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let X=1
X=1
X^2=1
X^2 -1 = 0
(X+1)(X-1) = 1 (difference of two squares)
(X+1) = 1 (divide both sides by (X+1))
put X=1 in and:
(1+1) = 1
2 = 1?????
Where is the error?
Thios one took me a while to figure out, then, when I saw the light, I kicked myself to hard I convinced myself never to kick myself again.
(X+1)(X-1) = 1
2*0<> 1
should be
(X+1)(X-1) = 0
well spoted, should be:
X=1
X^2=1
X^2 -1 = 0
(X+1)(X-1) = 0 (difference of two squares)
(X+1) = 0 (divide both sides by (X+1))
put X=1 in and:
(1+1) = 0
2 = 0?????
There is obviously still a mistake somewhere
Also... x+1=x should be obviously false. Unless x = infinity!
:rolleyes:
Ok.. so
(X+1) = 1
is wrong.
Especially when stated with(divide both sides by (X+1))
Should become...
(x-1) = 0/(x+1)
and if x = 1...
0 = 0
Damn it I'm too tired. Right, I'm gonna get the question right this time.
X=1
X^2=1
X^2 -1 = 0
(X+1)(X-1) = 0 (difference of two squares)
(X+1) = 0 (divide both sides by (X-1))
put X=1 in and:
(1+1) = 0
2 = 0?????
Damnit, I've gone wrong again, but I don't know where. When I did this myslef the last read 2=1. Damn it.:mad:
LOL.
You can divide anything by 0; 0/(x-1)=0/0=undef.
If you do, it's undefined.
So, really..
2=undef.
Which makes the equation bogus.
I hope you meant:Quote:
You can divide anything by 0; 0/(x-1)=0/0=undef.
You can't divide anything by 0; 0/(x-1)=0/0=undef.
as that would have been right. It fooled me for ages, I though of course you can divide by (X-1) as long as you do it to both sides. I never twigged that (X-1) was in fact 0. Maybe I need sleep.
You can... but like I said, it's undefined.
Like you... you did, and we saw what happened.
:lol:
OK, I got home from work, had a pint.
watched a Dvd while drinking 2.5 pints per hour
that was 3 1/4 hours ago, how many pints have i had to drink?
if you can be arsed to work it out then you can have a drink on me because you're not having a fun enough time!
The answer is: enough, i think. Puntcuation is suffering.