• • # Thread: 1 = -1?

1. ## 1 = -1?

 I posted this "proof" in the 1/0 thread yesterday. I followed it up with discussion about another strange issue in math that I thought in some abstract way might be related, and then the discussion went in the direction of the second issue. So, I am starting a new thread to discuss the "proof" below. I do not know its resolution, other than that it is a reductio ad absurdum of imaginary numbers. I am open to other perspectives. -1 = -1 1/-1 = -1/1 sqrt (1/-1) = sqrt (-1/1) sqrt 1 / sqrt -1 = sqrt -1 / sqrt 1 1/i = i/1 i(1/i) = i(i/1) 1 = i^2 1 = -1 What is the deal with that?  Reply With Quote

2.  What? You've already been given a painstaking explication of exactly this. It isn't a reductio ad absurdum. The inference from step 3 to step 4 is unfounded. The end.  Reply With Quote

3. Originally Posted by Xei What? You've already been given a painstaking explication of exactly this. It isn't a reductio ad absurdum. The inference from step 3 to step 4 is unfounded. The end. With the "painstaking explication," we discussed the legitimacy of squaring the square root of a negative number and getting a negative number. Well, in this proof, the square of sqrt -1 is -1, as you believe it should be. The inference from step 3 to step 4 involves getting the square roots of negatives (not squaring them), which has not been the topic of our discussion. The inference also involves the principle that the square root of a fraction is the square root of the numerator over the square root of the denominator, an issue we have not discussed. So what is the problem in the proof? What exactly goes wrong from step 3 to step 4?  Reply With Quote

4.  "the principle that the square root of a fraction is the square root of the numerator over the square root of the denominator" This. It's the same issue as before, sqrt(a/c) = sqrt(a)/sqrt(c) is almost the same identity as sqrt(ab) = sqrt(a)sqrt(b), put b = 1/c.  Reply With Quote

5. Originally Posted by Xei "the principle that the square root of a fraction is the square root of the numerator over the square root of the denominator" This. It's the same issue as before, sqrt(a/c) = sqrt(a)/sqrt(c) is almost the same identity as sqrt(ab) = sqrt(a)sqrt(b), put b = 1/c. I would not say it is the same identity, but it too involves an exception because an imaginary number is involved. I said in the other thread that the two principles may be connected, and I think you just helped me make the connection more concrete in my mind. So the resolution is that getting the square root of a fraction does not involve a ratio of the square roots of the numerator and denominator when at least one is negative. That answers the question. The situation is just another exception scenario that results from assuming the reality of imaginary numbers. It is a reductio ad absurdum until "exception" is declared? That exception is even more bizarre than the other one. Getting the square root of a fraction by putting the square root of the numerator over the square root of the denominator seems like it should always be the logical method because squaring a fraction involves getting the square of the numerator over the square of the denominator. So, going in the reverse direction, it would make sense to get the ratio of the square root of the numerator to the square root of the denominator. But what we have run into is a situation of, "Oh no, we are dealing with imaginary numbers, so we can't do that here." You answered my question, but now I am even more convinced that the "proof" is a reductio ad absurdum of imaginary numbers.  Reply With Quote

6. Originally Posted by Universal Mind It is a reductio ad absurdum until "exception" is declared? I don't know why you're repeating this given that I invested a lot of time resolving this misconception and you were fine with it. I'll just quote myself if you don't mind. This statement is not an axiom of arithmetic; it is proven from the axioms of arithmetic, and the proof only works for when a and b are non-negative reals. When a and b are not non-negative reals, the proof does not work and the identity is in fact incorrect. Therefore, I reiterate my reiteration: there was no creation of a special case. The truth for a and b non-negative and the falsity otherwise are both deduced from the axioms of arithmetic. There is nothing wrong with this and it is not unusual. To help show this, here is another identity: a^2 = a*|a| for a non-negative. |a| means the absolute value of a, which means its size irrespective of whether it is positive or negative. You can write it as sqrt(a^2) if you want. This is exactly analogous. The identity is not an axiom of arithmetic: it is proven from the axioms of arithmetic, and the proof only works when a is non-negative. If a is negative, the proof fails, and the correct identity is in fact a^2 = -a*|a| What your argument says is that we have invented a special case because the identity doesn't work for negative numbers; and therefore, negative numbers are somehow fake. As you don't believe that negative numbers have the same problem as complex numbers, hopefully you now see why your argument doesn't make sense. Originally Posted by Universal Mind That exception is even more bizarre than the other one. Getting the square root of a fraction by putting the square root of the numerator over the square root of the denominator seems like it should always be the logical method because squaring a fraction involves getting the square of the numerator over the square of the denominator. So, going in the reverse direction, it would make sense to get the ratio of the square root of the numerator to the square root of the denominator. But what we have run into is a situation of, "Oh no, we are dealing with imaginary numbers, so we can't do that here." You answered my question, but now I am even more convinced that the "proof" is a reductio ad absurdum of imaginary numbers. I am repeating myself yet again by saying this: the identity does not apply to matrices either. If it 'disproves the existence of complex numbers' then it disproves the existence of matrices too. You repeatedly insist on referring to it as if it is some kind of ancillary axiom necessitated by consistency when it has been clearly explained to be no such thing; like all mathematics, propositions are proven from the axioms. This particular proposition can be proven for non-negative integers, but it doesn't happen to go through for other entities. Saying 'it seems like it should make sense' is not a proof. By not delineating the actual mathematical argument properly you obfuscate the subtlety which here leads to the mistake you are making. The subtlety is that the square function is not injective. Two numbers having equal squares does not prove that they are equal.  Reply With Quote

7. Originally Posted by Xei I don't know why you're repeating this given that I invested a lot of time resolving this misconception and you were fine with it. I'll just quote myself if you don't mind. You are mixing up the two scenarios again. The proof for this thread is the only thing I have said is a reductio ad absurdum. The equation involving products of square roots demands an imaginary number exception, but I never said it was a reductio ad absurdum or proves anything. I have just made the point that it was an exception that had to be made when something new and fictitious was introduced to the system. Make sure you understand the difference. The topic of this thread stands out to me as a reductio ad absurdum because of the principle I explained regarding the purpose of getting the square root of the numerator and denominator. It is the logical thing to do in all situations that are rooted in reality. Multiplying square roots by getting the square root of the product of the radicands doesn't seem to follow such invincible logic, in my mind, so I have not declared the need for the exception a reductio ad absurdum. Understand? I am going to study the logic behind the product rule before ever declaring the other scenario a reductio ad absurdum. It might be one, but I have yet to claim that it is. Originally Posted by Xei I am repeating myself yet again by saying this: the identity does not apply to matrices either. If it 'disproves the existence of complex numbers' then it disproves the existence of matrices too. You repeatedly insist on referring to it as if it is some kind of ancillary axiom necessitated by consistency when it has been clearly explained to be no such thing; like all mathematics, propositions are proven from the axioms. This particular proposition can be proven for non-negative integers, but it doesn't happen to go through for other entities. A matrix is a man made construct. It's like a board game in that way. People made the rules for what they represent and what to do with them. I am going to repeat myself yet again too. The proposition cannot be proven for negative numbers because even roots of negative numbers are a crock of shit. Originally Posted by Xei Saying 'it seems like it should make sense' is not a proof. By not delineating the actual mathematical argument properly you obfuscate the subtlety which here leads to the mistake you are making. The subtlety is that the square function is not injective. Two numbers having equal squares does not prove that they are equal. I was being polite by saying "seems like it should." I'll put it another way-- it is what is logical. I explained my reasoning. Your argument that two numbers having equal squares does not prove they are equal does not disprove my argument. Yes, a number has two square roots, one positive and one negative. What does that have to do with this? The radical symbol represents principal square root, which is the positive one. When more than the principal square root can be involved, such as in an equation with a variable, you fix the problem by using +/-. That is not necessary here. There is no variable, and it is principal square root involved, and that is one number.  Reply With Quote

8. Originally Posted by Universal Mind You are mixing up the two scenarios again. The proof for this thread is the only thing I have said is a reductio ad absurdum. The equation involving products of square roots demands an imaginary number exception, but I never said it was a reductio ad absurdum or proves anything. I have just made the point that it was an exception that had to be made when something new and fictitious was introduced to the system. Make sure you understand the difference. The topic of this thread stands out to me as a reductio ad absurdum because of the principle I explained regarding the purpose of getting the square root of the numerator and denominator. It is the logical thing to do in all situations that are rooted in reality. Multiplying square roots by getting the square root of the product of the radicands doesn't seem to follow such invincible logic, in my mind, so I have not declared the need for the exception a reductio ad absurdum. Understand? I am going to study the logic behind the product rule before ever declaring the other scenario a reductio ad absurdum. It might be one, but I have yet to claim that it is. The misconception is about it being an 'exception', not about it being a reductio ad absurdum... as should be clear given that that is what the quoted response is all about. I was being polite by saying "seems like it should." I'll put it another way-- it is what is logical. I explained my reasoning. If it were logical it would have taken the form of a proof. You are yet to write one. It is necessary that you write one because you are currently relying on a heuristic description which turns out to be incomplete. The issue stems from where you vaguely said 'going in the reverse direction'. This is not actually well defined, because the function isn't injective, like I just said. Principal roots have nothing to do with it. You need to prove it properly and then you will see why the identity does not apply to all numbers, and that will fully answer the problem in the OP.  Reply With Quote

9. Originally Posted by Xei If it were logical it would have taken the form of a proof. You are yet to write one. It is necessary that you write one because you are currently relying on a heuristic description which turns out to be incomplete. The issue stems from where you vaguely said 'going in the reverse direction'. This is not actually well defined, because the function isn't injective, like I just said. Principal roots have nothing to do with it. You need to prove it properly and then you will see why the identity does not apply to all numbers, and that will fully answer the problem in the OP. I made an argument. That can be done in math without the writing of a proof. I will make my point another way. The square roots of the numerator and denominator are, based on the definition of "square," the numbers that are squared to get the numerator and denominator. Right? I mentioned principle square roots because of the off target issue you brought up. If a formal proof is all you will see, here you go. b = 1/c (valid everywhere c =/=0) sqrt(a)sqrt(b) = sqrt(ab) sqrt(a)sqrt(1/c) = sqrt(a*1/c) sqrt(a)*1/sqrt(c) = sqrt(a/c) sqrt(a)/sqrt(c) = sqrt(a/c) My complements to PhilosopherStoned for typing that in the other thread.  Reply With Quote

10.  Congrats! You've proved that when sqrt(a)sqrt(b) = sqrt(ab), we also have sqrt(a)/sqrt(c) = sqrt(a/c) for c =/= 0. Can you go the other way? Remember, you'll have to take b=0 as a special case.  Reply With Quote

11.  Also contingent upon sqrt(1/c) = 1/sqrt(c).  Reply With Quote

12.  Just note that (1/sqrt(c))^2 = 1/(sqrt(c)^2) = 1/c by the definition of multiplication.  Reply With Quote

13.  Has the same caveat as sqrt(a)sqrt(b) = sqrt(ab)and its proof, that's what I meant to highlight.  Reply With Quote

14.  Xei, I am thoroughly impressed by your perseverance with this topic. It seems like some people just want you to keep repeating yourself...  Reply With Quote

15. Originally Posted by Pensive Patrick Xei, I am thoroughly impressed by your perseverance with this topic. It seems like some people just want you to keep repeating yourself... I'm sure you have an extremely clear understanding of what this debate is about. However, prove it. Also, I am just one person, and I have not been repeating Xei's self. I have made some of the same points more than once because they have needed to be taken into account more than once. Is the "Pensive" in your screen name a reference to the high price of your gold teeth? Originally Posted by PhilosopherStoned Congrats! You've proved that when sqrt(a)sqrt(b) = sqrt(ab), we also have sqrt(a)/sqrt(c) = sqrt(a/c) for c =/= 0. Can you go the other way? Remember, you'll have to take b=0 as a special case. Why do I need to go the other way? I proved my point Xei challenged me to prove. To be a little more to the point, the next step in your proof that I cut and pasted could be wrtitten as sqrt(a/c) = sqrt(a)/sqrt(c). (Symmetric Property of Equality) I know that sqrt(a)sqrt(b) = sqrt(ab) doesn't apply to imaginary numbers, fantasy numbers (I just added that rule.), or the Flying Spaghetti Monster.  Reply With Quote

16.  UM, this reminds me of that clip that one of us posted one time for a creationist. Was it Noogah? Or maybe Ne-Yo. This time, you're the Black Night and Xei is Arthur. Enjoy!  Reply With Quote

17.  Ironically, your nonresponsiveness to a very detailed and on-point post suggests that it is you who has lost his arms and legs. I used your proof to show that the square root of a fraction is the square root of the numerator over the square root of the denominator and agreed that it cannot apply to imaginary numbers. We have different takes on why that is, but I would have to say something I have already said to explain why that is, and I would hate to make 'pensivePatrick think that reveals a flaw in my argument because he doesn't know enough about this stuff to get what is actually happening. However, I will be glad to talk about movies if you have had enough of this. My reaction to the Black Knight scene when I saw that movie for first time was literally one of the three best laughs of my life. Have you seen Kentucky Fried Movie? Man, it's up there with Holy Grail and Life of Brian, but it was written by the Zucker Brothers and Jim Abrahms. They wrote Airplane and the Naked Gun movies. Great stuff.  Reply With Quote

18.  No. There was no irony in my response. The irony here is that you claiming to have made an on-point post is equivalent to the Black Night claiming that his arm being off is only a flesh wound. Specifically, Xei has been asking you to prove that the square root is multiplicative. You posted my proof that the square root being multiplicative on a domain implies that it distributes over division on that domain as well. Of course this would work for any multiplicative function. So it already assumed what you were asked to prove. So your post was not just not on-point but it was not even wrong. You were asked to prove one thing and proved another. What time is it? Fifty kilograms.  Reply With Quote

19.  Oh, sorry. Here we go. Originally Posted by Universal Mind -1 = -1 1/-1 = -1/1 sqrt (1/-1) = sqrt (-1/1) sqrt 1 / sqrt -1 = sqrt -1 / sqrt 1 1/i = i/1 i(1/i) = i(i/1) 1 = i^2 1 = -1 What is the deal with that? Originally Posted by Xei The inference from step 3 to step 4 is unfounded. The end. Originally Posted by Universal Mind we discussed the legitimacy of squaring the square root of a negative number and getting a negative number. Well, in this proof, the square of sqrt -1 is -1, as you believe it should be. The inference from step 3 to step 4 involves getting the square roots of negatives (not squaring them), which has not been the topic of our discussion. The inference also involves the principle that the square root of a fraction is the square root of the numerator over the square root of the denominator, an issue we have not discussed. So what is the problem in the proof? What exactly goes wrong from step 3 to step 4? Originally Posted by Universal Mind Getting the square root of a fraction by putting the square root of the numerator over the square root of the denominator seems like it should always be the logical method because squaring a fraction involves getting the square of the numerator over the square of the denominator. So, going in the reverse direction, it would make sense to get the ratio of the square root of the numerator to the square root of the denominator. But what we have run into is a situation of, "Oh no, we are dealing with imaginary numbers, so we can't do that here." You answered my question, but now I am even more convinced that the "proof" is a reductio ad absurdum of imaginary numbers. Originally Posted by Xei Saying 'it seems like it should make sense' is not a proof. Originally Posted by Xei You need to prove it properly and then you will see why the identity does not apply to all numbers, and that will fully answer the problem in the OP. Originally Posted by Universal Mind b = 1/c (valid everywhere c =/=0) sqrt(a)sqrt(b) = sqrt(ab) sqrt(a)sqrt(1/c) = sqrt(a*1/c) sqrt(a)*1/sqrt(c) = sqrt(a/c) sqrt(a)/sqrt(c) = sqrt(a/c) My complements to PhilosopherStoned for typing that in the other thread. Originally Posted by Universal Mind To be a little more to the point, the next step in your proof that I cut and pasted could be wrtitten as sqrt(a/c) = sqrt(a)/sqrt(c). (Symmetric Property of Equality) I know that sqrt(a)sqrt(b) = sqrt(ab) doesn't apply to imaginary numbers, fantasy numbers (I just added that rule.), or the Flying Spaghetti Monster. Easy Rider is another really cool movie. The philosophy and drama of it are great enough, but the sound track and cinematography add to that and make it one of the best films of all time. The Graduate is right up there with it, for the same reasons. The comedy in the Graduate makes it a different sort of movie, though.  Reply With Quote

20.  You proved that sqrt(a/c) = sqrt(a)/sqrt(c) whenever sqrt(ab) = sqrt(a)sqrt(b) and sqrt(1/c) = 1/sqrt(c). So like I said at the start, this is the same issue as before. It depends on the proof for both of these things, and the proof of these things does not work for all numbers.  Reply With Quote

21.  Hey, did Xei just repeat himself again? Hey Universal Mind, the 'Pensive' in my username is actually referring to my fifteen year old opinion of what sounded like a good username... Thanks for asking! Here is my response to your insult:  Reply With Quote

22.  Holy crap patrick! You were 15 when you signed up? That means you are gonna be 22 this year...and Im gonna be 24 >< I cant believe Ive known about this place for so long. Its incredible this community remains so bright and vibrant after so long. It didnt hit me how long Ive known about this place until you just said that. Back on topic...this is getting quite circular. UM...Xei is saying that the proof you got from philosopher is not true for every number, just every positive number. So I guess I should ask, do you believe negative numbers in general are crap? Or just roots of negatives? Negative numbers are very useful and show the removal of real physical objects, just like positives show the adding of, or the state of those physical objects. I suppose you could say from the point of view of "being" negative numbers dont make sense. But they do make sense from a relativistic point of view (ex: motion). Which I think is all they are ever used for anyways arnt they?  Reply With Quote

23. Originally Posted by Xei You proved that sqrt(a/c) = sqrt(a)/sqrt(c) whenever sqrt(ab) = sqrt(a)sqrt(b) and sqrt(1/c) = 1/sqrt(c). So like I said at the start, this is the same issue as before. It depends on the proof for both of these things, and the proof of these things does not work for all numbers. It hit me earlier today what the miscommunication has been, and tkdyo's post spelled it out. I was not quite clear enough about my perspective on what I meant by an exception needing to be conveniently introduced with the introduction of imaginary numbers. When I said it was true for all numbers and then an exception had to be made when imaginary numbers were introduced, I meant that the principle was deemed for centuries to be true for all numbers that give the equations any real meaning. Back in the day, negative numbers were not even up for consideration as square root radicands. With the introduction of imaginary numbers, an exception had to be made to the rule in terms of numbers that are actually viable in the equations in the first place. It dawned on me that you were talking about excluded values for a and b. I was talking about the radical expression values. For a long time, the square root of a negative number was not even considered a number. That is why I said an exception had to suddenly be made. What had not been considered a number became considered a number. The rule works for all values a and b except those that give radical expressions imaginary values. We of course agree on that because we both agree that the equations do not work with negative radicands. We just disagree on the nature of negative radicands. Originally Posted by Pensive Patrick Hey, did Xei just repeat himself again? Hey, I have done that too. Don't ban me from your tree house just yet. Feel free to make arguments about the actual topic. Being a mere sideline heckler is kind of weak. The video guy's insults majorly lack creativity, but that makes the video hilarious. tkdyo, you figured out the miscommunication. I explained somewhere in the sea of posts in the other thread why I do think negative numbers are legitimate and non-fictitious. If I owe you 5 dollars, you owe me -5 dollars. If I am 4 feet in front of a line, I am -4 feet behind it, etc. Negative numbers denote values below zero, which I think makes perfect sense.  Reply With Quote

24. Originally Posted by Universal Mind tkdyo, you figured out the miscommunication. I explained somewhere in the sea of posts in the other thread why I do think negative numbers are legitimate and non-fictitious. If I owe you 5 dollars, you owe me -5 dollars. If I am 4 feet in front of a line, I am -4 feet behind it, etc. Negative numbers denote values below zero, which I think makes perfect sense. Even the positive integers didn't make sense to many people for the whole history of human culture before their introduction. There are people today that I've encountered in real life that don't accept even zero let alone the negative integers. Of course, as somebody that understands them, you think that that's stupid. They normally come up on some travesty of mathematical reasoning that depends upon taking a trivial statement out of context and then call it a proof. Most of them are fairly smart outside of mathematics and formal logic. They must be to care. I once saw a man, within the same hour no less, deny the "existence" of negative numbers and then point out that we could use peanut butter that we had at hand as a sealant for an impromptu bong. That's pretty smart. When I get around to it, I'll go back to the other thread and point out how your assumptions about the "fantasy" unit, together with the assumption that they satisfy field axioms, imply that the fantasy unit is a square root for -1. I'll then leave it for you as an exercise to show that the square root isn't multiplicative on the fantasy numbers. We will prove that the fantasy numbers are isomorphic to the complex numbers. That is to say that however you take the algebraic completion of the reals, one gets the complex numbers. They're determined by the question "How do I factor every polynomial into linear roots?" to the same degree that the reals are determined by the question "How do I take the limit of any convergent series of rationals?". The reals are the analytic completion of the rationals and the complex numbers are the algebraic completion of the reals. Even if one just wants the algebraic completion of the rationals, one is still looking at a subset of the complex numbers that isn't contained in the reals, i.e. one has to factor the equation x2 + 1. The way that I originally learned to construct the complex numbers (and this is kinda backward, one would normally first encounter it in terms of R2) is as the quotient ring R[x]/(x2 + 1). It can also be embedded in matrix rings as Xei has pointed out. It represents rotations and dilations. It can also be mapped onto a sphere where it represents rotations in three space. C is a very real thing. Or at least as real as N, Z, Q, and R. Pfft. These can only represent rotations and dilations in one space and only rotations in two space. It would be nice if we had something that satisfied field axioms on general n space. A lot of mathematics (related to matrices and tensors in general, i.e. grassman algebras and differential forms) can be regarded as attempting to build some useful approximation of this. It would be nice though it's been proven impossible. We only get numbers containing the reals and using archimedian norms on one and two dimensions. Plus they're apparently even useful for practical things on occasion and that's always nice.  Reply With Quote

25.  ur my hero It would be nice if we had something that satisfied field axioms on general n space. A lot of mathematics (related to matrices and tensors in general, i.e. grassman algebras and differential forms) can be regarded as attempting to build some useful approximation of this. It would be nice though it's been proven impossible. We only get numbers containing the reals and using archimedian norms on one and two dimensions. I don't know about this, want to talk about it some more? Also out of curiosity where and when do you get all of this learning?  Reply With Quote

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