Question

Suppose \(n\) (infinitely long) straight lines lie on a plane in such a way that no two of the lines are parallel, and no three of the lines intersect at a single point. Show that this arrangement divides the plane into \(\frac{n^{2}+n+2}{2}\) regions.

Short answer

Yes, it's true. With \(n\) straight lines, which are not parallel and don't intersect at a single point, the plane will be divided into \(\frac{n^{2}+n+2}{2}\) regions.

Step-by-step solution

Base Case

In combinatorics and induction, you often start from the very basic. Start from the case when there are no lines (\(n=0\)). It will create 1 region. When a single line (\(n=1\)) is added, it cuts the plane into 2 regions.

Induction Step: Hypothesis

In the next step, suppose the statement is true for \(n=k\) lines. That is, for \(k\) lines it is assumed there are \(\frac{k^{2}+k+2}{2}\) regions. This is called the induction hypothesis.

Induction Step: Show for \(n=k+1\)

Now, add line \(k+1\). Because it is an infinitely long straight line, it will cross all the \(k\) lines. Since this line is crossing all the other lines, and not meeting at a single point, it cuts the current figure into exactly \(k+1\) new regions. Therefore, the total regions with \(k+1\) lines becomes \(\frac{k^{2}+k+2}{2} + k + 1 = \frac{(k+1)^2+(k+1)+2}{2}\)

Conclusion

By mathematical induction, when there are \(n\) non-parallel lines on a plane, no three of which intersect at a single point, the plane is divided into \(\frac{n^{2}+n+2}{2}\) different regions.