try this, it works.
x^2 = (x+1)(x-1)+1
It only works for certain values of 1.
No, it doesn’t. Try it w/ anything, negative #'s, decimals, and “0”. You are wrong.
don’t know nuthin about that number stuff, but I’m pretty sure equasion is spelled with a “t” where that “s” is.
Lessee…
x^2=(x-1)(x+1)-1.
Using the FOIL method, we get…
x^2=x^2+x-x-1-1
x^2=x^2-2
0=-2
Hmmm. Number subsitution doesn’t seem to work either.
Which values of 1 don’t work?
No strainger - her said + 1
(x-1)(x+1) = x^2 - 1
So the equation is correct. Try it with anything you like.
eg (0 - 1) (0 + 1) + 1 = (-1)*1 + 1 = -1 + 1 = 0
pan
The larger ones. As 1 -> infinity the equation fails.
If you take the “+1” off the end, it works.
X^2 + x - x=X^2
the two Xs cancel out, leaving:
X^2=X^2.
You’re leaving off a term - (x+1)(x-1) is x^2+x-x-1
Crap, you’re right. I knew I shouldn’t have been looking back and forth between the post and the reply window while I was tired.
This won’t work in sub-space field harmonics.
You’ll just get a transverse polynomial with a sequential tri-phase declination.
But I guess everybody knows that.
I’m kind of confused as to why clayton_e decided to post this in the first place. Yeah, it works. So what? And did s/he spell “equation” incorrectly on purpose? My guess is that s/he is administering a test to see who would try to prove it wrong and sound like a smarty pants. It worked - s/he caught Strainger and iampunha. Fun. Although I am most definitely not on the same math level as Slip Mahoney, I can easily prove that equation to be true using very basic algebraic math skills. We’re talking 7th grade, here. Hmmm…
Yeah, I don;t get it. We all know it only works for small values of 1 and we all know it is spelled “equashon” so, what’s the deal?
Ok, been a few months, but lets try mathematical induction.
First, prove x is true for x=1
LS = x^2 RS = (x+1)(x-1)+1
= 1^2 = (1+1)(1-1)+1
= 1 = (2)(0)+1
= 0+1
= 1
Now, let us assume x is true for x=k
ie, assume k^2 = (k+1)(k-1)+1
Now, I will try to prove x is true for x=k+1
LS = (k+1)^2 RS = ((k+1)+1)((k+1)-1)+1
= k^2+2k+1 = (k+2)(k)+1
= k^2+2k+1
Since LS=RS, then the equation is true.
As said, it has been a while, so I’m not sure if this is right. I remember having to use step number two in number three, but again, not sure.
That only proves it true for positive real integers.
pan
I’m not sure if to take this seriously. Are you guys serious? Can anyone here multiply?
(x+1) * (x-1) = x^2 - x + x - 1 = x^2 -1
Tht is true for any value of x
now, x^2 = (x+1)(x-1)+1 is the same as x^2 - 1 = (x+1) * (x-1)
does anyone see a slight similarity?
(I feel I must be missing some joke here)
Yeah, Chum’s right, that’s why I started this thread, to have a few ppl try to prove that something true is false, and so I can trash them if I see them ever trying to sound smart in any other thread.
Well, it looks like your evil scheme has paid off, clayton_e. Thanks to my misreading of your “math equasion,” I no longer hold any credibility on the SDMB. I will be debunked in General Questions, discredited in Great Debates, mocked in IMHO, shunned in MPSIMS, and farted at in the BBQ Pit. Also, my cat will puke on my bedsheets and I’ll develop a screaming case of jock itch.
Now, when does school get started again? Soon, isn’t it? Bring some extra cash so you’ll still be able to eat after your lunch money gets stolen. You’ll have to hide it though (hint: rent Papillon).