What is the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis?

Question

Question: If $x \ge 0$ and $y \ge 0$, then the area bounded by the graph of $ \lfloor x \rfloor + \lfloor y \rfloor =2 $ with the $x$ and $y$-axis is? Answer provided: $3$ units$^2$.

My doubt: How do we graph this relation in order to find the area?

Answer 1

As pointed out, the "graph" of the relation is not a one-dimensional curve but a series of non-overlapping regions. Presumably the question is about finding the area of the regions in the first quadrant.

If $x,y\ge0$, then also $\lfloor x\rfloor,\lfloor y\rfloor\ge0$ and so $\lfloor x\rfloor,\lfloor y\rfloor=0,1,2$ are the only possible values.

First consider $\lfloor y\rfloor=0$ (i.e. $0\le y<1$). We have $\lfloor x\rfloor=2$, that is, $2\le x<3$; hence we have the unit square $$\{(x,y):2\le x<3,0\le y<1\}$$ Similarly $\lfloor y\rfloor=1$ gives rise to the unit square $$\{(x,y):1\le x,y<2\}$$ and $\lfloor y\rfloor=2$ to the unit square $$\{(x,y):0\le x<1,2\le y<3\}$$ These squares are non-overlapping; hence the total area is $3$.

Answer 2

When I graph via Desmos, here, the area is 6. Note that the points $(0,3), (1,3), (1,2), (2,2), (2,1), (3,1), (3,0)$ are not part of the bounded area but are all arbitrarily close to it. (In other words $(2,2)$ isn't part of bounded area, but $2-\epsilon,2-\delta$ (for arbitrarily small positive $0<\epsilon,\delta<1$) is.) Along with the point $(0,0)$ they outline the area.

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EDIT

Thanks to input from GFauxPas and AkivaWeinberger, I contend that the upper boundary looks like this (image created by AkivaWeinberger), with the upper boundary "open" in the sense that $(2,2)$ isn't part of relation, but $2-\epsilon,2-\epsilon$ (for arbitrarily small positive $0<\epsilon<1$) is. This assumes $x,y \in \mathbb{R}$, which isn't specified in OP.

IMO, this leads to ambiguity in determining whether the area occupied by the relation is to be included in the area bounded by the relation, the x-axis, and the y-axis. There are 4 possibilities I see:

1. $x,y$ are real And area bounded by relation (series of diagonally linked squares) is to be included in area asked for in OP.
2. $x,y$ are real And area bounded by relation is not to be included in bounds in area asked for in OP.
3. $x,y$ integer, which makes relationship a curve without area.
4. $x,y$ real, And interpret conditions as saying what is the area of the relation in Quadrant I. In other words the OP saying $x,y$ axis are boundaries i meant to bound the relation, not the area. That interpretation makes the condition $x,y \ge 0$ redundant, but nonetheless the use of "with" instead "and" lends weight to this interpretation.

Since OP says answer specifies an area of $3$, that would eliminate #1, but assuming the answer a priori isn't ideal.

Bottom line: question needs clarification.

Answer 3

$\lfloor x \rfloor$ and $\lfloor y \rfloor$ are nonnegative integers. So find the pairs of nonnegative integers $(A,B)$ so that $A + B = 2$. For each such $(A,B)$, find the set of $(x,y)$ values for which $\lfloor x \rfloor = A$ and $\lfloor y \rfloor = B$.

The result isn't a curve. For $(A,B) = (1,1)$, the set of all $(x,y)$ values is every point inside the square with bottom left corner $(1,1)$ and top right corner $(2,2)$. The square includes the bottom and left boundaries, but none of the top and right boundaries.

Answer 4

Note that $$ \left\lfloor x \right\rfloor = a\quad \Rightarrow \quad a \leqslant x < a + 1 $$ and therefore $$ \begin{gathered} \left\{ \begin{gathered} \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2 \hfill \\ 0 \leqslant \left\lfloor x \right\rfloor ,\left\lfloor y \right\rfloor \hfill \\ \end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered} 0 \leqslant a \leqslant x < a + 1 \hfill \\ 0 \leqslant 2 - a \leqslant y < 2 - a + 1 \hfill \\ \end{gathered} \right. \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} 0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\ a \leqslant x < a + 1 \hfill \\ 2 - a \leqslant y < 2 - a + 1 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$

Therefore the answer $3$ refers to the area of the region defined by the condition ${\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2}$ in itself. $$ \begin{gathered} R = \left\{ {(x,y):\;\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2} \right\}\quad \Rightarrow \quad \int\limits_{\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor = 2} {dxdy} = \hfill \\ = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{x = a}^{a + 1} {\int_{y = 2 - a}^{2 - a + 1} {dxdy} } } = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} 1 = 3 \hfill \\ \end{gathered} $$

The area bounded by the given condition and the $x$ and $y$ axes will be instead $$ \begin{gathered} S = \left\{ {(x,y):\;\left\{ \begin{gathered} \;0 \leqslant \text{integer}\;a \leqslant 2 \hfill \\ 0 \leqslant x < a + 1 \hfill \\ 0 \leqslant y < 2 - a + 1 \hfill \\ \end{gathered} \right.\quad } \right\}\quad \Rightarrow \quad \int\limits_S {dxdy} = \hfill \\ = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\int_{y = 0}^{2 - a + 1} {dy} } = \sum\limits_{0\, \leqslant \,a\, \leqslant 2} {\left( {2 - a + 1} \right)} = \hfill \\ = 3 + 2 + 1 = 6 \hfill \\ \end{gathered} $$