Square root, a reply to a thread in physics

[Klaynos]

From http://www.scienceforums.net/forums/showthread.php?t=20184&page=2

 

specify a negative number. it comes out positive.

squaring any real number gives a positive result' date=' the square root of a positive number gives a positive result.

if you still think otherwise, start a thread in mathematics.

however, taking the square root first will provide a negative number via [b']i[/b].

i've done a lot of work on index laws, and "not just being the inverse of the other" is exactly my point.

you cant simplify something unless there are EXACT inverse functions used

^2 is not the exact inverse of ^0.5 so it should not be simplified as such.

 

 

when my post went through, it was still 5. the refresh button may help.

 

my point is, reductionism can be taken too far and often is.

it should be used in moderation and all possibilities thouroughly checked.

 

the square root of a positive number gives a positive result.

 

I'm sorry but this is wrong, the squareroot of ANY number is a +/- result, it has 2 possibilities, that the situation dictates...

[ecoli]

true, because [imath](-x)^2 = (x)^2[/imath]

[matt grime]

The square root of a positive number is taken to mean the positive one. It is called picking a branch, and that is what we do.

[Dak]

Really quick question, but the square root of y is x or -x, but what about [imath]\sqrt{(x)^2}[/imath]? would that also be x or -x, or would it just be x? i.e., is it possible to cancell out the square root and the square against each other, leaving an unmodified x? or would it still transform x into [imath]\pm x[/imath]?

[MattC]

You still have +/-. Otherwise, you could easily use the squaring and square-rooting operations to create a paradox.

[Rocket Man]

a little back ground...

i was discussing simplification of

(x^2)^.5

turning into the function

|x|

because the squared term causes all outcomes to be positive.

a am aware that the square root term is a +/- answer, however, in a graph situation, |x| will return the same shape.

[s pepperchin]

you can only say that it is always positive if you take the square root then square it.

[Dave]

As matt says, we take the square root of a positive real number to be the positive branch. The reason? Well, if we take the function defined by [math]f(x) = \sqrt{x}[/math] then this is well-defined over the positive reals. However, the function [math]f(x) = \pm\sqrt{x}[/math] is not well-defined; it's a multi-valued function. Each value of x has two corresponding values for f(x).

 

I wouldn't go so far as to say that this is semantics, but the far more interesting case is taking the square root of a negative number. This brings things like Riemann surfaces into play, which is a highly interesting part of complex analysis.

[Rocket Man]

you can only say that it is always positive if you take the square root then square it.

 

i thought of that, but the square root of a negative number is a multiple of i, i^2 is -1.

so (x^.5)^2 = x

[matt grime]

The _convention_ is that [math]\sqrt(x)[/math], when x is positive, is the positive number whose square is x. Thus it is correct to say that (x^2)^{1/2}=|x| when x is a real number.

[Tom Mattson]

I'm sorry but this is wrong' date=' the squareroot of ANY number is a +/- result, it has 2 possibilities, that the situation dictates...[/quote']

 

Others have already pointed out that Rocket Man was right, but I'll add one thing. I think you've mistakenly conflated the idea of the square root of x with that of the solution of an equation such as [imath]x^2-2=0[/imath]. For sure there are two solutions to that equation: [imath]+/-\sqrt 2[/imath]. But [imath]\sqrt 2[/imath] is positive, by convention.

[Tom Mattson]

However' date=' the function [math']f(x) = \pm\sqrt{x}[/math] is not well-defined; it's a multi-valued function. Each value of x has two corresponding values for f(x).

 

Why does the fact that it is multi-valued make it not well-defined? The relation f is basically a set of points, and for any point P it is certainly the case that P is either in the set or not in the set. Also, you could turn it into a single-valued relation by a change of coordinates that rotates the graph by 90 degrees (so you have y=x^2). Does it make sense to say that a coordinate transformation can turn an ill-defined relation into a well-defined one?