The Number of Points on Two Line Segments

We say that a set is countably infinite if we can pair the elements with set of counting numbers 1, 2, 3, and so on. Believe it or not, the number of positive integers and the number of integers (both negative and positive including 0) have the same number of elements. It is because we can pair them in a one-to-one correspondence such as shown in the below.

As shown on the table, if we continue indefinitely, we know that we can pair each counting number with an integer in a one-to-one correspondence without missing any element.

counting numbers integers bijection

Using this concept, we show intuitively that the number of points on two line segments is equal even if they have different lengths. We can do this by showing that for each point on segment \overline{AB}, there is a corresponding point on segment \overline{CD}.

Given two line segments with different lengths, let us label the longer segment \overline{AB} and the shorter segment \overline{CD}. We place them in a horizontal position such that they do not overlap.

segments with different lengths

Next, connect the end points of the segment and extend them on the part of shorter segment to form a triangle. We call the point of intersection E.

segments with differerent lengths2

Notice that each ray drawn from E that passes through segment \overline{CD} will also pass through segment AB. That is, if ray \overline{EP} passes through segment \overline{CD} at P, then it also passes through \overline{AB} at P^\prime. If a ray passes through \overline{CD} at Q, then it also passes through \overline{AB} at Q^\prime. Clearly, this is true for all rays that can be drawn.

segments with differerent lengths3

 

Now since each point P on \overline{CD} has a corresponding point P^\prime on \overline{AB}, there is a one-to-one correspondence between the points in \overline{CD} and the points on \overline{AB}.  As a result, the number of points on the two segments is equal.

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