Substitution for |x|

[keebler_giant]

I was wondering, if when trying to prove something, are you allowed to let |x|=(x*x)^(1/2)

[homology]

Is |x| the modulus of x, a complex number and is the asterick to indicate conjugation?

[keebler_giant]

|x| is the absolute value of x, and x*x is just x times x:tumble:

[homology]

Hmmm, I don\'t know. I suppose as long as you only consider the positive square root, but just keep in mind the square root returns two possible answers.

[PritamDamania]

(x*x)^1/2 returns only positive values and not two values as stated by homology.
Therefore

(x*x)^1/2 = |x|

The two values homology
is talking about only arises when we take square root on both sides of equality. The square root radical is always positive.
For example:

If x*x = a*a
then now if we take square roots on both sides x=+or-a

But (4)^1/2 = 2 and not + or - 2.

I hope you have understood that the + or - arises when we take square root on both sides of equality.

This is a common misconcept which mostly develops in lower classes when you are introduced to the square root radical. I also understood the square root radical properly only in high school.

[homology]

Hmm, could you justify your argument more PritamDamania?

The square roots of four are certainly both 2 and -2.

[ket]

If x*x = a*a
then |x| = |a|. (x, a real)

The square root function, applied to positive real numbers, returns the positive root. If it returned multiple values, it would be much less useful in proofs. It wouldn\'t even be a function.

Applied to negative and complex numbers, it\'s harder to decide which square root to use.


[Edited on 2004-6-24 by ket]

[PritamDamania]

homology


I think i should illustrate a bit on this.
Let
x*x = 4
Now if we take square root on both sides

(x*x)^1/2 = 2

but (x*x)^1/2 = |x| and not x

therefore |x| = 2
and thus we get x=+or-2

Now consider

(4)^1/2 this can be written as
(2*2)^1/2
or
(-2*-2)^1/2
for the first case answer is
|2| = 2
and second case is
|-2| = 2
Now i think you must have understood where the +or- case comes. This is also used as a result as square root of any quantity is is positive. As i said earlier what you are talking about is a severe misconcept which develops in lower classes and is hardly sorted out by any teacher or professor in higher classes. When i was in about 7th or 8th standard i was also told that square root of 4 is +or-2 but then fortunately one of my professors cleared that out for me just about a year ago

[homology]

Huh, :) well I\'ll be, I\'ll have to think about that.

I can see the importance of the absolute value when dealing with functions. But if

|x|=sqrt[x*x] then x=+sqrt[x*x], certainly this is not a function, but it does return the same values and satisfies the original equation.

However, the form |x|=sqrt[x*x] is certainly the one to use in proofs and such. Verrryyyy tricky, very tricky indeed.

Thanks for the help

Kevin

[Edited on 6-24-2004 by homology]

[Larne]

Originally posted by keebler_giant
I was wondering, if when trying to prove something, are you allowed to let |x|=(x*x)^(1/2)

I wouldn\'t. They may seem to be equal but they really aren\'t the same function (for example, their derivatives are different).

Generally the safer thing to do on a proof would be to prove the statement for x >= 0, and again for x < 0.

[ket]

Originally posted by Larne

I wouldn\'t. They may seem to be equal but they really aren\'t the same function (for example, their derivatives are different).



Nope. They\'re identical functions (equal at all values x) so they have to have the same derivative. In this case, the derivative is x/|x|

[keebler_giant]

Yep...they do have the same derivatives...well if we allow (x*x)^(1/2) to equal |x|

y=(x*x)^(1/2) is the same as saying that
y*y=x*x...and when you find dy/dx you get 2y(dy/dx)=2x, so dy/dx=x/y, and since originally y=(x*x)^(1/2), dy/dx=x/[(x*x)^(1/2)]

And if y=|x|, dy/dx=x/|x|...so either way, the derivatives are equal

[Larne]

Originally posted by keebler_giant
Yep...they do have the same derivatives...well if we allow (x*x)^(1/2) to equal |x|

Yes, you\'re right -- in thinking about this I carelessly made the very substitution that\'s at issue here. I suppose the lesson is that many simplifications that are usually taken for granted really aren\'t valid over the real numbers.

But I\'ll stand by my main point. |X| has a perfectly valid definition, it equals x when x >=0 and -x when x < 0. Such piecewise functions are important in both math and physics, and I\'m not sure what\'s gained by trying to rewrite them in a closed form.

[ket]

Originally posted by Larne
I\'m not sure what\'s gained by trying to rewrite them in a closed form.

Oh yeaaaaahhh.... I was going to say that but I forgot. I\'m of the opinion that it won\'t help at all to rewrite |x|. :)