The "Proof" that 2=1
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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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vankusss wrote:What 'more time' means?
I'm going to buy some ham.
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Neil_Bryan wrote: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
To quote my Maths teacher here...ortsz wrote:It looks like you're dividing by (a-b), and since a = b that's division by zero.
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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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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x = 0 in this problem.
4x + 2x = 9x
Add shit.
6x = 9x
Divide by the constant (x)
6=9
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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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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.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.

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= 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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OH GOD MY DRIVING!!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.

vankusss wrote:What 'more time' means?
I'm going to buy some ham.
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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 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.[ - sec x cos x; 0, pi] = -1

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Why the heck did I put limits in!? I've edited them out to make it *look* more plausible.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 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.[ - sec x cos x; 0, pi] = -1
Though, I do understand what's wrong with it.
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What? If anything, every one of these invalid proofs just shows that maths works. 0! is just a matter of definition.Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...

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

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Tsukatu wrote:Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Incidentally, this is also proof that Y2k was equal to Y1k.
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Yet despite humanity's superior knowledge, better technology, and even despite Stephen Hawking, the reaction to both was about the same. Of course, back in the day when the knights and peons and whatnot saw a comet, they went and whipped themselves, so maybe we've gotten somewhere.SlappyMcGee wrote:Tsukatu wrote:Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
Incidentally, this is also proof that Y2k was equal to Y1k.
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Why should they fall from the sky, if physics is wrong? Why not explode, or cease to exist entirely? Our preconceptions run pretty deep.Tsukatu wrote:Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...
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I hope you realize how unreasonable this question is.Mute Monk wrote:Why should they fall from the sky, if physics is wrong? Why not explode, or cease to exist entirely? Our preconceptions run pretty deep.Tsukatu wrote:Math is broken. Physics is wrong. I expect planes to start falling from the sky once reality realizes this.Drathmoore wrote:Quite mind-blowing, going through school, only to find that Maths isn't perfect. I mean, 0! for example...

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∞ = 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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What this shows is that: You Don't Mess With Infinity.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.
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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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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It does and doesn't at the same time. Mainly a bad joke, really.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.
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