Why is sqrt(x²) not equal to x? - General

If a^b^c = a^bc = a^cb then I could write x^2^1/2 as x^2/2= x^1=x so that would prove that x^2^1/2 equals x, but it doesn't! Because (ab)^1/2 = (a^1/2)*(b^1/2) and a^b^c = a^bc = a^cb, you could write x^2^1/2 as (x^1/2)*(x^1/2), but in this case, x cannot be negative... So is a^b^c = a^bc = a^cb true? Is multiplication in exponents commutative or not? What am I doing wrong? [B]Edit:[/B] I think I'm not holding into account that there's -sqrt and +sqrt

:psyboom:

sqrt(x^2) [b]IS[/b] x, so long as x is positive.

X can be negative too, the function looks like this [img]http://gyazo.com/fd5b5b257847ef8c7308fc5f6e006f05.png[/img]

[QUOTE=Number-41;24517365]If a^b^c = a^bc = a^cb then I could write x^2^1/2 as x^2/2= x^1=x so that would prove that x^2^1/2 equals x, but it doesn't! Because (ab)^1/2 = (a^1/2)*(b^1/2) and a^b^c = a^bc = a^cb, you could write x^2^1/2 as (x^1/2)*(x^1/2), but in this case, x cannot be negative... So is a^b^c = a^bc = a^cb true? Is multiplication in exponents commutative or not? What am I doing wrong?[/QUOTE] The title tricked me into believing this was a simple math question. Now my brain is full of fuck.

sqrt((x²)) will be equal to x even if it's negative because -x =/= (-x)

Where is JohnnyMo when you need him :sigh: I'm sure that I'm just forgetting some extremely basic math rule... My math book states this: (∀ a ∈ R+0, z ∈ Z) ((sqrt(a)^z=sqrt(z)^a)) So multiplication in exponents is only associative when a is strictly positive... I came to this problem because when calculating infinity limits like lim sqrt(g(x)), you take the highest grade monomial (e.g. x^2) and bring it outside of the square root. When calculating positive infinity, it's just x*sqrt(function divided by x²) and when it's negative infinity, it's -x*sqrt(function divided by x²). I don't see how they get to the minus in the second case...

sqrt(x²) = X as long as X > 0 You cant take the sqrt out of a negative number (sqrt(-1) = inf)

[QUOTE=jimhowl33t;24517423]The title tricked me into believing this was a simple math question. Now my brain is full of fuck.[/QUOTE] [IMG]http://images.memegenerator.net/jackie-chan/ImageMacro/941156/MY-HEAD-IS-FULL-OF-FUCK.jpg[/IMG] same

[QUOTE=windwakr;24517750][code] sqrt(x²)= |x| for any x. [/code] [editline]...[/editline] You updated your post. About the sqrt(-1), yes you can, it's i. [b]But more importantly, x² is always positive.[/b] So like I said, sqrt(x²) = |x|[/QUOTE] and then sqrt(x²) if x<0 is -x ? If it is, then I can stop worrying...

Yeah pretty much but I thought everyone already knew that. What I'd like to know is why sqrt(x^2) and (x^2)^1/2 aren't the same thing

or -70 :eng101: because the function x²=4900 has two solutions

[QUOTE=PacificV2;24517887]Yeah pretty much but I thought everyone already knew that. What I'd like to know is why sqrt(x^2) and (x^2)^1/2 aren't the same thing[/QUOTE] I think with the latter, you bring in the rule that the powers multiply, making it the same as x^1, keeping it negative if it was originally negative.

Actually yeah now that you say it... [B]Edit:[/B] nope I lost it [B]Edit:[/B] or wait I think I get it again, I meant sqrt(x²) = sqrt(x)² The 2nd function WolframAlpha says that it's the same as x, but it can't be!

[QUOTE=Rasrap Smurf;24517929]I think with the latter, you bring in the rule that the powers multiply, making it the same as x^1, keeping it negative if it was originally negative.[/QUOTE] Pretty much, but from what I could tell, sqrt(x) = x^(1/2), so sqrt(x^2) = (x^2)^(1/2), which simplifies to x^2^(1/2) = x^(2*(1/2)) = x^1 = x, which isn't true for |R [editline]05:43PM[/editline] I've always thought about this problem as the exponent operation taking priority in regards to the square root, but if we put the whole operation in just exponents (like x^2^1/2) the situation becomes muddier. I ain't a math major but I'd like to know how this unravels.

WolframAlpha isn't helping either... It makes the same mistake as I do

[QUOTE=PacificV2;24517887]Yeah pretty much but I thought everyone already knew that. What I'd like to know is why sqrt(x^2) and (x^2)^1/2 aren't the same thing[/QUOTE] They are the same. [code] sqrt(10^2) = 10 (10^2)^.5 = 10 sqrt((-10)^2) = 10 ((-10)^2)^.5 = 10 sqrt(-10) = i (-10)^.5 = i [/code]

Yeah the problem is now if a^b^c = a^bc = a^c^b then why is sqrt(x²) not equal to (sqrt (x))²

[QUOTE=ColdFusion;24518243]They are the same. [code] sqrt(10^2) = 10 (10^2)^.5 = 10 sqrt((-10)^2) = 10 ((-10)^2)^.5 = 10 sqrt(-10) = i (-10)^.5 = i [/code][/QUOTE] Way to read the thread

-snip- Nevermind

I thought this was going to be an easy question I could answer :C

[QUOTE=ColdFusion;24518482][code] sqrt(10^2) = 10 (sqrt(10))^2 = 10 [/code] I dont see it EDIT: Ofc that wont work with a negative number. But that's just the order you do the calculation in[/QUOTE] stop using your calculator we KNOW that it doesn't work like that but we can't find an analytical reason (x^2)^(1/2) = x^2*(1/2) = x which is NOT true for |R

Oh right. [url]http://en.wikipedia.org/wiki/Real_number[/url] |R (or apparently just R) represents the set of all real numbers. when I mean it's not true for R it means that for every element of R (aka for every number) the equation in question isn't correct. the equation being sqrt(x^2) = (x^2)^(1/2) that simplifies to sqrt(x^2) = (x)^(2*(1/2))

[QUOTE=windwakr;24518630]Excuse me, but what is |R?[/QUOTE] Real numebers

This is a better question than it seemed from the title. I can't think of a reason and I've been looking for one. If you can wait until about 7:00 EST I'll be back from my class and I'll ask my professor. Shit you're in belgium it will be like 1 am. Well, I'll post the reason here anyway.

this is why i dont like math

I feel like it has something to do with the domain of x^(1/2)

[QUOTE=JohnnyMo1;24518851]I feel like it has something to do with the domain of x^(1/2)[/QUOTE] Probably, that's the only explanation I can think of. If you don't consider domains it makes no sense and has no obvious solution. If you limit the domain though it becomes much simpler

Why are you people making his maths homework for him?

[QUOTE=koekje4life V2;24518916]Why are you people making his maths homework for him?[/QUOTE] Any math teacher who would assign this question for homework is a dick. [editline]01:24PM[/editline] Or actually pretty cool, but it would have to be worth extra credit.