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alright, here goes...
x squared = x squared yes?
that means that...
x(x-x)=(x+x)(x-x)
the 2 (x-x)'s cancel each other out...
so it becomes....
x=(x+x)
now we expand it...
x=2x...
so therefore...
1=2
how can this be possible? try and work it out...
if it didn't make sense to you at all, then ill edit it for you...
I have no clue. I suck at the math. Also, all I do in math is draw.
x(x-x) = x into 0 = 0
(x+x)(x-x)= (x+x) into 0 = 0
(x-x) = 0. So you are dividing by 0 here:Quote:
Originally posted by Bad Neighbour
alright, here goes...
x squared = x squared yes?
that means that...
x(x-x)=(x+x)(x-x)
the 2 (x-x)'s cancel each other out...
so it becomes....
x=(x+x)
now we expand it...
x=2x...
so therefore...
1=2
how can this be possible? try and work it out...
if it didn't make sense to you at all, then ill edit it for you...
x(x-x)=(x+x)(x-x)
You may NEVER divide by 0.
reason:
To divide a number by another number, means to find another number such that the first number times the second number equals the third number.
This also means that if:
12/4 = 3, that 3*4 = 12, which it is.
0 * 0 = 0
1 * 0 = 0
2 * 0 = 0
etc...
The first part of what I said earlier is true with 0/0 = 0, but the problem lies with the 2nd part which has more then just one answer ;).
edit:
Just found something nice on google.
Source: http://home.ubalt.edu/ntsbarsh/zero/ZERO.HTM#rdivzerotrQuote:
Dividing by Zero Can Get You into Trouble! If we persist in retaining such errata in our educational texts, an unwitting or unscrupulous person could utilize the result to show that 1 = 2 as follows:
(a).(a) - a.a = a2 - a2
for any finite a. Now, factoring by a, and using the identity
(a2 - b2) = (a - b)(a + b) for the other side, this can be written as:
a(a-a) = (a-a)(a+a)
dividing both sides by (a-a) gives
a = 2a
now, dividing by a gives
1 = 2, Voila!
This result follows directly from the assumption that it is a legal operation to divide by zero because a - a = 0. If one divides 2 by zero even on a simple, inexpensive calculator, the display will indicate an error condition.
Again, I do emphasis, the question in this Section goes beyond the fallacy that 2/0 is infinity or not. It demonstrates that one should never divide by zero [here (a-a)]. If one does allow oneself dividing by zero, then one ends up in a hell. That is all.
Ace is right. You can't do (x-x)/(x-x). That is dividing by zero. So it is 0=0.
err.yeah /0=not possible.
sheeesh,can't believe i forgot that.just had a maths exam today.
sin90/cos90=1/0=tan90=not possible:wacko:
When I first looked at that the way it was explained and the notation bad due to the computer made it VERY confusing.
It looks like everyone figured it out before I got here...Oh well.
You can't divide by zero because if you were to do so it would change the value to infinity (although infinity isn't really a value) and then the equation wouldn't be true because you did something irreversable...
Actually, technically, any number divided by zero is infinity.
Because zero*infinity=a finite number.
Multiplying a finite by zero, is, ofcourse, zero.
I don't know what you were trying to pull here, but you got it wrong...Quote:
Originally posted by Bad Neighbour
alright, here goes...
x squared = x squared yes?
that means that...
x(x-x)=(x+x)(x-x)
the 2 (x-x)'s cancel each other out...
so it becomes....
x=(x+x)
now we expand it...
x=2x...
so therefore...
1=2
how can this be possible? try and work it out...
if it didn't make sense to you at all, then ill edit it for you...
x ( x - x ) = ( x * x ) - ( x * x ) NOT ( x + x )( x - x )
x ( 0 ) = x^2 - x^2
0 = 0
How hard can that be...? :S
Im pretty good at math...
What means
??????????Quote:
x squared = x squared yes? that means that... x(x-x)=(x+x)(x-x)
x^2=x^2 true
but
x(x-x)=(x+x)(x-x)[/QUOTE]
^^^ bullshit. x <> (x+x)
BTW. U MAY delete by zero. But result is infinity, endlessness, eternity (dunno how in english).
x*(x-x) = (x+x)*(x-x)Quote:
Originally posted by Urn
Im pretty good at math...
What means
??????????
x^2=x^2 true
but
x(x-x)=(x+x)(x-x)
^^^ bullshit. x <> (x+x)
BTW. U MAY delete by zero. But result is infinity, endlessness, eternity (dunno how in english).
x * 0 = 2x * 0
which is true, seeing how anything times 0 = 0.
n / 0 = Undefined...
Any computer/calculator will tell you that...
Any mathematician will tell you that...
It's like time travel... IT CAN'T BE DONE!
Now, what Ace posted reminded me of this:
1/2 > 1/4
log(1/2) > log(1/4)
log(1/2) > log((1/2)^2)
log(1/2) > 2*log(1/2)
1 > 2
Anyone...? :)
dog = dog
1*dog=1*dog
0*dog=0*dog
0*dog=0*15*dog ; without zeros
dog=15*dog ; bullshit. !!not identical transformation!!
:) :) :) :)
Someones sig:
'Cats have 4 legs, dogs have 4 legs, therefore, all dogs are cats'
Funny, isnt it?
PS. Ill think about logs.
vagrantcharly, log for fractions with 0 as the characteristic, are inversely proportional to the fractions as they're negative.
Basically, log1/2 = log0.5 = -1.6990
And log1/4 = log0.25 = -1.3979
Evidently, log1/4 > log 1/2
So the first step in your series is incorrect.
Also, this can be verified the following way as well:
1/2 > 1/4
log 1/2 > log 1/4 [Not true ;) ]
log1 - log2 > log1 - log4 [Now, subtracting log1 from both sides..]
- log2 > - log4
log2 < log4 [Sign change, inequality changes]
Which is true.. =D
One more thing log is an exponential function, so if you take:
e1/2 < e1/4
You ultimately end up with the same result, not true. :)
About the logs...
Base of the log must be >= 1
i mean
log(1/2) > 2*log(1/2)
is actually true. Think about it, with base=2 -1>-4
and
log(1/2) > 2*log(1/2) <=>
log(1/2) - 2*log(1/2) > 0 <=>
- log(1/2)> 0 <=>
log(1/2) < 0 ; with base >= 1 it is true, else the very first
; log(1/2) > log(1/4) will be wrong
Again not identical transformation :) :) :) :)
U cant here log(1/2) > 2*log(1/2) delete on log(1/2)
Indeed... :)
That's a little something our Numerical Analysis teacher told us to solve a little while ago (along with a few others)... I wanted to see if anyone here got it right... :P
ahhhhhhhhhhhhh too much math to comprehend!!
I got losed around the point you guys started doing log(). We haven't done that in my math class yet...
well i for one have absolutely no idea what is going on here. i was about to spew out my knowledge of multiplication properties to end this early, then i saw odd looking equations popping up, each more strange than the last. i am lost and have been lost since before i was halfway done with Ace's first post. and i have math tomorrow. good luck to whoever gets it.
how can you add something to one side of an equation without adding it to the other side@_@?Quote:
Originally posted by vagrantcharly
Now, what Ace posted reminded me of this:
1/2 > 1/4
log(1/2) > log(1/4)
log(1/2) > log((1/2)^2)
log(1/2) > 2*log(1/2)
1 > 2
Anyone...? :)
i dunno anything about logs though:P
He didn't add anything, LoL.
He just took log on both sides :P.
Also, Urn, when logs are generally taken, they're assumed to be to the base, e, that is, natural or napieran logarithms, commonly, we use base 10 logarithms for all our calculations, and in both those cases, the above result is false :D
ooook.as long as logs aren't there in the Xth boards,i don't mind.but they give you the log tables,don't they?i mean,you don't have to memorise it.......do you:S?
LoL ofcourse not, they always give you the log tables. Only in the IIT entrances you're not allowed log tables, you have to use the napieran formula and there's another algorithm, to calculate the logs.. Or they'll give you the log you need to use in the calculation, as well as the anti log.
Eek, for any purpose I've had to use logs, it is either acceptable to leave it in log form or you can use a calculator for it.
That was acheived with the logarithm rules
ln(a^b) = b*ln(a)
ALL THESE NUMBERS! Shit, I feel like a blonde stuck in a lost-pink shoes situation! *wonders what it would feel like since she's not a blonde and HATES pink!*
bottom line-I SUCK AT MATHS!
HEEEEEEEEEYYYYYYY, MasJ !!
ln - log with base=e
lg - log with base=10
just log(x) is a wrong notation.
Thats common math rules.
At least in my country :)
In what I've learned log(x) is assumed to be base 10, just as ln(x) is base e, guess it's different here. Just out of curiosity what math are you guys taking? I'm in calc 2 right now and it pretty much sucks.
2xaenn: what do u mean?
Also base 10 here if no other base has been specified.Quote:
Originally posted by Xaenn
In what I've learned log(x) is assumed to be base 10, just as ln(x) is base e, guess it's different here. Just out of curiosity what math are you guys taking? I'm in calc 2 right now and it pretty much sucks.
Advanced Math here, highest math available at my school ><
why do i get the feeling that altavista won't be able to translate that:wacko:?Quote:
Originally posted by MasJ
LoL ofcourse not, they always give you the log tables. Only in the IIT entrances you're not allowed log tables, you have to use the napieran formula and there's another algorithm, to calculate the logs.. Or they'll give you the log you need to use in the calculation, as well as the anti log.
Urn: If just log(x) is written it's generally assumed that it's base 10. The base 10 logarithm is written that way on calculators and it's just kinda a standard thing.
Hmm, i didnt know that.
So screw it. Im tired of it.........
Yeah, I don't really care either, just thought I'd try to clear it up