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

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