• Proof that a = -a
    89 replies, posted
I believe what you've proven there is that the absolute value of a is the same as the absolute value of -a, which is of course true.
[QUOTE=Doom14;24257416] You don't do SqRt with negatives unless you get into imaginary (i) numbers which tend to just fuck everything up. They also, if I remember right, don't equal 0.[/QUOTE] Which is why this fails, because, but even if you count in imaginary numbers, I x I = -1
The equation is wrong. You're over-expanding it. [img]http://img715.imageshack.us/img715/5193/firstx.png[/img] The square root of x^2 should just be x. The square root and the squared cancel out. So square root of x^2 equals itself, but for the purpose of substitution, it equals x.
[QUOTE=kevn150;24257467]The equation is wrong. You're over-expanding it. [IMG]http://img715.imageshack.us/img715/5193/firstx.png[/IMG] The square root of x^2 should just be x. The square root and the squared cancel out. So square root of x^2 equals itself, but for the purpose of substitution, it equals x.[/QUOTE] Hes trying to 'prove it', 'simplifying it down' means you miss the point entirely.
I think everything I used in the OP was pretty standard stuff apart from possibly the imaginary numbers. But all those steps should be acceptable. I confirmed them from a book we're supposed to be using in our matriculation exam - which is problematic since the book's been checked by quite a few math professors. Can you find any source that explicitly states that sqrt(x*x) =/= sqrt(x)*sqrt(x) for negative x?
Next OP is gonna tell us 2+2=4.
[QUOTE=ThePuska;24257536]I think everything I used in the OP was pretty standard stuff apart from possibly the imaginary numbers. But all those steps should be acceptable. I confirmed them from a book we're supposed to be using in our matriculation exam - which is problematic since the book's been checked by quite a few math professors. Can you find any source that explicitly states that sqrt(x*x) =/= sqrt(x)*sqrt(x) for negative x?[/QUOTE] [url]http://en.wikipedia.org/wiki/Imaginary_unit[/url]
Please note that this discussion has no point at all.
sqrt(x) * sqrt(x) = x
Math is silly.
[QUOTE=Neolk;24257566][URL]http://en.wikipedia.org/wiki/Imaginary_unit[/URL][/QUOTE] This: The imaginary number [I]i[/I] is defined SOLELY by the property that its [URL="http://en.wikipedia.org/wiki/Square_%28algebra%29"]square[/URL] is −1: [img]http://upload.wikimedia.org/math/4/4/b/44b33da6be905320353c4c1906ac7d29.png[/img] The link given describes the proper use of imaginary numbers.
[QUOTE=TH89;24257213][img]http://img265.imageshack.us/img265/4333/mahts.jpg[/img] I think this is your problem.[/QUOTE] Oh shit a red circle check out the teacher man.
So your theory fails X and Y X = (Y*-1) Y > 0 √(Y*Y) = √(Y) * √(Y) = Y, this is true BUT √((Y*-1)*(Y*-1) =/= √(Y*-1) * √(Y*-1). So Instead, we substitute I in for the -1, since it is impossible to have a -1 in a square root. √((Y*-1)*(Y*-1) = I√(Y) * I√(Y) The Ys cancel into Y, and the I's become -1. Thus, -Y, or as we established before, X. X = Y*-1 and thus, X =/= Y
what the fucking hell am i looking at
[QUOTE=Neolk;24257824]So your theory fails X and Y X = (Y*-1) Y > 0 √(Y*Y) = √(Y) * √(Y) = Y, this is true BUT √((Y*-1)*(Y*-1) =/= √(Y*-1) * √(Y*-1). So Instead, we substitute I in for the -1, since it is impossible to have a -1 in a square root. √((Y*-1)*(Y*-1) = I√(Y) * I√(Y) The Ys cancel into Y, and the I's become -1. Thus, -Y, or as we established before, X. X = Y*-1 and thus, X =/= Y[/QUOTE] I'm not sure you understand it yourself either. You can't do the step √((-Y)*(-Y)) = i√Y * i√Y because the former is equal to √(Y*Y) and the latter is equal to -√(Y*Y) Thank you for the Wiki link, it clarified this a bit. I'll have to complain to the publisher of that book because they're extremely ambiguous about when sqrt(ab) = sqrt(a)*sqrt(b) holds true.
[QUOTE=ThePuska;24257128][img]http://img715.imageshack.us/img715/5193/firstx.png[/img] So when [img]http://img825.imageshack.us/img825/6500/second.png[/img] (and a is a real number) [img]http://img693.imageshack.us/img693/6130/thirdd.png[/img] [img]http://img843.imageshack.us/img843/1254/fourthu.png[/img] Therefore a = -a when a > 0. What am I doing wrong?[/QUOTE] -a^2 is NOT EQUAL to (-1)^2 * a^2. (-1)^2 * a^2 is a^2
Fucking [B][I][U]what is going on.[/U][/I][/B]
[QUOTE=Mister B;24257954]Fucking [B][I][U]what is going on.[/U][/I][/B][/QUOTE] This is an old bunch of nonsense everyone learns in one form or another for proving that 1 = 0 or 1 = -1 or whatever else you want by hiding mistakes where people don't tend to notice them (sign errors being proven to be, what, 70% of all errors by students? It's the simple shit.) It doesn't hold, it's just funny to pore over looking for that really obvious flaw you're missing.
My brain :psyduck:
Just because the square root is the same doesn't mean they're the same number
[QUOTE=Xenocidebot;24257967]This is an old bunch of nonsense everyone learns in one form or another for proving that 1 = 0 or 1 = -1 or whatever else you want by hiding mistakes where people don't tend to notice them (sign errors being proven to be, what, 70% of all errors by students? It's the simple shit.) It doesn't hold, it's just funny to pore over looking for that really obvious flaw you're missing.[/QUOTE] Actually no, this wasn't a deliberate attempt to hide an error. It's an attempt to find it.
[QUOTE=TH89;24257213][img]http://img265.imageshack.us/img265/4333/mahts.jpg[/img] I think this is your problem.[/QUOTE] Nothing wrong with that. sqrt(x^2) = x, and sqrt(x)^2 = x
Basic ITT: Someone youtubes "math trick" and finds a stupid trick that is actually a mathematical error and thinks "LOLLOLO FACEPUNCH THREAD IM SO CUUL" and makes a thread. :woop:
[QUOTE=ThePuska;24257937]I'm not sure you understand it yourself either. You can't do the step √((-Y)*(-Y)) = i√Y * i√Y because the former is equal to √(Y*Y) and the latter is equal to -√(Y*Y) Thank you for the Wiki link, it clarified this a bit. I'll have to complain to the publisher of that book because they're extremely ambiguous about when sqrt(ab) = sqrt(a)*sqrt(b) holds true.[/QUOTE] Ya I fucked up, you can't make the jump period, √(x*x) =/= √(x) * √(x) not always anyway
All it really proves is that -a2 = still equals a2, and the rest is irrelevant for i thought we (I) already knew this, a has the same mathematic destination as its opposite.
This is basic midschool maths and anyone who fails to notice the mistake is pretty much dumb. Simplest way to put this: the number, x, can only be >=0, but its square root can be either positive or negative. Example: 2*2=4 (-2)*(-2)=4 So sqrt(4)=(+/-)2. That doesn't, and can never (if you follow unanimously accepted algebra and not some stupid The Big Bang Theory shit) mean that 2 equals -2. Another example of this is, you can prove that any number equals to any other number by dividing them both by zero.
Fuck your math, school hasn't started yet. D:
[QUOTE=overdark;24258075]This is basic midschool maths and anyone who fails to notice the mistake is pretty much dumb. Simplest way to put this: the number, x, can only be >=0, but its square root can be either positive or negative. Example: 2*2=4 (-2)*(-2)=4 So sqrt(4)=(+/-)2.[/QUOTE] That has nothing to do with this. Besides, sqrt(4) = 2. There are two roots for equations such as x^2 = 4 but there's only one principal square root of 4.
As been said before, the Sqrt of anything is +/- that number. You don't know which it is, but it can only be one, so either -a = -a or a=a.
[QUOTE=ThePuska;24258147]That has nothing to do with this. Besides, sqrt(4) = 2. There are two roots for equations such as x^2 = 4 but there's only one principal square root of 4.[/QUOTE] But it does. He's using the square roots to prove that a = -a.
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