It appears from the comments that the actual question is, why is the given formula a good way to see whether two vectors are nearly equal?
The $\sum_1^n(x_i-y_i)^2$ will be small if and only if the vectors $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ are close --- indeed, it measures the square of the Euclidean distance between the two vectors. One could use $\sum_1^n|x_i-y_i|$ instead, but the formula with the squares in it has the advantage of being differentiable, which you lose when you go to absolute values.
Now, $\sum_1^n(x_i-y_i)^2$ gives the same value to $x=(2,0,0),y=(1,0,0)$ and to $x=(2,1,0),y=(1,1,0)$ as it gives to $x=(101,100,100),y=(100,100,100)$. This may be OK for some purposes, but for other purposes, it's absurd. The first pair of vectors, one is twice the other, so it seems strange to say they are nearly equal. The second pair, the vectors don't point at all in the same direction (they differ by about $18$ degrees). The third pair are nearly the same length, and point in nearly the same direction, so they better fit the notion of "nearly equal". And the way to take that into account is to divide by the square of the length of the vector $x$, and that's what the $\sum_1^nx_i^2$ in the denominator does.
Then, when we take the square root of the quotient, we are measuring the distance between two vectors, divided by the length of one of the vectors. So, it's how much they differ, relative to their length. Seems like a pretty good measure of whether they are nearly equal.