:

(i) $x_{1} + y_{1} < 1$

and

(ii) $x_{1} + y_{1} \geq 1$

Case (i):

$\left \lfloor x + y \right \rfloor = \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$

Case (ii):

$\left \lfloor x + y \right \rfloor = x_{0} + y_{0} + 1 > \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor $

Therefore $\forall x,y \in \mathbb{Q}$, $\left \lfloor x + y \right \rfloor \geq \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor$ as required

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    Thanks for your nice comment. I am also a fan of rigorously defined work, and I like yours as well. – 2017-02-24
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If $x$ and $y$ are rational (or real) numbers, then there must exist numbers $h$ and $k$ such that $0 \le h,k < 1$, $x = \left \lfloor x \right \rfloor + h$, and $y = \left \lfloor y \right \rfloor + k$.

So: \begin{align} \left \lfloor x + y \right \rfloor &= \left \lfloor \left \lfloor x \right \rfloor + h + \left \lfloor y \right \rfloor + k \right \rfloor \\ &= \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor + \left \lfloor h+k \right \rfloor\\ &\ge \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor \\ \end{align}

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    Your last inequality symbol is backwards, FYI. – 2017-02-24
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    I fixed it awhile. – 2017-02-24
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    @TheCount - The older I get, the more I do this. – 2017-02-24
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    I feel you. The other day I declared with confidence that $1/3+1/7=1$. I felt so dirty. – 2017-02-24
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For any $x$ then there is an integer $n$ so that $n \le x < n + 1$. We call $n = [x]$.

so $[x] \le x < [x]+ 1; [y] \le y < [y] + 1$ so $[x]+ [y] \le x + y < [x] + [y] + 2$.

so there are two possibilities:

$[x] + [y] \le x + y < [x] + [y] + 1$ and so $[x+y] = [x]+ [y]$. This will happen if $(x -[x]) + (y - [y]) < 1$. e.g. $x = 2.3$ $y = 7.4$.

of $[x]+ [y] + 1 \le x+y < [x]+ [y]+ 1 + 1$ and $ [x+y] = [x]+[y] + 1$. This will happen if $(x -[x]) + (y - [y]) \ge 1$. e.g $x = 2.5$ and $y=7.6$.

So $[x+y] = \{[x]+[y], [x]+[y]+1\}$.

In either case... $[x]+ [y] \le [x+y]$.