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View Full Version : Wow I found a useless equation!!

Lurker
11-03-2003, 06:17 PM
Here is the equation:

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 ?

joshdick
11-03-2003, 06:34 PM
Proof:

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 :)

Lurker
11-03-2003, 06:37 PM
Well at least its one of the first equations that wasn't burned to the ground on these forums :D .

joshdick
11-03-2003, 06:45 PM
And now because I'm taking a course called Techniques of Math Proof, here's another proof. This one uses case analysis:

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.

joshdick
11-03-2003, 06:46 PM
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.

XSquared
11-03-2003, 07:55 PM
For any x, y and a, where they are all elements of R, and y = x - a, x^2 - ax = y^2 + ay.

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

akirakun
11-03-2003, 09:24 PM
a lot of discoveries in the world of mathematics are useless, so don't feel too bad. ;)

major_small
11-04-2003, 07:58 AM
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

Lurker
11-04-2003, 04:23 PM
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