[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.
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.[/QUOTE]
Haven't you heard of imaginary numbers? Though I'm guessing you're still in mid-school
[QUOTE=FunnyBunny;24258162]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]
Really, the square root of for example 4 is the root to the equation
x^2=4
With the lowest angle (in real numbers, the positive one). So, 2. While -2 is a root to the equation, it's not the principal square root
[QUOTE=GunsNRoses;24258201]Haven't you heard of imaginary numbers? Though I'm guessing you're still in mid-school[/QUOTE]
Of course I heard but OP's theory is based on the numbers being in the uh.. I don't know what's it in English.. real numbers?
[QUOTE=overdark;24258306]Of course I heard but OP's theory is based on the numbers being in the uh.. I don't know what's it in English.. real numbers?[/QUOTE]
In the sense that hes wrong yes.
[QUOTE=thf;24258257]Really, the square root of for example 4 is the root to the equation
x^2=4
With the lowest angle (in real numbers, the positive one). So, 2. While -2 is a root to the equation, it's not the principal square root[/QUOTE]
OP's theory cannot work in the lowest 'angle' because then there wouldn't be any '-a'.
Damn you, broke mah automerge!
OP, nobody understands your asspie grade mathematics.
[QUOTE=ThePuska;24257985]Actually no, this wasn't a deliberate attempt to hide an error. It's an attempt to find it.[/QUOTE]
Well to start with you fucked up hard here:
[img]http://img594.imageshack.us/img594/1254/fourthu.png[/img]
sqrt -1 = i
Remember, if it's not in parentheses, the negative sign won't distribute.
So you're saying the square root of negative A squared is equal to the square root of X squared which is inherently absurd. Run it in your head with A being 4 quick. Negative four squared equals sixteen, square root of sixteen equals four, cool, four squared equals sixteen, the root of negative sixteen is four i. You can't equate imaginaries to real numbers.
FP, I request that we have more of these threads more often.
[QUOTE=Xenocidebot;24258529]Well to start with you fucked up hard here:
[img]http://img594.imageshack.us/img594/1254/fourthu.png[/img]
sqrt -1 = i
Remember, if it's not in parentheses, the negative sign won't distribute.
So you're saying the square root of negative A squared is equal to the square root of x squared which is inherently absurd. Run it in your head with A being 4 quick. Negative four squared equals sixteen, square root of sixteen equals four, cool, four squared equals sixteen, the root of negative sixteen is four i. You can't equate imaginaries to real numbers.[/QUOTE]
Notice that the exponent is outside of the square root.
It's sqrt(-a)^2
Not sqrt(-a^2)
[QUOTE=TH89;24257213][IMG]http://img265.imageshack.us/img265/4333/mahts.jpg[/IMG]
I think this is your problem.[/QUOTE]
Nope, every single one of those is equivalent.
And that post having 23 agrees and 4 disagrees makes me loose hope in facepunch.
[QUOTE=ThePuska;24258552]Notice that the exponent is outside of the square root.
It's sqrt(-a)^2
Not sqrt(-a^2)[/QUOTE]
woah shit what nonsense are you using that draws them that close
[QUOTE=Xenocidebot;24258572]woah shit what nonsense are you using that draws them that close[/QUOTE]
I entered them into Wolfram Alpha and screencapped :P
Does anyone else agree that these pointless posts of people going:
"OH EM GEE I DON'T GET THIS MATHS STUFFS"
are getting in the way of people correcting the OP's mistakes?
[QUOTE=overdark;24258326]OP's theory cannot work in the lowest 'angle' because then there wouldn't be any '-a'.
Damn you, broke mah automerge![/QUOTE]
Well, the definition of the square root is exactly that, lowest angle
a isnt a number its a letter
Huh
you still have a problem where TH89 noted in which root whatever times whatever = root whatever times root whatever because that doesn't hold when whatever is negative
only true for non-negative reals
but anyway
so instead it's
sqrt negative 4 which is 2i, 2i^2 is 4*-1 so -4
sqrt negative 4 which is 2i we just did this
the next one fails because of what I just said
sqrt negative one * sqrt A = sqrt negative A doesn't hold
nerds
this is like the time during GCSE maths class I though I proved 3 equals 5.
I still managed to get an A
What is so difficult about it? This just proofs what they taught us in school. Root equals absolute value of the number (+/-).
[QUOTE=FunnyBunny;24258560]Nope, every single one of those is equivalent.
And that post having 23 agrees and 4 disagrees makes me loose hope in facepunch.[/QUOTE]
X= -5, tell me how they are fucking equivalent.
Mind = Blown to fucking pieces.
Also :page3:
since [b]sqrt(x^2)=abs(x)[/b] you've proven that [b]abs(-a)=a[/b]...amazing
[QUOTE=RP.;24259491]since [b]sqrt(x^2)=abs(x)[/b] you've proven that [b]abs(-a)=a[/b]...amazing[/QUOTE]
No.
sqrt(x^2) = |x|
sqrt(x)^2 = x
If sqrt(x^2) was sqrt(x)^2 then I'd have proven that x = |x| - the implication of which would be that every negative number is actually positive and vice versa.
[QUOTE=RP.;24259491]since [b]sqrt(x^2)=abs(x)[/b] you've proven that [b]abs(-a)=a[/b]...amazing[/QUOTE]
This.
that's 1/3 of what i said in the OP
[QUOTE=DrLuke;24257651]sqrt(x) * sqrt(x) = x[/QUOTE]
For positive numbers only.
[img]http://math.daggeringcats.com/?\sqrt{x^2} \neq \sqrt{x}^2[/img] in general. That's what he is doing incorrectly.
[img]http://math.daggeringcats.com/?\sqrt{x^2} = x[/img] is valid for positive and negative values of x.
[img]http://math.daggeringcats.com/?\sqrt{x}^2 = x[/img] is only valid for positive values. Taking into account negative values of x, you need to make use of the imaginary unit and then you get
[img]http://math.daggeringcats.com/?\sqrt{-x}^2 = (\pm i\sqrt{x})^2 = -x[/img]. And that's what he is doing wrong. So actually he should have written
[img]http://math.daggeringcats.com/?\sqrt{x}^2 = \pm x[/img] (for real numbers).
[editline]10:56AM[/editline]
[QUOTE=Xenocidebot;24258529]...
sqrt -1 = i
...[/QUOTE]
i is defined by [img]http://math.daggeringcats.com/?i^2 = -1[/img] only. [img]http://math.daggeringcats.com/?\sqrt{-1} = \pm i[/img] is the proper use to deal with sqrt of negative numbers. Since this is loosely written the only definition is for i is the first one.
I remember I saw OPs calculations in my math book. It was in the introduction of imaginary units I believe.
The thing you've done is one of the mathematical fallacies.
Much easier proof stolen from Wikipedia.
[img]http://upload.wikimedia.org/math/b/e/c/bece7a4ff69a1ca64899e8693b697dc3.png[/img], which is wrong...
because
[img]http://upload.wikimedia.org/math/b/a/2/ba207ebfb441f09f40ca7937a12215df.png[/img] " is generally valid if at least one or two numbers are positive".
You CAN take the square root of a negative number, you just get an imaginary one, which is actually often used in maths.
[QUOTE=aVoN;24259856]i is defined by [img]http://math.daggeringcats.com/?i^2 = -1[/img] only. [img]http://math.daggeringcats.com/?\sqrt{-1} = \pm i[/img] is the proper use to deal with sqrt of negative numbers. Since this is loosely written the only definition is for i is the first one.[/QUOTE]
it was four fucking AM and I was trying to give him the quick straight dope not lay it out in full he ain't three ease up
besides we were both pretty fucking late to the party considering this was finished on page 1
-snip-
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