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wordcross

 Posted: Sun Apr 25, 2004 3:54 am    Post subject: 1 My brother just showed me the "1=2" proof using simple algebra. For those of you who haven't seen it, it goes something like this: code:``` a=b a^2=ab a^2-b^2=ab-b^2 (a+b)(a-b)=b(a-b) a+b=b and since a=b 2b=b 2=1 and as we all know, .99999...=1, so .99999...=2 ``` So my question is, why exactly does this work? I know there's the whole divide by zero thing when you cancel out what's been factored, but why does that matter, in this case, when the algebraic representation works out? ------------------ GLer Sheep! Googlepic game
borschevsky
Chessnut

Posted: Sun Apr 25, 2004 5:16 am    Post subject: 2

 Quote: when the algebraic representation works out?
The algebraic representation doesn't work out. You can only go from xz = yz to x = y when z is not zero. People often skip over this check for zeros, but this rule is always true. You can't "cancel x from both sides" unless you know that x is not zero.
Antrax
ESL Student

 Posted: Sun Apr 25, 2004 6:42 am    Post subject: 3 That's true, but in this case the only problem lies when you divide by (a-b), as a=b -> a^2=ab would pass the zero test (it remains true). Antrax ------------------ "Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time
Daedalian Member

 Posted: Sun Apr 25, 2004 2:52 pm    Post subject: 4 Then you go and subtract b^2 from both sides and get 0=0. You can factor anything out from that: 756*0=12*0. So, I guess that means 756=12. Although, that's not really necessary. From 1=2, you can prove all integers are equal. 1=2 1+1=2+1 2=3 1=3 n=n+1 n+1=n+1+1 n+1=n+2 n=n+2 (OK, I'm a bit rusty on my proofs: You prove it for a specific case and you prove it for the general case and your done, right?)
Antrax
ESL Student

 Posted: Sun Apr 25, 2004 7:29 pm    Post subject: 5 Okay, so what are you saying? Antrax ------------------ "Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time
Legion
Daedalian Member

Posted: Sun Apr 25, 2004 7:52 pm    Post subject: 6

There's also the multiply by zero bit
 Quote: (a+b)(a-b)=b(a-b)

just because 2*0=35*0 it don't mean that 2=35

The Cruciverbalist
Lucrative Britches

 Posted: Mon Apr 26, 2004 2:10 am    Post subject: 7 The implications of this go way beyond mere math. I mean, George W. Bush and Sadaam Hussein are two people, right? So, substituting one for two, that means George W. Bush and Sadaam Hussein are one person! Of course, this also proves John Kerry is Charles Manson, so I probably wouldn't vote for either one of them. (Not that I can vote, since the 2=1 proof clearly shows that I'm the Taco Bell chihuahua.)
Daedalian Member

 Posted: Mon Apr 26, 2004 3:27 am    Post subject: 8 Not really. I mean you're one person (Which means you're two people) and its one dog. Nowhere have we proven people=dogs (although we probably could).
Hitchhiker
Finally got a ride.

 Posted: Mon Apr 26, 2004 4:37 am    Post subject: 9 There was a great proof once, using similar logic, which proved that Donald Rumsfeld is actually a carrot.
wordcross

 Posted: Mon Apr 26, 2004 5:14 am    Post subject: 10 you need a proof for that?
Jedo the Jedi
Paragon in Training

 Posted: Mon Apr 26, 2004 5:24 am    Post subject: 11 Well no, you could just watch The Apprentice.
Jedo the Jedi
Paragon in Training

 Posted: Mon Apr 26, 2004 5:25 am    Post subject: 12 Sorry, that's Donald Trump. I read it wrong. But the same is true. [This message has been edited by Jedo the Jedi (edited 04-26-2004 01:25 AM).]
Antrax
ESL Student

 Posted: Tue Apr 27, 2004 9:40 am    Post subject: 13 Look. There is nothing wrong, mathematically, in writing: 35*0=2*0 It is true. Nowhere in that proof is anyone multiplying by zero. The only problem is when you try to divide by zero, since this means you're multiplying by 0's opposite, which is nonexistent in R, C, or any field I know of. Antrax ------------------ "Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time
Beartalon
'Party line' kind of guy

 Posted: Tue Apr 27, 2004 2:51 pm    Post subject: 14 If you're not multiplying by zero, why are you using *0 ? You mean you're only showing the factoring and just not doing the math?
Antrax
ESL Student

 Posted: Tue Apr 27, 2004 8:23 pm    Post subject: 15 You have a=b. That means a^2=b^2 (though not the other way around). Since a=b. you get a^2=ab. Nobody, in any point, did anything like: a=b |*a Ergo, no multiplication by zero. Antrax ------------------ "Look, that's why there's rules, understand? So that you think before you break 'em" - Lu-Tze, Thief of Time
borschevsky
Chessnut

Posted: Tue Apr 27, 2004 9:43 pm    Post subject: 16

 Quote: That's true, but in this case the only problem lies when you divide by (a-b)
Yeah, that's what I'm saying. wordcross was asking why there's a problem, when the algebra works out. I'm saying that the algebra doesn't work out, because as you say, we can't go from

(a+b)(a-b)=b(a-b) to
a+b=b

What I was trying to explain is that people often think that there's a rule that says
xz=yz implies x=y, when the rule really says
xz=yz and z <> 0 implies x=y.

It just struck me from the way the question was asked that wordcross was thinking there's a "hole" in algebra, where you can use a valid string of algebra and come up with a wrong result.

Short answer: don't divide by zero
+1

Posted: Tue Apr 27, 2004 9:56 pm    Post subject: 17

 Quote: since this means you're multiplying by 0's opposite inverse
Daedalian Member

 Posted: Wed Apr 28, 2004 12:53 am    Post subject: 18 a^2-b^2 is zero (As is ab-b^2). So, really, you can't go any further than that point, because there's not much math you can do with 0=0. You can factor out (a-b) from both sides, but you could just as easily factor out a^171+3b^155-27c^0.5.
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