# Graphing Inequalities in Two Variables

When we have to graph an inequality in two variables, we usually graph the corresponding equality, i.e. the straight line on the coordinate plane, which divides the plane into two parts. Then, we use a so-called 'test point' to decide which of the two parts represents the inequality.
So, here we clearly assume that one of the two parts formed by the straight line represents the '>' inequality, whereas the other part represents the '<' inequality.
Is there any rigorous proof that can show that this is indeed the case? Thanks.

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Well, you need that the functions involved are continuous. Consider for example the inequality $f(x,y)>0$, where $f(x,y)=xy$ if $x$ and $y$ are rational and $f(x,y)=-xy$ otherwise. – Hans Lundmark Jan 9 '12 at 11:30
Moreover, for nonlinear $f$, the set of negative values may not be connected. Take for instance $f(x,y)=y^2-x^3+x$. – lhf Jan 9 '12 at 11:45

It's just the interpretation of the graph of a function:

$$\text{ the point (x,y) is on the graph of the function f if and only if y=f(x).}$$

As an example, consider the inequality $$y>2x-4$$

If you set $$f(x)=2x-4,$$ then the graph of $f$ is precisely the solution set to $y=2x-4$.

Now, if you fix $x=x_0$ and draw a vertical line thorough $x=x_0$, then one portion of the line will be above the graph of $f$ and one below. We have for any point $(x_0,y_1)$ on the portion that's above that $y_1> 2x_0-4=f(x_0)$.

So any point on the line $x=x_0$ above the graph of $f$ is a solution to the inequality. Similarly any point on the line $x=x_0$ below the graph of $f$ is not a solution to the inequality.

The same can be said for other vertical lines. The $y$-coordinate of any point $(x,y)$ above the graph of $f$ will be greater than the $y$ coordinate of the corresponding point $(x, f(x))$ on the graph of $f$; and from this you can say any point above the graph of $f$ is a solution to the inequality.

If $\color{maroon}{(x_0 ,y_1)}$ is on the maroon line segment, then $y_1>f(x_0 )=2x_0-4$.

If $\color{darkgreen}{(x_0 ,y_2)}$ is on the green line segment, then $y_2<f(x_0 )=2x_0-4$.

The above applies to any inequality in $x$ and $y$ that can be written as $y>f(x)$. The reason why the "pick a test point method" works in this case is that the graph of a function does divide the plane into three mutually disjoint and exhaustive sets: the graph of the function (where $y=f(x)$), the region "above the graph" (where $y\gt f(x)$), and the region "below the graph" (where $y\lt f(x)$). So if the test point is in one of these particular regions, then every point in that region will satisfy the same inequality that the test point satisfies.

Of course, things go awry when you have an inequality not of the above form...

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Thank you so much! – AdityaGhosh Jan 9 '12 at 16:42

The assumption you talk about clearly follows from the trichotomy law for real numbers.

http://en.wikipedia.org/wiki/Trichotomy_%28mathematics%29

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Trichotomy states that any two numbers x and y are related as x<y or x=y or x>y. Suppose, we have an inequation 3x + 13y < 7 to graph. Then, clearly, we have either of the cases which holds true: 3x + 13y < 7 or 3x + 13y = 7 or 3x + 13y > 7, which comes from trichotomy. Also, the straight line, divides the coordinate plane into three regions, which again hints at trichotomy. But, how are these two things related??? – AdityaGhosh Jan 9 '12 at 11:09
the value of a function, say $3x+13y$ will always be a real number for $x,y \epsilon R$, so the trichotomy law will apply to the equation of line as well. – NikBels Jan 9 '12 at 11:13