Trying to understand the property of absolute value inequality

[mark2142]

Homework Statement

$|x|<c$ is equivalent to $-c<x<c$

Relevant Equations

I tried to understand it by breaking each part.

First lets focus on $|x|$ which is defined as distance between $x$and $0$. But if we look into it closely
$$13=|-11-2|$$ which is distance between -11 and 2 but $$13=|11-(-2)|$$ which means this is distance between 11 and -2. Which is it?
In the same way $$x=|x-0|$$ is distance between 0 and x but $$x=|0-(-x)|$$ is distance between 0 and -x. Which is it?

Second lets focus on $c$ which is defined as distance between 0 and c and distance between 0 and -c.
$$|c-0|=c$$ and $$|-c-0|=c$$
$c$ mean two things but shouldn't we take one meaning?
Either we take distance between 0 and c or between 0 and -c.

 

[FactChecker]

Homework Statement: $|x|<c$ is equivalent to $-c<x<c$
Relevant Equations: I tried to understand it by breaking each part.

First lets focus on $|x|$ which is defined as distance between $x$and $0$. But if we look into it closely
$$13=|-11-2|$$ which is distance between -11 and 2 but $$13=|11-(-2)|$$ which means this is distance between 11 and -2. Which is it?
In the same way $$x=|x-0|$$ is distance between 0 and x but $$x=|0-(-x)|$$ is distance between 0 and -x. Which is it?

It is both. The absolute distance between -11 and 2 is 13 and also the absolute distance between 11 and -2 is 13.

Second lets focus on $c$ which is defined as distance between 0 and c and distance between 0 and -c.
$$|c-0|=c$$ and $$|-c-0|=c$$
$c$ mean two things but shouldn't we take one meaning?

No. The absolute distance is a number that can be used many different ways. Just because the distance between one house and another is one block does not mean that the distance between any other two houses can never be one block.

Either we take distance between 0 and c or between 0 and -c.

|c-0| is the absolute distance. It is positive no matter what the sign of the number inside is.

 

[PeroK]

@mark2142 draw a graph!

 

[mark2142]

The absolute distance is a number that can be used many different ways. Just because the distance between one house and another is one block does not mean that the distance between any other two houses can never be one block

Ok. 13 can be written in infinite ways. $|0-13|$ , $|0-(-13)|$ , $|5-(-8)|$ , $|-2-11|$. Similarly if $x$ is a variable in place of a number then $|x|$ is the distance between 0 and x. But how do we add $-x$ to the story to prove $-c<x<c$?

 

[Mark44]

Ok. 13 can be written in infinite ways. $|0-13|$ , $|0-(-13)|$ , $|5-(-8)|$ , $|-2-11|$. Similarly if $x$ is a variable in place of a number then $|x|$ is the distance between 0 and x. But how do we add $-x$ to the story to prove $-c<x<c$?

You don't need to prove it, because it's a definition. |x| < c means -c < x < c. Here c is assumed to be a positive constant.

As was already suggested, draw a section of the number line, with points at -c and c. x can be any number between these two points.

 

[Petek]

Instead of saying that the absolute value of a number is its distance from zero, I suggest that you use the following definition:

If $x \geq 0$ then $|x| = x$. If $x < 0$ then $|x| = -x$.

Then consider the two cases in which either $x \geq 0$ or $x < 0$.

 

[PeroK]

Ok. 13 can be written in infinite ways. $|0-13|$ , $|0-(-13)|$ , $|5-(-8)|$ , $|-2-11|$. Similarly if $x$ is a variable in place of a number then $|x|$ is the distance between 0 and x. But how do we add $-x$ to the story to prove $-c<x<c$?

It's not clear what you are trying to do here. What's not clear about the modulus function $|x|$?

 

[mark2142]

It's not clear what you are trying to do here. What's not clear about the modulus function |x|?

$|x|$ means distance between x and 0. But I am thinking we can also write $|x|=|0-(-x)|$ which means it is the distance between 0 and -x. Am I right?

 

[mark2142]

Instead of saying that the absolute value of a number is its distance from zero, I suggest that you use the following definition:

If $x \geq 0$ then $|x| = x$. If $x < 0$ then $|x| = -x$.

Then consider the two cases in which either $x \geq 0$ or $x < 0$.

If $x>0$ then $|x|<c$ means $x<c$
If $x<0$ then $|x|<c$ means $x>-c$
So for all $x$ , $|x|<c$ means $-c<x<c$ ($x$ lies between $-c$ and $c$).

 

[Petek]

Your argument looks good* (we assume that c > 0). You've shown that if |x| < c, then -c < x < c. To show the statements are equivalent, use a similar argument to show that -c < x < c implies that |x| < c.

*Except you didn't cover the case x = 0, which is very simple. If you're writing this solution up to be graded, you should include that possibility.

 

[mark2142]

If $x \geq 0$ , $-c<x<c$ means $|x|<c$
If $x<0$ , $-c<x<c$ means $|x|<c$
So $|x|<c = -c<x<c$

means distance between x and 0. But I am thinking we can also write |x|=|0−(−x)| which means it is the distance between 0 and -x. Am I right?

Can you answer this please (post # 8) ?

 

[YouAreAwesome]

$$|c-0|=c$$ and $$|-c-0|=c$$
$c$ mean two things but shouldn't we take one meaning?
Either we take distance between 0 and c or between 0 and -c.

Where exactly are you having trouble? It seems to me you are asking why two distinct yet equivalent statements compute the same result. Is this question akin to asking, "Is $x$ the distance from $0$ to $x$, or from $x$ to $0$"?

 

[PeroK]

$|x|$ means distance between x and 0. But I am thinking we can also write $|x|=|0-(-x)|$ which means it is the distance between 0 and -x. Am I right?

$|x| = |-x|$ is how I would think about it. Or, in general, the distance from $x$ to $y$ is thr same as the distance from $y$ to $x$:
$$|x-y| = |y-x|$$

 

[mark2142]

Where exactly are you having trouble? It seems to me you are asking why two distinct yet equivalent statements compute the same result. Is this question akin to asking, "Is $x$ the distance from $0$ to $x$, or from $x$ to $0$"?

Hello! Good to see you.
I don’t know why but I have a desire to write $|x|$ as $x$ which is completely wrong since $|-5| \neq -5$. If we are not careful it becomes $x<c$. On a number line $x$ would be left of $c$ up to $-\infty$ , if $c>0$. This doesn’t match with the book. This may seem absurd. But sometimes this makes sense which I know is not right But I can’t disprove it. So I mentioned this. Haha…
Now, after thinking I think what I am doing is I am modifying |x| by applying $| |$ on $x$. $|5|=5<c$ but say $|-5|<c$ then $5<c$. So in both cases $x=5,-5$ are in the left of $c$. But what I should be doing is to go back, reverse engineer to find what is $x$. Like in case of equation or linear inequality or non linear, we try to get the values of $x$. Solve the equation!
So $|x|$ means if $x \geq 0$ it would be $x$. $x>c$.
If $x<0$, $|x|$ means $-x$ and $-x<c$.
Does this all makes sense? I think I am in the middle of trying to learn the language of maths. Somethings makes sense. Something don’t. Something is deducible but of no use. We can’t reach to any conclusio. Like that.

 

[YouAreAwesome]

Does this all makes sense?

Hello! Yes, keep learning the language of maths. Formulating questions can often lead to solving the questions you are formulating.

As for the absolute value, you really can think about it as "length" or "distance", because the distance from me to you is the same distance as you to me. So if I travel $x$ metres to give you a high five, my return home is also $x$ metres. Yet if direction is taken into account, and my travelling to you is $+x$, then my return trip is $-x$.

 

[mark2142]

Hello! Yes, keep learning the language of maths. Formulating questions can often lead to solving the questions you are formulating.

Is $| |$ a notation on $x$ which means $x$ or $-x$ depending on the value of $x$ ?
Like the notation $x^n$ means $x.x…x$ (n terms).

 

[YouAreAwesome]

Is $| |$ a notation on $x$ which means $x$ or $-x$ depending on the value of $x$ ?
Like the notation $x^n$ means $x.x…x$ (n terms).

Well if we are focussing in on language, you should say "n times" or "n factors" etc. but not "n terms" as this expression has one term, $x^n$ and each $x$ is a factor not a term.

When we take the $||$ of $x$ we are taking the distance of $x$ from zero. Distance is always a positive value, a natural number that we can easily relate to in the natural world. We don't take negative distances, or if we do, we use the word displacement. But with distance, direction doesn't matter, only the length matters. And this is why negative values work in the same way as positive values, because distance doesn't need to account for positive or negative direction.

I don't think defining it the way you have is correct. $||$ is not "a notation on $x$ which means $x$ or $-x$". Rather, it takes the length of $x$ from zero, irrespective of whether $x=c$ or $x=-c$.

 

[mark2142]

Instead of saying that the absolute value of a number is its distance from zero, I suggest that you use the following definition:

If $x \geq 0$ then $|x| = x$. If $x < 0$ then $|x| = -x$.

Then consider the two cases in which either $x \geq 0$ or $x < 0$.

Hello! Yes, keep learning the language of maths. Formulating questions can often lead to solving the questions you are formulating.

As for the absolute value, you really can think about it as "length" or "distance", because the distance from me to you is the same distance as you to me. So if I travel $x$ metres to give you a high five, my return home is also $x$ metres. Yet if direction is taken into account, and my travelling to you is $+x$, then my return trip is $-x$.

Why are you guys only using the definition of $|x|$ ? Why can’t we prove the property $|x|<c$ is equivalent to $-c<x<c$ by considering the fact that $|x|$ is distance between $0$ and $x$ or as I am speculating it as distance between $0$ and $-x$ ?

$|x| = |-x|$ is how I would think about it. Or, in general, the distance from $x$ to $y$ is thr same as the distance from $y$ to $x$:
$$|x-y| = |y-x|$$

It seems like you are saying the way I am thinking will take me nowhere and I am only complicating a simple thing?

 

[PeroK]

Why are you guys only using the definition of $|x|$ ? Why can’t we prove the property $|x|<c$ is equivalent to $-c<x<c$ by considering the fact that $|x|$ is distance between $0$ and $x$ or as I am speculating it as distance between $0$ and $-x$ ?

For the real numbers, the usual definition of "distance" between $x$ and $y$ is $x-y$ (if $x \ge y$) and $y - x$ (if $x < y$). "Distance" is, therefore, equivalent to $|x - y|$.

For two dimensions, the usual definition of distance from the origin to a point $(x, y)$ is $\sqrt{x^2 + y^2}$. If we write that as $|(x,y)|$, then:
$$|(x,y)| < c \ \Leftrightarrow \ x^2 + y^2 < c^2$$
You must have a definition of distance.

It seems like you are saying the way I am thinking will take me nowhere and I am only complicating a simple thing?

Yes!

 

[TonyStewart]

Assume that |x| >=0 is true and |x|<0 is false and reciprocity a+b=b+a

 

[mark2142]

For the real numbers, the usual definition of "distance" between $x$ and $y$ is $x-y$ (if $x \ge y$) and $y - x$ (if $x < y$). "Distance" is, therefore, equivalent to $|x - y|$.

For two dimensions, the usual definition of distance from the origin to a point $(x, y)$ is $\sqrt{x^2 + y^2}$. If we write that as $|(x,y)|$, then:
$$|(x,y)| < c \ \Leftrightarrow \ x^2 + y^2 < c^2$$
You must have a definition of distance.

So? Show me how can I prove the property by this definition.

 

[PeroK]

So? Show me how can I prove the property by this definition.

You can't. It's not the only possibility for "distance". It's just the usual one. Although, you could look for proofs of Pythagoras' theorem. Of which there are dozens, if not hundreds.

 

[mark2142]

You can't. It's not the only possibility for "distance". It's just the usual one. Although, you could look for proofs of Pythagoras' theorem. Of which there are dozens, if not hundreds.

Can you be a bit more clear in what you are saying ?

 

[PeroK]

Can you be a bit more clear in what you are saying ?

I think you need to step back from this a bit. You may have developed a mental block.

 

[vela]

$|x|$ means distance between x and 0. But I am thinking we can also write $|x|=|0-(-x)|$ which means it is the distance between 0 and -x. Am I right?

Yes, so what you've discovered is that for a given value of $x$, the distance between $0$ and $x$ is the same as the distance between $0$ and $-x$. For example, both 1 and -1 are a distance of 1 away from the origin.

 

[YouAreAwesome]

Why can’t we prove the property |x|<c is equivalent to −c<x<c by considering the fact that |x| is distance between 0 and x or as I am speculating it as distance between 0 and −x ?

Yes, I think we can do exactly that.

Consider the distance from $x$ to $0$ where $x<0$.

$|-x|=|x|<c$

If $|-x|<c$ then $-x<c$ and this implies $x>-c$.

Consider the distance from $x$ to $0$ where $x>0$.

$|x|<c$

If $|x|<c$ then $x<c$.

Combining we have that if $|x|<c$ then $-c<x<c$.

 

[Mark44]

Consider the distance x<0, in other words consider −x.

What do you mean by this? Distances are ordinarily nonnegative.
Also, the phrase "in other words" is confusing. If x < 0, then "in other words consider -x" seems to be a non sequitur.

 

[YouAreAwesome]

What do you mean by this? Distances are ordinarily nonnegative.
Also, the phrase "in other words" is confusing. If x < 0, then "in other words consider -x" seems to be a non sequitur.

Thanks, I've edited.

 

[mark2142]

You can't. It's not the only possibility for "distance". It's just the usual one. Although, you could look for proofs of Pythagoras' theorem. Of which there are dozens, if not hundreds.

You are saying that we cannot prove the property by this definition of distance $|x-y|$ but there are other definitions by which we can.

 

[MidgetDwarf]

You are saying that we cannot prove the property by this definition of distance $|x-y|$ but there are other definitions by which we can.

it is taken as an axiom/definition. You can prove properties of this axiom, ie., triangle inequality and other stuff. But one does not prove an axiom.

Yes, the notion of distance gets a bit "weird" when we talk about things like metric spaces, but that belongs to topology.

 

[PeroK]

You are saying that we cannot prove the property by this definition of distance $|x-y|$ but there are other definitions by which we can.

I don't understand what you mean by that statement. The number line isn't a physical thing you can measure. We're not talking about physical distance here. There are no physical units involve, like inches or centimeters. So, you have to define the "distance" between $x$ and $y$ in some abstract mathematical way. The usual definition is $|x - y|$.

This makes some physical sense, because if you have a physical length of something like a rope and you measure $x$ units from the end and $y$ units from the end, then the distance you measure between these two points is $|x - y|$. That can't be proved either, in the sense that it's an experiment you must do.

 

[mark2142]

I

I don't understand what you mean by that statement. The number line isn't a physical thing you can measure. We're not talking about physical distance here. There are no physical units involve, like inches or centimeters. So, you have to define the "distance" between $x$ and $y$ in some abstract mathematical way. The usual definition is $|x - y|$.

This makes some physical sense, because if you have a physical length of something like a rope and you measure $x$ units from the end and $y$ units from the end, then the distance you measure between these two points is $|x - y|$. That can't be proved either, in the sense that it's an experiment you must do.

There is some miss understanding. Maybe my English.

it is taken as an axiom/definition. You can prove properties of this axiom, ie., triangle inequality and other stuff. But one does not prove an axiom.

Do you mean ‘$|x|<c$ is equivalent to $-c<x<c$’ as axiom?

 

[PeroK]

Do you mean ‘$|x|<c$ is equivalent to $-c<x<c$’ as axiom?

There's a difference between an axiom and a definition. Axioms are more fundamental. The modulus function must have a definition. That's been given above: $|x| = \pm x$, depending on whether $x$ is positive or negative. From that definition, you can prove that $|x|<c$ is equivalent to $-c<x<c$, where $c$ is positive..

What you can't do in mathematics is introduce some undefined notion of "distance". You must define what you mean by distance. In this case, we define the distance between $x$ and $y$ as $|x - y|$.

 

[mark2142]

it is taken as an axiom/definition. You can prove properties of this axiom, ie., triangle inequality and other stuff. But one does not prove an axiom.

Translation: Axioms are not that difficult to understand.
But I don’t know why this one doesn’t feel obvious.
By ‘it’ you mean property of absolute value inequality just to be clear.

 

[mark2142]

There's a difference between an axiom and a definition. Axioms are more fundamental. The modulus function must have a definition. That's been given above: |x|=±x, depending on whether x is positive or negative. From that definition, you can prove that |x|<c is equivalent to −c<x<c, where c is positive..

Ok. We define things and combine them to make fundamental statements which can be proved by the definition itself like here.
But I am not satisfied that distance definition doesn’t work to prove axiom.