I. We probably noticed long ago that
2 × 2 = 2 + 2,
and no other pair of identical numbers would – when multiplied together – equal the sum of those same two numbers.
(That is, unless you consider zero. But let's just deal with numbers greater than zero as we proceed, and in an addendum coming later, we'll see how zero doesn't compute.)

II. So, let's express 2 × 2 as 2^{2},
and let's express 2 + 2 as 2(2).
Then our initial equation can be represented as
2^{2} = 2(2).

III. Doing some substitution:
Let the number that we are adding or multiplying by itself be expressed as X,
and let the number of times X is involved in each operation as n.
Then, we are looking at this equation:
X^{n} = n(X).
Using this equation (and a minimum of trial and error), we indeed find that X = 2 if n = 2.

IV. If n = 3, what would X be in order to have
X × X × X = X + X + X?
Using our equation, X^{n} = n(X),
we have X^{3} = 3(X).
Then X^{2} = 3, having divided both sides by X.
So X = the square root of 3 which I can only type as 3^{(1/2)},
and this we got by taking the square root of both sides.
Working it out, we get what we expect which is
3^{(1/2)} × 3^{(1/2)} × 3^{(1/2)} = 3^{(1/2)} + 3^{(1/2)} + 3^{(1/2)},
and both sides are indeed the same: 3 × 3^{(1/2)} = 3 × 3^{(1/2)}.

V. One more integer for n:
If n = 4, what would X be in order to have
X × X × X × X = X + X + X + X?
Using our equation, X^{n} = n(X),
we have X^{4} = 4(X).
Then X^{3} = 4, having divided both sides by X.
So X = the cube root of 4 which I can only type as 4^{(1/3)},
and this we got by taking the cube root of both sides.
Working it out, we get what we expect which is
4^{(1/3)} × 4^{(1/3)} × 4^{(1/3)} × 4^{(1/3)} = 4^{(1/3)} + 4^{(1/3)} + 4^{(1/3)} + 4^{(1/3)},
and both sides are indeed the same: 4 × 4^{(1/3)} = 4 × 4^{(1/3)}.

VI. We intended to make our formula universal for determining X when given n, and so far the examples have looked good.
Let's simplify the formula to solve directly for X:
X^{n} = n(X).
Then X^{(n1)} = n, having divided both sides by X.
Now, let's take the "n1" root of both sides:
(n1)root of X^{(n1)} = (n1)root of n.
Then X = (n1)root of n.

VII. So, as seen above:
With n = 3, X = 3^{(1/2)} which is the (n1)root of n – i.e., the square root of 3.
With n = 4, X = 4^{(1/3)} which is the (n1)root of n – i.e., the cube root of 4.
And we could go on and on with whatever n we choose.
