You proved the statement "1 = 2" implies "1 = 1" -- that is trivial because "1 = 1" follows from "things are equal to themselves" and in no way supports the notion that "1 = 2" is true. "1 = 2 ⇒ 1 = 1" does not mean "1 = 2" is true. You multiplied by sides by a number (2), and added a number (-3) to both sides. You also squared both sides which is legitimate. But if you perform the same three operations in other orders you prove that: "1 = 2 ⇒ 1 = 13" "1 = 2 ⇒ -4 = -2" "1 = 2 ⇒ 4 = 1" "1 = 2 ⇒ 2 = 8" "1 = 2 ⇒ -2 = 1" In general a false assumption may be connected with any consequent. http://en.wikipedia.org/wiki/Principle_of_explosion This does nothing to demonstrate the truth of either side.
The square function, \(f(x) = x^{2}\) is 'many to one', ie multiple inputs can give the same output. Thus it is not invertible so just because f(x) = f(y) doesn't mean x=y. This is why a function must be both injective and surjective in order to be bijective.
You never proved that 1 = 2. What you did was to assume that 1 = 2 and use that to prove 1 = 1. That doesn't prove that 1 = 2 whatsoever. You could have started with -1 = 1 and prove that 1 = -1 by squaring both sides like you did as part of the proof. That too is not a proof that -1 = 1.
sounds like that algorithm used to prove woman = evil. Please Register or Log in to view the hidden image!
1. True premise+correct reasoning=> True conclusion (not your case) 2. False premise+correct reasoning=> False conclusion (your case above)
Hmm…looks good to me. In fact, I’d be willing to trade all my onesies for your twosies. A guy like you must have hoards of two-dollar bills, eh? Please Register or Log in to view the hidden image!
The way proofs work is that you start out with an equality which you know is true. You then apply a series of operations on both sides of the equal sign. You willthen end you end up with an expression which is gaurenteed to be true. This is not what was done in post #1. Therefore there was no proof that 1 = 2 presented.
Technically, nothing. The equals sign is supposed to be transitive, and you could exploit this fact to rework the OP in a more proper manner: \(sqrt(1) = -1 (TRUE) sqrt(1) = 1 (TRUE) -1 = 1; Add 3... 2 = 4; Divide by 2... 1 = 2; QED\) This was done in the proper order and if you want to say it's improper then you must claim that the equals operator is not transitive. Please Register or Log in to view the hidden image!
If you say \(\sqrt{1} = \pm 1\) then your definition of \(\sqrt{x}\) isn't a function and therefore you have no right to say \(+1 = -1\). Case solved.
In case it's not clear, for reals the square root is always defined to be positive, so \(\sqrt{1} = 1\) and not \(-1\). If you use \(\sqrt{1}\) to denote either root of \(x^{2} = 1\) then you're using ambiguous notation, and there's no reason the two roots should be equal just because you give them the same name. RJ's "proof" is a bit like saying "Let \(x = 2\) and \(x = 3\). Therefore transitivity implies \(2 = 3\).". (For the same reason, in mathematics we much prefer to say \(i^{2} = -1\) rather than \(i = \sqrt{-1}\).)
Perhaps there should be a different equals operator to distinguish between \(sqrt{1} = \pm 1\) and \(sqrt{1} = -1\) ?
But if you're going to go this route then... \(-1^2 = 1\) yet \(sqrt{1} = 1\) so now you've broken the square function's inverse?? hehe I'm just trolling here...I just found it comical that rpenner and others felt compelled to point out that 1 does not in fact equal 2. Please Register or Log in to view the hidden image!
