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AlexEh
May 15th, 2010, 07:50 AM
Lol I was expecting something more explicit with all the trouble I had but I was wrong. Still awesome. Not sure if I should post this though seems like I'm defeating the purpose...

http://www.itchstudios.com/psg/main.php?id=misc

5, 12, 17, 29, 46, ?

Solution
5+12=17
12+17=29
17+29=46
29+46=75

...followed by:
5, 12, 21, 32, 45, 60, ?
12-5=7
5+7=12

21-12=9
12+9=21

32-21=11
21+11=32

45-32=13
32+13=45

60-45=15
45+15=60

X-60=(15+2) (Note the subtraction equations have always equaled 2+ the previous answer so this one is 15+2=17
X=60+17
X=77

(This one was hard I thought it was 75 for a while)

...and finish with:
oboe, fobo, ofob, cofo, ocof, ?

This ones hard but got it faster then second one...
obo(e)---> (e)obo---->(f)obo (Next letter in alphabet is f)
fob(o)---> (o)fob (Last letter is moved to front)
ofo(b)---> (b)ofo ----> (c)ofo
cof(o)---> (o)cof
oco(f)---> (f)oco----> (g)oco= goco

Crash
May 15th, 2010, 09:07 AM

looooooooooooooool

Baron Impossible
May 15th, 2010, 10:21 AM
1
1,1
2,1
1,2,1,1
1,1,1,2,2,1
..?

Reymus
May 15th, 2010, 10:52 AM
I see you've played knifey spooney before.

If a = x [true for some a's and x's]
then a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]

Camilla
May 16th, 2010, 07:09 AM
@ Reymus: You divide by zero. That's a no go:
http://lgo.mit.edu/blog/drewhill/files/divide-by-zero-blog-safe.jpg

Flashback
May 16th, 2010, 08:07 AM
"@ Reymus: You divide by zero. That's a no go:"

Except, if you use imaginary numbers.

Randis
May 16th, 2010, 08:29 AM
exciting

Prometheus|ANJ
May 16th, 2010, 09:49 AM
I seems like I left some older versions of the problems in the source code of that page.

That gallery itself is kind of embarrassing, I need to prune it.

hippl5
May 16th, 2010, 10:03 AM
I seems like I left some older versions of the problems in the source code of that page.

That gallery itself is kind of embarrassing, I need to prune it.

I actually figured the code out a little while ago before this thread was posted.

YOU ARE A MONSTER

Camilla
May 18th, 2010, 07:10 AM
"@ Reymus: You divide by zero. That's a no go:"

Except, if you use imaginary numbers.

Err ... perhaps! Now, I haven't worked with imaginary numbers for ages, so I am very rusty in this aspect. Could you enlighten me? (Just bring on the formulas :-))

Flashback
May 18th, 2010, 08:13 AM
Correction: The simplest sense, divide by zero is under defined, so as it approaches the zero it curves violently, getting close to zero but not touching it creating an Asymptote. This kind of creates infinity.

http://thepublicinterest.freedomblogging.com/files/2009/07/asymptote.png

In computer, divide by zero would cause a program to crash. Sort of creating a infinite loop.

real numbers would be 5, -4.1, 54/2, 0, 3.3434343434343434343434.........

Imaginary number are when you square root -1 or non-positive numbers.
http://rlv.zcache.com/imaginary_numbers_tshirt-p235311307141194890t5hl_400.jpg

SusanMart
May 18th, 2010, 08:43 AM

lol, still can't get over it......so to the point!

:sungod:

Camilla
May 18th, 2010, 08:59 AM
Correction: The simplest sense, divide by zero is under defined, so as it approaches the zero it curves violently, getting close to zero but not touching it creating an Asymptote. This kind of creates infinity.
...
Imaginary number are when you square root -1 or non-positive numbers.

If a=x then a-x=0. That is a straight, round zero, not something that approaches zero.

I realize that imaginary numbers enable you to square root negative numbers. What I don't remember is, what they "say" about dividing by zero.
However, I really doubt that any creative juggling with imaginary numbers would make the 2=1 end result of Reymus' proof correct, since this would go against one of the most basic axioms of math, 1+1=2.

Please show me how imaginary numbers would solve Reymus' riddle and make 2=1 a valid result.

Flashback
May 18th, 2010, 11:46 AM
Nope, hence the correction.

Besides, the problem fallacy is apparent here:

2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)] <--- here factoring problem????
2(a-x) = a-x [x-2x = -x]

I used to use imaginary numbers mostly to solve imaginary equation problems or quadratic formula problems.

Imaginary numbers were invented for mathematicians to solve negative square root problems, among other things.

I * 0 = 1,

but since

(I + 1) * 0 = I * 0 + 1 * 0 = 1 + 0 = 1

so that

(I + 1) * 0 = 1,

we would have to say that

I + 1 = I

By subtracting I from both sides, we would find that

1 = 0

Source: http://mathforum.org/library/drmath/view/53872.html

Arshes Nei
May 18th, 2010, 12:45 PM
I seems like I left some older versions of the problems in the source code of that page.

That gallery itself is kind of embarrassing, I need to prune it.

Eh old artwork is embarrassing when you know you have improved, but it's still fun to see the progress!