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### Insufficient of real numbers
Real numbers are numbers that people could written in decimals, where we allow infinitly many decimal digit and only finitely many integer digit. For example
$$
4,\qquad 3.12,\qquad 1.010010001\cdots,\qquad 5040,\qquad e,\qquad \pi
$$
are all real numbers. The infinite-digit-integer
$$\cdots1111$$ is not a real number

However, real numbers are **insufficient** for our daily use. One phinomenon is that people can not solve the following problem.

> **Ex** Consider two circles $C_1,C_2$ of radius 1 in $x,y$-coordinate plane, the coordinate for the center of $C_1$ is $O_1=(0,0)$, the coordinate for the center of $C_2$ is $O_2=(3,0)$. Find the equation of the line linking two intersection points of the two circle.

One found it impossible to find such a line, because it seems has no intersections. However, such a line **do exists**. See the black line in the following picture.

**Solution with traditional geometry:** This line is **the verticle bisector line** for the two centers $O_1$,$O_2$. The verticle bisector line is the collection of all points that has the same distance with $O_1$ and $O_2$. This line solves the problem with a simple logic: Given that the intersection point of the two circle must have the same distance to $O_1$ and $O_2$, the two intersection points must be on this line. So this is the line linking two intersection points. The equation is $x=3/2$.

**Solution with analytic geometry:** But indeed, this problem can also be solved by analytic geometry. Suppose the two intersection points is $(x_1,y_1)$ and $(x_2,y_2)$. Since it is the intersection points, they all satisfies the following equation

$$
\left\{\begin{array}{ll}x^2+y^2&=1\\x^2-6x+9+y^2&=1\end{array}\right.
$$
So it must satisfies the equation
$$
(x^2+y^2)-(x^2-6x+9+y^2)=1-1=0
$$
which simplifies as
$$
6x-9=0 \iff x=\frac32
$$
So $x=3/2$ is the line where both points are located.

**What's wrong**: Let us try to find the intersection points. We are trying to solve the equation $x^2+y^2=1$, $(x-3)^2+y^2=1$. Now use the first equation we have $y^2=1-x^2$. Substitude this into second equation we have the equation for $x$:
$$
(x-3)^2+(1-x^2)=1\implies -6x+9=0\implies x=\frac32
$$
Yes that is correct, both intersection points **do have** x-coordinate equal to $\frac32$, no wonder why the line $x=3/2$ link them! However, if we try to find the $y$-coordinate
$$
\left(\frac32\right)^2+y^2=1
$$
we stuck with
$$
y^2=-\frac54\qquad y=\frac{\sqrt 5}2\cdot\sqrt{-1}
$$
Here $\sqrt{-1}$ is not a real number, we can not see intersections of the two circle in real coordinates.

This example explains the non-real number **is useful** and without it we can not see a lot of important phinomenon.