The "Proof" that 2=1

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Postby Neil_Bryan » 2009.10.17 (03:43)

Okay. I think you have seen this lately.

a=b
a²=ab
a²+b²=ab+b²
(a+b)(a-b)=b(a-b)
a+b=b
b+b=b
2b=b
2=1

I was just wondering: what is wrong with the theory? I know it's wrong, but I haven't found out what the heck is wrong with this sort-of "proof".

Can you please help? Math isn't up for debate so I posted it here.
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Postby 乳头的早餐谷物 » 2009.10.17 (03:51)

It looks like you're dividing by (a-b), and since a = b that's division by zero.
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Postby otters » 2009.10.17 (03:52)

Yeah. After that step, both a and b are undefined values, so I guess two a's equals one a in that case.
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Postby Kablizzy » 2009.10.17 (04:28)

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Postby T3chno » 2009.10.17 (04:33)

By the way, to go from (a²+b²) to (a+b)(a-b) is not possible. The result if a difference of squares so it should've been (a²-b²).
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Postby Drathmoore » 2009.10.17 (08:41)

Neil_Bryan wrote:Okay. I think you have seen this lately.

a=b
a²=ab
-b²=ab-b²
(a+b)(a-b)=b(a-b)
a+b=b
b+b=b
2b=b
2=1
ortsz wrote:It looks like you're dividing by (a-b), and since a = b that's division by zero.
To quote my Maths teacher here...

IT'S A MASSIVE CLUDGE!


Yes. Point is, as soon as you divide by 0, everything goes out of the window. You might as well say 2+2=5 under the same proof. Nothing makes sense anymore when you've divided by 0.

Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...

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Postby lord_day » 2009.10.17 (12:42)

Assertion:
The smallest positive real number is 1.

Proof:
Let r be the smallest positive real number. Either r > 1 or r = 1 or r > 1

If r > 1 then 0 < sqrt(r) < r and r is not the smallest positive real number.
If r < 1 then 0 < r^2 < r and r is not the smallest positive real number.

Therefore r = 1.
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Postby squibbles » 2009.10.17 (13:31)

Another one that fucks up with Divide by 0.

x = 0 in this problem.
4x + 2x = 9x
Add shit.
6x = 9x
Divide by the constant (x)
6=9
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Postby jinxed_07 » 2009.10.17 (13:42)

a+b=b
b+b=b
2b=b
a=b
a²=ab
a²+b²=ab+b²
only possible when both a and b are = to 0
Otherwise, every algebra teacher in the US is going to drive by your house

a²+b²=ab+b²
You added a 'b' and are missing a 'a' wtf!

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Postby t̷s͢uk̕a͡t͜ư » 2009.10.17 (13:56)

lord_day wrote:Assertion:
The smallest positive real number is 1.

Proof:
Let r be the smallest positive real number. Either r > 1 or r = 1 or r > 1

If r > 1 then 0 < sqrt(r) < r and r is not the smallest positive real number.
If r < 1 then 0 < r^2 < r and r is not the smallest positive real number.

Therefore r = 1.
That conclusion is a total non-sequitur. What you've actually shown is that there's no such thing as the smallest positive real number.
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Postby 999_Springs » 2009.10.17 (16:08)

integrate(tan x)
= integrate(sec x sin x)
= [ - sec x cos x] - integrate( - cos x sec x tan x) {Integrating by parts with u = sec x and v' = sin x}
= -1 + integrate(tan x)

Therefore integrate(tan x) = -1 + integrate(tan x)
=> 0 = -1
=> 3 = 2
=> 5 = 2 + 2. QED.

[Edited to make it seem more plausible]
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Postby Chase » 2009.10.17 (16:32)

Now this children, is why you don't divide by 0. It makes everything you've just learnt over the past 11 years useless.
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Postby Kablizzy » 2009.10.17 (16:35)

Chase wrote:Now this children, is why you don't divide by 0. It makes everything you've just learnt over the past 11 years useless.
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Postby Spawn of Yanni » 2009.10.17 (16:48)

999_Springs wrote:integrate(tan x; 0, pi)
= integrate(sec x sin x; 0, pi)
= [ - sec x cos x; 0, pi] - integrate( - cos x sec x tan x; 0, pi) {Integrating by parts with u = sec x and v' = sin x}
= -1 + integrate(tan x; 0, pi)

Therefore integrate(tan x; 0, pi) = -1 + integrate(tan x; 0, pi)
=> [ln |sec x|; 0, pi] = -1 + [ln |sec x|; 0, pi]
=> 0 - 0 = -1 + 0 - 0
=> 0 = -1
=> 3 = 2
=> 5 = 2 + 2. QED.
[ - sec x cos x; 0, pi] = -1
- sec x cos x evaluated from 0 to pi equals 0, not -1. It equals -1 evaluated from 0 to pi, which is not just -1, it's 0. Therefore the integral of tan x between 0 and pi equals the integral of tan x between 0 and pi and everything from that point on is invalid.
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Postby 999_Springs » 2009.10.17 (16:58)

Spawn of Yanni wrote:
999_Springs wrote:integrate(tan x; 0, pi)
= integrate(sec x sin x; 0, pi)
= [ - sec x cos x; 0, pi] - integrate( - cos x sec x tan x; 0, pi) {Integrating by parts with u = sec x and v' = sin x}
= -1 + integrate(tan x; 0, pi)

Therefore integrate(tan x; 0, pi) = -1 + integrate(tan x; 0, pi)
=> [ln |sec x|; 0, pi] = -1 + [ln |sec x|; 0, pi]
=> 0 - 0 = -1 + 0 - 0
=> 0 = -1
=> 3 = 2
=> 5 = 2 + 2. QED.
[ - sec x cos x; 0, pi] = -1
- sec x cos x evaluated from 0 to pi equals 0, not -1. It equals -1 evaluated from 0 to pi, which is not just -1, it's 0. Therefore the integral of tan x between 0 and pi equals the integral of tan x between 0 and pi and everything from that point on is invalid.
Why the heck did I put limits in!? I've edited them out to make it *look* more plausible.

Though, I do understand what's wrong with it.
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Postby Spawn of Yanni » 2009.10.17 (18:52)

With your new version, you've ignored the arbitrary constant brought about by indefinite integration. Which brings me on to
Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
What? If anything, every one of these invalid proofs just shows that maths works. 0! is just a matter of definition.
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Postby t̷s͢uk̕a͡t͜ư » 2009.10.17 (22:29)

Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.
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Postby SlappyMcGee » 2009.10.17 (23:11)

Tsukatu wrote:
Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.

Incidentally, this is also proof that Y2k was equal to Y1k.
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Postby otters~1 » 2009.10.18 (01:14)

SlappyMcGee wrote:
Tsukatu wrote:
Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.

Incidentally, this is also proof that Y2k was equal to Y1k.
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Postby Mute Monk » 2009.10.18 (01:34)

Tsukatu wrote:
Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.
Why should they fall from the sky, if physics is wrong? Why not explode, or cease to exist entirely? Our preconceptions run pretty deep.
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Postby t̷s͢uk̕a͡t͜ư » 2009.10.18 (01:45)

Mute Monk wrote:
Tsukatu wrote:
Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.
Why should they fall from the sky, if physics is wrong? Why not explode, or cease to exist entirely? Our preconceptions run pretty deep.
I hope you realize how unreasonable this question is.
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Postby Drathmoore » 2009.10.19 (17:23)

Solution!

∞ = An infinately big number.
2∞ = 2 x an infinately big number.
Yet ∞ is infinately big, so...

∞ = 2∞
1 = 2


Reality:

Aleph Numbers, and that sort of stuff. Orders of Infinity, etc.

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Postby 999_Springs » 2009.10.19 (18:15)

Drathmoore wrote:Solution!

∞ = An infinitely big number.
2∞ = 2 x an infinitely big number.
Yet ∞ is infinitely big, so...

∞ = 2∞
1 = 2 <==You can't "cancel" infinities like that.
What this shows is that: You Don't Mess With Infinity.

Here are some more to decipher:

exp(i*pi) + 1 = 0 (A well-known fact...)
=> exp(i*pi) = -1
=> exp(2i*pi) = 1
=> ln(exp(2i*pi)) = ln(1)
=> 2i*pi = 0
=> 1 = 0
=> 3 = 2
=> 5 = 2 + 2. QED.

Alternatively:

ln 2
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8...
= 1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + 1/5 - 1/10 - 1/12 + 1/7...
= (1 - 1/2) - 1/4 + (1/3 - 1/6) - 1/8 + (1/5 - 1/10) - 1/12 +...
= 1/2 - 1/4 + 1/6 - 1/8 + 1/10 - 1/12 +...
= 1/2 ln 2

Therefore ln 2 = 1/2 ln 2
=> 2 = 1
=> 3 = 2
=> 5 = 2 + 2. QED. Zap.
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Postby Donfuy » 2009.10.19 (18:17)

Drathmoore wrote:Solution!

∞ = An infinately big number.
2∞ = 2 x an infinately big number.
Yet ∞ is infinately big, so...

∞ = 2∞
1 = 2


Reality:

Aleph Numbers, and that sort of stuff. Orders of Infinity, etc.

Does infinity equal to infinity?


edit: hah, 999_springs beat me to it.
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Postby Drathmoore » 2009.10.19 (19:03)

Donfuy wrote:
Drathmoore wrote:Solution!

∞ = An infinately big number.
2∞ = 2 x an infinately big number.
Yet ∞ is infinately big, so...

∞ = 2∞
1 = 2


Reality:

Aleph Numbers, and that sort of stuff. Orders of Infinity, etc.

Does infinity equal to infinity?


edit: hah, 999_springs beat me to it.
It does and doesn't at the same time. Mainly a bad joke, really.


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