No real number x such that x<|0| where

[SW VandeCarr]

I'm not sure this is a proper statement since zero is not signed. However, I want to say that there is no real number x such that x<|0| where || contains only absolute valued real numbers and zero.

 

[Wizlem]

|a| is defined as a if $$a \geq 0$$ and -a if a<0

 

[SW VandeCarr]

|a| is defined as a if $$a \geq 0$$ and -a if a<0

I'm not sure I follow this. If -a=|a| and a=|a| then one could say -a=a which is clearly not true.

I want to say there is no real number x less than |0|.

 

[Outlined]

|0| = 0
it is not more complicated than that

 

[Petek]

The statement that there is no real number x such that x < |0| is false. For example, -1 < |0| since, as was pointed out, |0| = 0. Perhaps you meant to write that there is no real number x such that |x| < 0?

 

[SW VandeCarr]

The statement that there is no real number x such that x < |0| is false. For example, -1 < |0| since, as was pointed out, |0| = 0. Perhaps you meant to write that there is no real number x such that |x| < 0?

No. But |0|=0 does answer my question. So |x|= x, -x unless x=0.

EDIT: So for the function f(x)=|x|, where x is any real number; how do we describe the point f(0)? It is a singular point and clearly is a minimum point. So can we say that f(0) is smaller than any f(x) where x is not 0? I know this was suggested earlier, but given the function f(x)=|x|, then f(0)=|0|, so this in a sense defines |0| in the context of f(x).

 

[Studiot]

I'm not sure what you are driving at, but the function

f(x) = mod(x) is bounded below.

The point zero forms the greatest lower bound.

The function is not differentiable at x=0

 

[Outlined]

To add some information to the discussion: $$-0 = 0$$, also $$0 \le 0$$.

 

[Mark44]

So |x|= x, -x unless x=0.

A better way to say this is
|x| = x, if x >= 0
|x| = -x, if x < 0

The way you wrote it suggests that |x| has two values, which isn't true.

 

[SW VandeCarr]

A better way to say this is
|x| = x, if x >= 0
|x| = -x, if x < 0

The way you wrote it suggests that |x| has two values, which isn't true.

|x| does not have two values. I defined a function f(x)=|x|. For each pair x, -x there exists a unique value of f(x). For x=0 there is a singular point and a minimum on the function f(x) such that f(0)=|0|.

I just wanted to show that while |0|=0, |0| has the property of being a singular point and a minimum on the V shaped function f(x)=|x| while 0 is a nonsingular and not a minimum on the real number line.

 

[Studiot]

An interesting question for you to consider:

What is the difference between

$$\begin{array}{l}<br /> f(x) = |x| \\ <br /> f(x) = \sqrt {{x^2}} \\ <br /> \end{array}$$

 

[SW VandeCarr]

An interesting question for you to consider:

What is the difference between

$$\begin{array}{l}<br /> f(x) = |x| \\ <br /> f(x) = \sqrt {{x^2}} \\ <br /> \end{array}$$

Different functions, but the same result except possibly at f(0). Is $$\sqrt 0^{2}$$ defined? As you approach zero from the positive side for real x less than 1, the squared values of x are smaller than x. Obviously this pattern cannot hold at x=0.

EDIT: On rethinking this, the ratio $$x^{2}/x$$ does seem go to zero at the limit so I guess 0*0=0 is valid.

 

[Petek]

An interesting question for you to consider:

What is the difference between

$$\begin{array}{l}<br /> f(x) = |x| \\ <br /> f(x) = \sqrt {{x^2}} \\ <br /> \end{array}$$

The only difference I can think of is that the square root function, when applied to a positive real number, has two possible values (one positive and the other negative). The usual convention is to use the positive value. With that convention the two functions have the same value for all reals. If we chose the negative square root, then the functions would not be the same. Is that what you had in mind?

 

[SW VandeCarr]

The only difference I can think of is that the square root function, when applied to a positive real number, has two possible values (one positive and the other negative). The usual convention is to use the positive value. With that convention the two functions have the same value for all reals. If we chose the negative square root, then the functions would not be the same. Is that what you had in mind?

I have to give Studiot credit. It seems we have an X shaped function centered on the origin with the upper half plane having the same V form as f(x)=|x| and the lower half plane also having the same form, but inverted. I completely missed this.

 

[Studiot]

Sorry I thought I would not need to tell a bunch of mathematical physicists that the positive root only is defined to be the root function.
Don't forget that to qualify as a function f(x) has to be single valued. So we consider the whole graph as two functions stitched together.

The hint about any differences lies in my previous post.

|x| is continuous at 0, but not differentiable there - you cannot draw a tangent.

What about the (positive) root function?

 

[Petek]

Are you saying that $\sqrt{x^2}$ is differentiable at x = 0?

 

[Outlined]

Are you saying that $\sqrt{x^2}$ is differentiable at x = 0?

No, he says it is NOT differentiable in 0.

 

[Petek]

No, he says it is NOT differentiable in 0.

Studiot seems to claim that there's a difference between |x| and $\sqrt{x^2}$. Perhaps he's just trying to get someone to say that there is no difference.

 

[g_edgar]

What is the difference between

$$\begin{array}{l}<br /> f(x) = |x| \\ <br /> f(x) = \sqrt {{x^2}} \\ <br /> \end{array}$$

When x is real, no difference. When x is complex, much difference.

 

[SW VandeCarr]

When x is real, no difference. When x is complex, much difference.

Well, I've been specifying real x throughout this thread.

 

[mathman]

|x| is a well defined function. Since the square root has two values, it becomes |x| by convention to take only the positive value.

 

[HallsofIvy]

|x| is a well defined function. Since the square root has two values, it becomes |x| by convention to take only the positive value.

Oh, dear, mathman. Surely you know better than that. The equation $x^2= a$ has two solutions but only one of them is $\sqrt{x}$. The square root, like any real valued function, has only one value for each x.

 

[Studiot]

My little interjection is generating some definite head scratching and good discussion.

I'm glad I didn't just pontificate.

 

[Coin]

If we're nitpicking, then I want to nitpick this:

To add some information to the discussion: $$-0 = 0$$

Except in computer floating point arithmetic! ^_^

 

[Hurkyl]

If we're nitpicking, then I want to nitpick this:



Except in computer floating point arithmetic! ^_^

Sure, but those aren't real numbers. (or complex numbers!)

 

[CRGreathouse]

Except in computer floating point arithmetic! ^_^

That depends on what the meaning of "is" is. They do compare as equal...

 

[mathman]

Oh, dear, mathman. Surely you know better than that. The equation $x^2= a$ has two solutions but only one of them is $\sqrt{x}$. The square root, like any real valued function, has only one value for each x.

Basically you are saying that $\sqrt{x}$ is always the positive value for positive x. All I am asserting is that it is by convention. For other x (negative or complex) there is no convention, so both roots are equally valid.

 

[Hurkyl]

Basically you are saying that $\sqrt{x}$ is always the positive value for positive x.

Nonnegative.

All I am asserting is that it is by convention. For other x (negative or complex) there is no convention, so both roots are equally valid.

Actually, there is a convention -- the principal square root, I think, is the first or fourth quadrant. (the positive imaginary axis, for the square roots of negative numbers)

All square roots of a number are "equally valid" square roots, even the negative square root of a positive number, for what seems to me to be the natural interpretation of those English words.