Distance

Now we return to two-dimensional space, where the notion of distance is a familiar one. In the figure below, $(x_1,x_2)$ and $(y_1,y_2)$ are two points in the plane, and the distance between them is the hypotenuse of the triangle with vertices
$$(x_1,x_2),(y_1,x_2)\text{ and }(y_1,y_2)$$
or
$$\sqrt{{(y_1-x_1)}^2+{(y_2-x_2)}^2}$$
by the theorem of Pythagoras.

It is of some interest to give a very simple and often unfamiliar proof of the theorem of Pythagoras, as follows: We wish to prove that if $a$ and $b$ are the legs of a right triangle and $c$ is its hypotenuse, then $a^2+b^2=c^2$. Let us form a square of side $a+b$, and decompose it in two ways:

From (1), the area of the square is
$$c^2+4\mathrm{\Delta }$$
where $\mathrm{\Delta }$ is the area of a right triangle with legs a and b. From (2), its area is
$$a^2+b^2+4\mathrm{\Delta }$$
Thus
$$a^2+b^2+4\mathrm{\Delta }=c^2+4\mathrm{\Delta }$$
or$$a^2+b^2=c^2$$
which was to be proved.

By applying the theorem of Pythagoras twice, we see that the distance between two points $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ in 3-space is
$$\sqrt{{(y_1-x_1)}^2+{(y_2-x_2)}^2+{(y_3-x_3)}^2}$$

Thus we are led to define the distance between two points $(x_1,x_2,\dots ,x_n)$ and $(y_1,y_2,\dots ,y_n)$ in n-space to be$$\sqrt{{(y_1-x_1)}^2+{(y_2-x_2)}^2+\dots +{(y_n-x_n)}^2}$$

We observe that this is equal to
$$\sqrt{{(x_1-y_1)}^2+{(x_2-y_2)}^2+\dots +{(x_n-y_n)}^2}$$

We also note that the distance between two points is 0 only if the points are not actually distinct, i.e., only if $x_1=y_1,x_2=y_2,\dots ,x_n=y_n$. (Why?)