Mathematics of today

Remember the HKCEE question of finding the value of sqrt(6+sqrt(6+sqrt(6+sqrt(6…..)?

sqrt means square root of.

To solve it, we first notice that the square of it is equal to 6 added to itself.

Let x be the summation of square roots(I suppose we steal this step via stealth application of the concept of limits), we have: x²=x+6

Solve this quadratic equation give x=3

Similarly, we could have

sqrt(n(n+1)+sqrt(n(n+1)+sqrt(n(n+1)+sqrt(n(n+1)+….)=n+1

i.e.

sqrt(2+sqrt(2+sqrt(2+….)=2
sqrt(6+sqrt(6+sqrt(6+….)=3
sqrt(12+sqrt(12+sqrt(12+….)=4
sqrt(20+sqrt(20+sqrt(20+….)=5
sqrt(30+sqrt(30+sqrt(30+….)=6

therefore,

sqrt(n(n+1)+sqrt(n(n+1)+sqrt(n(n+1)+sqrt(n(n+1)+….)-sqrt(n(n-1)+sqrt(n(n-1)+sqrt(n(n-1)+sqrt(n(n-1)+….)=1

We could reach the conclusion that every natural number is a result of summation of square root. What is the implication on the theory of numbers?

Added 4.08.08:
Therefore, also:
sqrt(n+sqrt(n+sqrt(n+sqrt(n+sqrt(n+sqrt(n+….)=(1+sqrt(4n+1))/2

How about sqrt(n-sqrt(n-sqrt(n-sqrt(n-sqrt(n-sqrt(n-….)?

Using similar technique, we could have: x²=x-n

Therefore the value=(1±sqrt(1-4n))/2

If n>1, this is an imaginary number, suppose we allow that,
then we will have 2 possible values of a single mathematic operation, is that something uncanny?

Next question, what if n is itself an imaginary number?