If the absolute value of a number cannot be negative...then

[Whitney in KY]

how can the abso. val. of A be -A?????

p!

I need math help.

Mucho Thanks!

Whitney

[Caitilin]

Can you give us the whole problem to work with?

[Jane in NC]

how can the abso. val. of A be -A?????

p!

I need math help.

Mucho Thanks!

Whitney

 

If A is a negative number, then its absolute value is -A.

 

Example:

 

Let A = -3

 

Then the absolute value of A is -(-3) or 3.

 

You know that the absolute value should be positive or zero. You were erroneously assuming that A was positive.

[Whitney in KY]

whitney

[Jane in NC]

You are welcome!

 

Mathematically yours, ;)

Jane

[hornblower]

whoa = now I'm confused.

 

I thought absolute value was the distance of a # from 0 on the number line & that it has no sign.

 

So |-3| = 3

 

&

 

|3| = 3

[edithcrawley]

whoa = now I'm confused.

 

I thought absolute value was the distance of a # from 0 on the number line & that it has no sign.

 

So |-3| = 3

 

&

 

|3| = 3

 

 

Yes, that is correct. This site has a good explanation: http://www.purplemath.com/modules/absolute.htm

[LisaKinVA]

whoa = now I'm confused.

 

I thought absolute value was the distance of a # from 0 on the number line & that it has no sign.

 

So |-3| = 3

 

&

 

|3| = 3

 

We just did this today... and this was the definition and explanation we were given.

[Dana]

Absolute value as distance from 0 is one of the easiest ways to understand the concept. It also can really help with picturing inequalities.

 

It can be defined algebraically though (generally written as a piecewise function... which I can't do easily here)...

 

abs(x) = x (if x>=0)

abs(x) = -x (if x<0)

 

So if x is positive or 0, abs(x) = x. (Thus abs(3) = 3.)

If x is negative, abs(x) = -x. (So abs(-3) = -(-3) = 3.)

[Jane in NC]

whoa = now I'm confused.

 

I thought absolute value was the distance of a # from 0 on the number line & that it has no sign.

 

 

 

Using absolute value as distance on a number line is an application in geometry. Dana provided a more general definition. The usual way of writing it is found in Wikipedia here.

[hornblower]

Ok, I admit my eyes glazed over about midway down the page, but

 

"As can be seen from the above definition, the absolute value of a is always either positive or zero, but never negative. "

 

 

So I just don't understand what's being said in this thread. It CAN be negative? But the wiki says it can't?

 

With things like this I often don't understand if people are agreeing & just adding more info, or contradicting, or ? :001_huh: And I don't really understand where geometry comes into it as it was part of dd's Algebra course that talked about this .....

 

(hey, I did make it through 1st year Calc, both differential & integral many, many years ago but obviously I'm not a mathy person :D)

 

So color me increasingly more confused. :confused:

[Jane in NC]

Ok, I admit my eyes glazed over about midway down the page, but

 

"As can be seen from the above definition, the absolute value of a is always either positive or zero, but never negative. "

 

 

So I just don't understand what's being said in this thread. It CAN be negative? But the wiki says it can't?

 

With things like this I often don't understand if people are agreeing & just adding more info, or contradicting, or ? :001_huh: And I don't really understand where geometry comes into it as it was part of dd's Algebra course that talked about this .....

 

(hey, I did make it through 1st year Calc, both differential & integral many, many years ago but obviously I'm not a mathy person :D)

 

So color me increasingly more confused. :confused:

 

Just because I write -A does not mean that -A is a negative number. We can let A equal any real number. If we choose A to be negative, then -A gives us a positive result.

 

Choose A = -7. Then -A = -(-7) = 7.

 

What we are all agreeing on is that absolute value is positive or zero--not negative. What you are having trouble seeing is the mathematical notation which expresses this.

 

As I noted to the OP, there is an assumption being made that A is positive so that -A is negative. Not true! We are letting A be any real number and real numbers are not necessarily negative.

 

Regarding distance: that is a geometric concept. It may be presented in an algebra class, but the idea is geometric. We use absolute value for many more ideas than just distance which is why the general definition is preferable.

[regentrude]

the absolute value can NOT be negative.

However, the absolute value of a negative number is MINUS that number:

if A=-3 then absolute value of A equals +3 which is equal to -A. because -A=-(-3)=+3.

[Tiberia]

I think of the minus sign as "opposite."

 

-A is the opposite of A.

 

If A is (-3), then -A is the opposite of A.

 

The opposite of (-3) is 3.

 

Hope that helps.

[KidsHappen]

Ok, I admit my eyes glazed over about midway down the page, but

 

"As can be seen from the above definition, the absolute value of a is always either positive or zero, but never negative. "

 

 

So I just don't understand what's being said in this thread. It CAN be negative? But the wiki says it can't?

 

With things like this I often don't understand if people are agreeing & just adding more info, or contradicting, or ? :001_huh: And I don't really understand where geometry comes into it as it was part of dd's Algebra course that talked about this .....

 

(hey, I did make it through 1st year Calc, both differential & integral many, many years ago but obviously I'm not a mathy person :D)

 

So color me increasingly more confused. :confused:

 

I am with you on this one. I understand the OP to be saying that the answer to the problem she is working with is 3 = |-3|. I am not following the part where Jane is saying that the answer is -A because the first number is not positive? :confused: From everything I have taught, read or learned, it does not matter whether the original number is positive or negative the answer will always be positive (or zero).

[Jane in NC]

I am with you on this one. I understand the OP to be saying that the answer to the problem she is working with is 3 = |-3|. I am not following the part where Jane is saying that the answer is -A because the first number is not positive? :confused: From everything I have taught, read or learned, it does not matter whether the original number is positive or negative the answer will always be positive (or zero).

 

Right.

 

So if A = -3, do you not agree that the Abs(A) = 3?

 

Thus Abs A = -A = - (-3) = 3.

[Snickerdoodle]

Does this help at all?

 

 

For x < 0, | x | = –x

 

This is a case in which the "minus" sign on the variable does not indicate "a number to the left of zero", but "a change of the sign from whatever the sign originally was". This "–" does not mean "the number is negative" but instead means that "I've changed the sign on the original value".

 

[hornblower]

Does this help at all?

 

no :confused: :)

 

Honestly, I'm not getting this at all. I don't see how what you quoted is true.

 

|-3| = 3

 

|3| = 3

 

I thought we agreed on this way back? :001_huh:

 

so what difference what the sign is in front of the numer inside the vertical slashes? No matter if it's >0 or <0, if it's -x or +x, shouldn't the answer still be x?

 

 

 

I don't see what I would put in between those two vertical slashes that would yield an answer with a - in front of it.

 

The only think I can think of is writing a negative sign in front of the vertical slash..... like :

 

-|3|= -3

 

is this what we're talking about here?

 

(oh btw, most keyboards have that vertical line thingy - it's on top of the backslash on mine & looks like it's dotted, but it's not when it displays)

[Snickerdoodle]

Let A = -3

 

What value of A must A be in order to obtain the absolute value of A?

 

Well, if A = -3 then the absolute value of A is -A.

[hornblower]

All is not lost - I've accomplished one thing: I've remembered what it feels like to be so frustrated that you're close to tears over something that you're sensing others are repeating.slower.and.louder.because.it's.so.simple

but you're just not getting it. It'll come in handy when I need to empathise with my kids' meltdowns..... :D

 

Sickerdoodle -

 

I managed to follow line 1.

 

It went downhill after that. :lol:

 

Line two? Sounds like something that google translate translated from Norwegian to Cantonese to French. I don't understand the question at all:confused:

 

And line 3 - well. I'm sure it follows. You guys obviously know what you're talking about so I'm not debating it at all - but I just don't get what happened.

 

This is what it sounds like in my brain:

 

Let white= white, ok? (Ok, nod head.)

 

Then mysterious static, in at least one foreign language (increasing confusion)

 

Therefore, white = black. (total panic)

 

 

I'm going out now with a friend where I'll take deep breaths & try to not panic. School starts for my kids on Sep 20. I hope by then I'll have regained some composure about my ability to teach them math. Maybe I'll stick to charts & plotting navigation courses. I can do that.....(on good days). :D

[Dana]

All is not lost - I've accomplished one thing: I've remembered what it feels like to be so frustrated that you're close to tears over something that you're sensing others are repeating.slower.and.louder.because.it's.so.simple

but you're just not getting it. It'll come in handy when I need to empathise with my kids' meltdowns..... :D

 

 

And that's a really good thing to remember! Goodness knows I need it when I'm trying to figure out if my son is really not understanding and needs a different approach or if he's just not trying.

 

If you want to make another attempt...

You're correct with |3| = |-3| = 3.

 

The idea is that with variables, we don't know what our number is.

We may be trying to solve an equation like |x| = 3.

In that case, we can't get rid of the absolute value bars because we don't know if x is positive or negative or zero (without sign).

That's where we need the piecewise definition.

 

When I teach this, I use the "distance from 0" as an illustration and a way to remember... so here we're looking for all numbers that are a distance of 3 from 0. That happens at x=3 and at x=-3.

 

The piecewise definition (formal algebraic one) lets us rewrite for ANY input (let's say "blob"), the absolute value of that blob without absolute value bars.

 

If "blob" is positive or 0, |blob| = blob. (There's no change to the number.)

If "blob" is negative, |blob| = - (blob) (It's the opposite of the "blob". This is where if we had |-3|, since our input is a negative number, |-3| = the opposite of (-3) = -(-3) = 3.

 

We're still getting the same answer you were: |-3| = 3, but the algebraic approach lets us generalize.

 

Thus |x-5| = x-5, if x>=5, but |x-5| = -(x-5) if x<5.

[Here the x-5 is our "blob".]

 

Clear as mud? ;)

Purplemath has pretty good explanations as well - and happy to discuss if you want to continue or drop completely otherwise! :D

[Laurie4b]

What if, instead of "negative" you read the - sign as "the opposite of"?

 

So if A is a negative number (-3), then the absolute value of A is the opposite of A ( -A or - (-3)?

 

The absolute value of a negative number is the opposite of the negative number, which is a positive number.

[Dana]

I was sharing this thread with dh and he wanted everyone trying to explain the problem to divert themselves by graphing the following polar equation:

 

r = | (| sec x| + |csc x|) | - | (|sec x| - |csc x|)|

 

I'm interested in the follow-up and how he sees it relating to the discussion.

 

I'm definitely weak in graphing in polar coordinates (other than extremely basic functions). My approach is all too often brute force... for instance, here, plotting points which can't be best.

 

Clearly this is undefined when x = 0, pi/2, pi, ...

 

Each quadrant (unit circle) has the same output, so graphing QI should be enough to get the pattern.

 

Using Wolfram Alpha for graphs, on the rectangular plane, I can see the repetition I expect, but the polar graph is showing up as empty.

 

I haven't done anything since Calc I in over 14 years, so rusty is a mild description for me. :001_smile:

What am I not seeing here?

[hornblower]

Clear as mud? ;)

Purplemath has pretty good explanations as well - and happy to discuss if you want to continue or drop completely otherwise! :D

 

:svengo:

 

Wait - I understood that!!!!!!!!

 

:lol:

 

OMG. Thank you Dana! You're my math goddess.

 

How shall I worship you? What offerings shall I send?

 

 

Everyone -thank you. I think it's hilarious that the OP is all 'oh yeah, thx!' & checked out ages ago & I was still here with the dazed & confused look. You've all been very patient & nice but it wasn't until Dana that I finally got it. Eliana, you're right it was like an optical illusion because once you see it, it seems so obvious but it took a lot of squinting on my part to get there :)

 

 

 

Now I'm going to steel myself to crack open the new TI-84 Plus calculator that arrived in the mail recently. Good grief that's a whole lot of scary looking buttons.....:001_huh:

[Jane in NC]

I was sharing this thread with dh and he wanted everyone trying to explain the problem to divert themselves by graphing the following polar equation:

 

r = | (| sec x| + |csc x|) | - | (|sec x| - |csc x|)|

 

Cool!

[Dana]

Cool!

 

Tell me what I'm missing, please!

 

I just know I'm overlooking something!

 

Frustrated here. :glare:

[Jane in NC]

Tell me what I'm missing, please!

 

I just know I'm overlooking something!

 

Frustrated here. :glare:

 

Hint: Start by considering r = sec x and r = csc x in polar coordinates.

[Dana]

Hint: Start by considering r = sec x and r = csc x in polar coordinates.

 

:confused: not helping me.

 

I'm very weak with understanding with polar coordinates. Here, I'm getting hung up on x vs theta... So the best I'm getting is then sec theta = r/x .. But that's not clarifying things for me.

[Jane in NC]

:confused: not helping me.

 

I'm very weak with understanding with polar coordinates. Here, I'm getting hung up on x vs theta... So the best I'm getting is then sec theta = r/x .. But that's not clarifying things for me.

 

Eliana used x for theta since most of us don't have a theta on our keyboard. So in her equation, you want to replace x with θ.

 

Let me sort of walk you through the interesting equation that was given above (again, assuming the x's are θ's.)

 

In polar coordinates, we use the following conversions:

 

x = r cos θ

y = r sin θ

 

The hint I gave was to look at r = sec θ and r = csc θ.

 

Using the first conversion equation, we replace r with sec θ:

 

x = (sec θ) (cos θ) = 1

 

Using the second equation,

 

y = (csc θ) (sin θ) = 1

 

So the hint I provided was leading you to two lines, one vertical and one horizontal.

 

Things have to be analyzed quadrant by quadrant in the more complicated equation that Eliana's husband provided, but what happens is that you get the graph of a square of side length two, centered at the origin.

 

I usually think of polar graphs as being pretty rosettes, circles or spirals, so it was cool to get a polygon!

 

Wishing I had one of those tablet things to draw out it out for you.

 

Jane

[JeneralMom]

Okay, see if this helps - it seems that many of you are thinking of the negative numbers on 1 single plane, so on a horizontal line going from <0>. However, with absolute numbers we are also talking about a vertical line and "absolute" means that the values are always 0> on the vertical line. Therefore, even if we are traveling negatively on the horizontal line, we are still positive on the vertical line and that is the number we are interested in.

 

There is a good graph here.

[Dana]

Thanks!

 

I'll still need to play with it a bit, but I think I'm seeing it.

I think of polar graphs as things you would draw with a spirograph :001_smile: so that does make this one nifty.

 

Eliana used x for theta since most of us don't have a theta on our keyboard. So in her equation, you want to replace x with θ.

 

Let me sort of walk you through the interesting equation that was given above (again, assuming the x's are θ's.)

 

In polar coordinates, we use the following conversions:

 

x = r cos θ

y = r sin θ

 

The hint I gave was to look at r = sec θ and r = csc θ.

 

Using the first conversion equation, we replace r with sec θ:

 

x = (sec θ) (cos θ) = 1

 

Using the second equation,

 

y = (csc θ) (sin θ) = 1

 

So the hint I provided was leading you to two lines, one vertical and one horizontal.

 

Things have to be analyzed quadrant by quadrant in the more complicated equation that Eliana's husband provided, but what happens is that you get the graph of a square of side length two, centered at the origin.

 

I usually think of polar graphs as being pretty rosettes, circles or spirals, so it was cool to get a polygon!

 

Wishing I had one of those tablet things to draw out it out for you.

 

Jane