What? You've already been given a painstaking explication of exactly this. It isn't a reductio ad absurdum. |
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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. |
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Last edited by Universal Mind; 03-25-2012 at 09:57 PM.
You are dreaming right now.
What? You've already been given a painstaking explication of exactly this. It isn't a reductio ad absurdum. |
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Last edited by Xei; 03-26-2012 at 01:34 AM.
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? |
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You are dreaming right now.
"the principle that the square root of a fraction is the square root of the numerator over the square root of the denominator" |
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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. |
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You are dreaming right now.
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. |
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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? |
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You are dreaming right now.
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. |
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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. |
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You are dreaming right now.
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. |
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Previously PhilosopherStoned
Also contingent upon sqrt(1/c) = 1/sqrt(c). |
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Just note that (1/sqrt(c))^2 = 1/(sqrt(c)^2) = 1/c by the definition of multiplication. |
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Previously PhilosopherStoned
Has the same caveat as sqrt(a)sqrt(b) = sqrt(ab)and its proof, that's what I meant to highlight. |
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Xei, I am thoroughly impressed by your perseverance with this topic. It seems like some people just want you to keep repeating yourself... |
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You are dreaming right now.
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! |
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Previously PhilosopherStoned
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. |
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Last edited by Universal Mind; 03-28-2012 at 05:01 AM.
You are dreaming right now.
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. |
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Previously PhilosopherStoned
Oh, sorry. Here we go. |
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You are dreaming right now.
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. |
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Hey, did Xei just repeat himself again? |
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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. |
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A warrior does not give up what he loves, he finds the love in what he does
Only those who attempt the absurd can achieve the impossible.
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. |
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Last edited by Universal Mind; 03-28-2012 at 09:56 PM.
You are dreaming right now.
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. |
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Previously PhilosopherStoned
ur my hero |
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