# Wow I found a useless equation!!

• 11-03-2003
Lurker
Wow I found a useless equation!!
Here is the equation:

Code:

```high * high - high = low * low + low where high and low are any real number, and high = low + 1```
I was very bored today, and messed around with my calculator until I found this. Completely uninteresting huh :D ?
• 11-03-2003
joshdick
Proof:
Code:

```Given:  x = y + 1.  Now, prove that x^2 - x = y^2 + y. First, square both sides: x^2 = (y + 1)^2 Expand: x^2 = y^2 + 2y + 1 Subtract x from both sides: x^2 - x = y^2 + 2y + 1 - x We were given that x = y + 1, so substitute on the right side of the equation: x^2 - x = y^2 + 2y + 1 - (y + 1) Simplify: x^2 - x = y^2 + y```
Yup :)
• 11-03-2003
Lurker
Well at least its one of the first equations that wasn't burned to the ground on these forums :D .
• 11-03-2003
joshdick
And now because I'm taking a course called Techniques of Math Proof, here's another proof. This one uses case analysis:
Code:

```x^2 - x ?= y^2 + y Factor: x (x - 1) ?= y (y+1) Substitute: x (x - 1) ?= xy Case 1: x != 0 Divide both sides by x:  x - 1 ?= y Add 1 to both sides:  x = y + 1 Case 2:  x = 0 Given that x = y + 1 and x = 0, y = -1. x^2 - x ?= y^2 + y Now, just plug in x = 0 and y = -1: 0 = (-1)^2 + -1 0 = 1-1```
Yeah, I dig proofs.
• 11-03-2003
joshdick
Quote:

Originally posted by Lurker
Well at least its one of the first equations that wasn't burned to the ground on these forums :D .
Well, there are now two proofs supporting your conjecture, so one can't argue with the truth. It still remains useless as you said.
• 11-03-2003
XSquared
For any x, y and a, where they are all elements of R, and y = x - a, x^2 - ax = y^2 + ay.

Code:

```x^2 - ax = (x - a)^2 + a(x-a) x^2 - ax = x^2 - 2ax + a^2 - a^2 + ax x^2 - ax = x^2 - ax```
• 11-03-2003
akirakun
a lot of discoveries in the world of mathematics are useless, so don't feel too bad. ;)
• 11-04-2003
major_small
look at it this way: you can use it as a really long way to find out if there are two consecutive numbers... or to prove that high-low=1, if high is low+1 and they're both real numbers :D
• 11-04-2003
Lurker
Quote:

Originally posted by major_small
look at it this way: you can use it as a really long way to find out if there are two consecutive numbers... or to prove that high-low=1, if high is low+1 and they're both real numbers :D
:D