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

Originally posted by alienfy
x+1=1 (divide by x-1)

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...

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

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

Originally posted by muzzleflash
im not a math genious by any means

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

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]

Originally posted by DragonsDemesne
Off-topic: Arcane Demesne, love the name!

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

Originally posted by radiant

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

x=1

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

Originally posted by radiant
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]

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

Originally posted by Arcane Demesne

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

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]

Originally posted by I_s_i_s

Originally posted by Arcane Demesne

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

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

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]

Originally posted by Arcane Demesne
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.

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

Originally posted by I_s_i_s

Originally posted by Arcane Demesne
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.

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.