1=2 ???

1=2 I know it. My very super genious computer engineer teacher, Elwin, helped prove it.

Let a =b

a^2 + a^2 = ab + a^2
2a^2 = a^2 + ab
2a^2 - 2ab = a^2 + ab -2ab
2a^2-2ab=a^2-ab
2 (a2-ab)=1 (a^2-ab)
2=1
????

amazing huh

here second proof represented by the congress of the skulls

Let x =1

x=1
x^2=2 (multiply each side by two)
x^2-1=2-1 (subtract one from each side)
(x+1) (x-1)=1 (expand)
x+1=1 (divide by x-1)
1+1 =1
2=1

hahah

psst. You can't divide by 0
(if x=1, then x-1=0)
(if a=b, then a^2=ab, then a^2-ab=0)

Still, the equations are always amusing to see, and it often takes me a bit to remember why an invalid result is returned..

[edited to include the divide-by-0 error in the first equation]

[edit on 1/11/2006 by Whiskey Jack]

Yes, you're factoring out terms that equate to zero, which will mess up your division steps.

Also 2 times x is not x^2 (2nd line of x=1 problem) in general, but it is if x is only 1. Then its not really an algebraic expression, just an equality.

[edit on 11-1-2006 by glastonaut]

Well, could just change the base...

Your very super genius computer engineer teacher, Elwin, helps you make wrong equations.

Edit:

Let a =b

a^2 + a^2 = ab + a^2

replace 'a' and 'b' with the same number...

(1)^2 + (1)^2 = (1)(1) + (1)^2
1 + 1 = 1 + 1
2 = 2

amazing huh

here second proof represented by the congress of the skulls

Let x =1

x=1
2x=2 (multiply each side by two)
2(1)=2
2=2

Ok.

[edit on 1/11/2006 by Arcane Demesne]

im not a math genious by any means

but i gave this equation he showed a try; and i respectfully disagree

you cant just pull things out of nowhere like that, at least , if you can and im just confused; i still think you shouldnt be Allowed *in decent math* to do such an uncalled for operation

now; perhaps 2=1 could be shown as true , but i do not believe this is the accurate way to describe it

i could be wrong; but i have a gut feeling on this one

Indeed. In the post above yours...I corrected his 'equations'. And nowhere did I get a '1' on either side of either equation.

I've seen several of these 'proofs' that 1=2 or 2+2=5 or that sort of thing. They all rely on dividing by zero, usually in such a way that it is not immediately obvious that you are dividing by zero.

Off-topic: Arcane Demesne, love the name!

Ha ha, I like it.

I've only looked at the first one and I believe it is valid.

You basically end up with:

2 x 0 = 1 x 0
ie 0 = 0.

You can make any two numbers equal by saying:

let a=0
7a = 42a
Cancel out the 'a's and your left with
7 = 42.

So the cancelling out is effectively dividing by 0.

x=.999999...
10x=9.99999...
subtract the original equation
-x=-.999999...
9x=9.000000...
divde both sides by 9 and.......

x=1

math is cool

(clarity)

[edit on 12-1-2006 by radiant]

[edit on 12-1-2006 by radiant]

Haha. Yes, I see we both dwell in a Demesne. Cheers!



That one messes with my head, and it follows all the laws. Good job!

Clever!

x is not equal to 1

Let start with

x = .9999...

10(.9999...)= 9.9999...

10(.9999...)-(.9999...) = 9.999... - .9999...

8.9991 = 8.9991

Actually it doesnt...

Say you start small and take only 10 digits of 9's

x = .9999999999

10 x = 10 ( .9999999999) = 9.999999999 ( see there's only 9 digits after the decimal?)

When he subtracted again by x he's really doing this:

9.9999999990
- .9999999999



[edit on 12-1-2006 by I_s_i_s]

Oh ok...I see. But I was under the impression that he meant .9999 with the bar over it (infinity 9's). Maybe not. I dunno.

edit: I can't spell impression, ha.

[edit on 1/12/2006 by Arcane Demesne]

I'm under the same impression.

Bottom line you're subtracting .99999( infinite digits) from 9.9999(infinity - 1 digits) You'll end up with 8.9999999999999999999999999999999991

The last underlined 9 repeats

Now I get it. Thanks!

We did a lot of these problems in algebra with change of base in logrithmic equations.

I was lead to believe that division by zero is undefined and that 9x would equal 8.9999 not 9.00000.