I Found a Way to Simplify √(a² + b²) — Here’s the Proof

In May 2020, I sought to tackle one of Algebra’s unspoken rules: √(a² + b²) cannot be simplified any further.

This assumption is OFFICIALLY DEBUNKED: I found a way to refactor this expression while highlighting a new way of representing values through recursion:

• where … represents a recursive series of b² — 2a*sqrt(…)

Proof

It may seem to good to be true, but notice that squaring both terms gives you a² + b² because the recursive identity when squared “forces” the cancellation of the middle term.

Theorem in Action

Furthermore, this recursive identity does seem to approximate many a,b pairs:

1. Let a = 5, b = 12, s qrt(a²+b²) = 13

after including the recursive term 9 times

✅ Correct Approximation

2. Let a = 3, b = 9, sqrt(a²+b²) = 9.486

✅ Correct Approximation

However, for other pairs the term in the middle becomes negative during approximation and thus the term becomes imaginary.

3. Let a = 1, b = 1, sqrt(a²+b²) = sqrt(2) ≈ 1.414

❌ Does not converge

This does not negate the recursive identity, but highlights how the “approximation” does not converge in the real plane for all pairs.

For the interior term to not be negative we want b >>> a. It may be possible to bypass this issue by refactoring a and b so that they represent the same squared sum, but have a larger difference.

Applications

It is interesting to see what applications this new perspective entails. For example, with regards to the Pythagorean Theorem, it is quite interesting how the hypotenuse between two numbers sqrt(a²+ b²) is equal to a point length from a + a recursive difference of a and b.

There are potential valuable applications for geometry (calculating distances) and numerical methods (approximations for radical sums)

Please let me know if you have any thoughts. Feel free to cite this, and let me know if anything like this has been discovered before. I believe it hasn’t.

Thanks 👊

https://www.linkedin.com/in/carlos-hernandez2/

https://github.com/carloshernandez201