Mathematical problem (attempt 2)

[x42bn6]

Yeah, I was wrong, so...

0=1-1+1-1+1-1+1-1+1-1+1-1+1-1+...
Right-hand side (RHS)=1-(1-1)-(1-1)-(1-1)-(1-1)-(1-1)-(1-1)-...
RHS=1-0-0-0-0-0-0-...
RHS=1
But we started with LHS=RHS.
Therefore LHS=1.
Therefore 0=1.

*

[raver_lost]

huh?? dude that aint cool... i'm way too gone to try and solve that!

[x42bn6]

Cool people won't fade in the eyes of a challenge.*

[raver_lost]

high people will...

[TrongaMonga]

It would depend on the signal of the last element of the sum.

But we'd have to have (how do you write polynomion, or whatever , btw?) a pair of +1 with -1 for that to equal zero.

That is, we'd need

0=1-1+1-1+1-1+...1-1+1-1+1-1
RHS=1-(1-1)-(1-1)-(1-...1)-(1-1)-(1-1)-1

[ddn2004]

0=1-1+1-1+1-1+1-1+1-1+1-1+1-1+...
RHS=(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+(1-1)+...
RHS=0+0+0+0+0+0+0+...
RHS=0

LHS=0

Therefore LHS=RHS which is bullshit somehow ><

[EEX_ca_aok]

Besides, you cannot assume LHS=RHS beforehand, that's flawed math.

[Master.America]

You can assume LHS=RHS in the first step... there's just the matter of proving it, which is why he uses this proof. However, the prove does make assumptions, making it untrue.

Because LHS=0, the last number in series RHS must be -1. The second step assumes that the last number in series RHS is 1, but it isn't.

0=1-1+1-1+1-1+1-1+1-1+1-1+1-1+...-1
RHS=1-(1-1)-(1-1)-(1-1)-(1-1)-(1-1)-(1-1)-...1
RHS=1-0-0-0-0-0-0-...1
RHS=0
But we started with LHS=RHS.
Therefore LHS=0.
Therefore 0=0. Who knew? I really thought 0=42.



What's the point of this? Are you trying to trick us?

What's the square root of 1000?

[EEX_ca_aok]

What I'm saying is you can "assume" LHS=RHS, but you cannot use that as a form of proof because it's an assumption, not a fact. It's a circular argument if you prove LHS=RHS by assuming LHS=RHS. That's like I could say I can prove George Bush is the anti-christ by assuming that as true.

Oh, and the square root of 1000 is 10√10, or roughly 31.6.

[Venice_native]

The answer is obviously 3.

[Master.America]

Originally Posted by EEX_ca_aok

What I'm saying is you can "assume" LHS=RHS, but you cannot use that as a form of proof because it's an assumption, not a fact. It's a circular argument if you prove LHS=RHS by assuming LHS=RHS. That's like I could say I can prove George Bush is the anti-christ by assuming that as true.

Oh, and the square root of 1000 is 10√10, or roughly 31.6.

Yeah, I know that. But it's not what you said. Sorry, I just felt like being annoying.

It's funny how many people think it's 100. Funny, yet sad.

[Bret]

Sq. Root of 1000 =

31.6227766016837933199889935444327.....

[x42bn6]

Originally Posted by TrongaMonga

It would depend on the signal of the last element of the sum.

But we'd have to have (how do you write polynomion, or whatever , btw?) a pair of +1 with -1 for that to equal zero.

That is, we'd need

0=1-1+1-1+1-1+...1-1+1-1+1-1
RHS=1-(1-1)-(1-1)-(1-...1)-(1-1)-(1-1)-1

There is no end of the sequence - it is infinite.*

[Chaos Marine]

Whats the point of this?

[lollercoasterzster]

he moved the bracket,

0 = (1-1) + (1-1) + (1-1) ... FIRST LINE yeah this is true
0 = 1 - (1 - 1) - (1 - 1) - (1 - 1) ... SECOND if the amount of numbers in this series hasn't changed (and it can't) it should end with -1, not -(1-1) which is -1 + 1.

You can't really say LS = RS if RS =/= RS

[Master.America]

Originally Posted by x42bn6

There is no end of the sequence - it is infinite.*

Then isn't the answer undefined?

[x42bn6]

I don't think undefined is the answer, but I think you hit the nail on the head but it slipped.

And the brackets moving is mathematically correct - a+b+c=a+(b+c)=(a+b)+c and so on.*

[Master.America]

So you are trying to trick us?

[x42bn6]

Of course - how can zero equal one?*