Why isn't the square root is cancelled in this formula?

Question

$\sqrt{\sum\limits_{i=1}^M \vec{V^2_d}(d)}$

This is the formula of the Euclidean length of a vector in the vector space. The vector $V$ has a power of 2 so it is $V^2$. Why isn't the square root of this formula cancels out by the square on this vector?

Answer 1

Let's look at a simple example to see the reason that the proposed way forward gives an incorrect result.

For a right triangle, the Pythogorean Theorem says that the sum of the squares of the lengths of the sides that form a right angle with each other is equal to the square of the length of the hypotenuse. So, we have the well-known $a^2+b^2 = c^2$ result. Taking square roots gives

$$c=\sqrt{a^2+b^2}=\sqrt{\sum_i a_i^2}$$

where $a_1=a$ and $a_2=b$.

Now, if we were to proceed as suggested, then we would have after removal of the square root

$$c=a+b$$

But, this would say that the sum of the lengths of two sides of a right triangle equals the length of the hypotenuse. We know intuitively that this is wrong and of course have a rigorous theorem to confirm that intuition.