Oops I broke the laws of math.

[revolutionomega]

either 2 or 3.
Duh. Haven't really learned about series and such yet.

[PhaeL v2]

Originally Posted by revolutionomega

either 2 or 3.

Yeah. Parentheses are like commas in mathematics.

[ieatyourlvllol]

Originally Posted by MrRubix

15.67999... = 15.68

Not "approximately."

Not "close"

EQUAL

DISCUSSSsss

Originally Posted by MrRubix

round one


fight

IEAT USES SUPER - MATHRYUKEN

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Firstly, remember that we can't really interpret 0.000...1 as a finite decimal value, but rather, the point that by definition bounds the limit of 1/n as n gets infinitely large. This point is of course 0 (which will be subsequently proven).

For any two values x1 and x2, x1/n can be made arbitrarily close to x2/n by taking n to be a sufficiently large number. Thus,

lim (n→∞) x1/n = lim (n→∞) x2/n, for all x

Since x is a constant, we can extend the notion and say that

lim (n→∞) x1/n = lim (n→∞) x2/n for all x =>
x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] for all x

Thus, taking x2 to be 0,

x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] =>
x1[lim (n→∞) 1/n] = 0[lim (n→∞) 1/n] =>
x1[lim (n→∞) 1/n] = 0 =>
lim (n→∞) x1/n = 0

And that's why lim (n→∞) x/n = 0 for all x. Proceeding with the proof...

Taking x1 to be 1,

lim (n→∞) x1/n = 0 =>
lim (n→∞) 1/n = 0 =>

From the aforementioned interpretation, 0.000…1 is equivalent to lim (n→∞) 1/n. Thus,

lim (n→∞) 1/n = 0 =>
0.000...1 = 0

And finally getting back to the problem at hand...

Since lim (n→∞) 1/n = 0.000...1 = 0 = lim (n→0) n,

15.68 - lim (n→∞) 1/n = 15.68 - lim (n→0) n =>
15.68 - 0.000...1 = 15.68 - 0 =>
15.67999... = 15.68

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IT'S SUPER EFFECTIVE!

[Ghakimx]

Originally Posted by PhaeL v2

Yeah. Parentheses are like commas in mathematics.

But its only 1 answer, unless stated otherwise.

[bluguerrilla]

Originally Posted by bLasTamos

4:
There's one "-1" missing in the end.

7:
The sum S is infinite, it doesn't make any sesne to say S-A is 4S because nothing can be the 4 times more infinite than something else.

No, you just can't rearrange infinite series that don't conditionally converge because you can make it equal whatever you want (not to mention divergent series).

ieat wouldn't 10^-n have been easier? :P

[Ghakimx]

If x is any number,

x ^ 0 = 1

Explain please.

[leonid]

The reason why I suggested we please not get into this is that we were going to discuss a thing that's of the same level as "IS ONE PLUS ONE TWO OR THREE"

[PhaeL v2]

Originally Posted by Ghakimx

If x is any number =/= 0,

x ^ 0 = 1

Explain please.

fixed, because you just can't divide by 0.

x² x x² = x^(2+2) = x^4
x² / x² = x^(2-2) = x^0
so, x^0 = x / x, it's like dividing x by itself, or just 1.

[bLasTamos]

Originally Posted by ieatyourlvllol

lim (n→∞) 1/n = 0 =>
0.000...1 = 0

What the ****?

[Ghakimx]

But can I say that 1 = 2?

0 x 1 = 0
0 x 2 = 0

Thus

0 x 1 = 0 x 2

and then

0/0 x 1 = 0/0 x 2

so finally

1 = 2

[bLasTamos]

Originally Posted by bluguerrilla

No, you just can't rearrange infinite series that don't conditionally converge because you can make it equal whatever you want (not to mention divergent series).

Proof 4 converges. All terms are 0.
But yeah whatever. The only reason you can't rearrange them is because they are already infinite and it doesn't make sense.

[leonid]

@hakim: You just divided by zero

[PhaeL v2]

Originally Posted by Ghakimx

But can I say that 1 = 2?

0 x 1 = 0
0 x 2 = 0

Thus

0 x 1 = 0 x 2

and then

0/0 x 1 = 0/0 x 2

so finally

1 = 2

Well, I think if you eliminate the 0's (in case, 0/0), you eliminate the whole line. Or in a different point of view:

Priorizing multiplication, instead of division, using parentheses, you have 0 / (0 x 1) = 0 / (0 x 2),

then, 0 / 0 = 0 / 0

but you can just ignore the parentheses using and priorize the multiplication, it will be right anyway. That's why I think if you eliminate the 0 division, you eliminate the whole line.

[MrRubix]

Originally Posted by ieatyourlvllol

0.000...1 = 0

[bLasTamos]

Originally Posted by MrRubix

ahahahaha
yeah
now THAT was mind blowing.

[bluguerrilla]

Originally Posted by bLasTamos

Proof 4 converges. All terms are 0.
But yeah whatever. The only reason you can't rearrange them is because they are already infinite and it doesn't make sense.

The problem is he takes an absolutely convergent series and changes into a conditionally convergent series, which cannot be rearranged without changing the outcome of the series (to whatever you want :P).

And I meant to say absolutely in my above post, sry.

[stargroup]

Originally Posted by Ghakimx

But can I say that 1 = 2?

0 x 1 = 0
0 x 2 = 0

Thus

0 x 1 = 0 x 2

and then

0/0 x 1 = 0/0 x 2

so finally

1 = 2

actually, this is typically the proof used to define 0/0 as an indeterminate form

[ieatyourlvllol]

yup pretty much

0*whatever = 0 =>
0*whatever1 = 0*whatever2 =>
0 / 0 = (whatever1 / whatever2)^sqrt(1) = whatever

wala you did math

[x After Dawn x]

Originally Posted by stargroup

Mind blown yet?

No, they're all wrong. gg

[Azafreak]

Are you all in math enriched? God damn, I dont understand anything of those things and I'm ''ok'' in Math( im not in enriched)