How do I prove |xy| = |x| * |y|?

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  • #1
athena810
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Homework Statement


Prove:
|xy| = |x| * |y|?

Homework Equations


|x| = sqrt(x^2)


The Attempt at a Solution


(|xy|)^2 = (|x| * |y|)^2
x^2y^2 = x^2 y^2

---I think this is incorrect though. Really, I don't understand proofs and the concept of proofing.
 

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  • #2
Office_Shredder
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One way you can do it is to break it down into cases

Case 1: If either x or y is zero....

Case 2: If x > 0 and y > 0

Case 3: If x > 0 and y < 0

etc. This is how things with absolute values are usually handled.

Alternatively, what you have done is very close to being complete (and pretty clever). Starting from x2 y2 = x2 y 2 you can get to (|xy|)2= (|x|*|y|)2. Now, in general if I have a2 = b2 then I know that a = b OR a = -b. So you have that |xy| = |x|*|y| OR |xy| = -|x|*|y|. You should be able to discount the second one fairly easily
 
  • #3
epenguin
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Homework Statement


Prove:
|xy| = |x| * |y|?

Homework Equations


|x| = sqrt(x^2)


The Attempt at a Solution


(|xy|)^2 = (|x| * |y|)^2
x^2y^2 = x^2 y^2

---I think this is incorrect though. Really, I don't understand proofs and the concept of proofing.

You don't understand proofs? You have never seen any mathematical proof?

Forget those attempts.

The truth statement itself is, I hope, blindingly obvious. But that's not enough for mathematicians. What you have to do, what they are asking you to do, is to take the definition of |x| ( i.e. |anything|) and show that the statement follows logically.

Actually there is more that needs defining. In principle you might need to define * but I don't think you are being asked to go that far back. You surely have learnt some rules about things like what is x* -y , -x * -y . You are required to combine these with the definition of | | to prove the statement.
 
  • #4
athena810
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One way you can do it is to break it down into cases
Case 1: If either x or y is zero....

Case 2: If x > 0 and y > 0

Case 3: If x > 0 and y < 0

If I do it this way, then do I just say something like:
If x, y = 0, then 0*0 = 0*0?

Now, in general if I have a2 = b2 then I know that a = b OR a = -b. So you have that |xy| = |x|*|y| OR |xy| = -|x|*|y|. You should be able to discount the second one fairly easily
How do I know that it's not -a = b, or does it not matter which one I make negative?
As for that case, in order to disprove it, do I just say that I can't sqrt negatives? Like: "sqrt(-b) is not a Real number."?
Thanks

You don't understand proofs? You have never seen any mathematical proof?

Forget those attempts.

The truth statement itself is, I hope, blindingly obvious. But that's not enough for mathematicians. What you have to do, what they are asking you to do, is to take the definition of |x| ( i.e. |anything|) and show that the statement follows logically.

Actually there is more that needs defining. In principle you might need to define * but I don't think you are being asked to go that far back. You surely have learnt some rules about things like what is x* -y , -x * -y . You are required to combine these with the definition of | | to prove the statement.

I've seen proofs, and I kind of understand that it's basically stating the obvious using theorems and whatever. I've done simpler proofs like proving a^2 > a if a > 1 but I'm not sure what to do with | |.
 
  • #5
vela
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(|xy|)^2 = (|x| * |y|)^2
x^2y^2 = x^2 y^2

---I think this is incorrect though. Really, I don't understand proofs and the concept of proofing.
You're right. It's not correct. A major mistake is that you're starting off using what you're trying to prove. In other words, don't start off the proof with |xy|=|x||y|.

One common technique is to start with one side and then get to the other, e.g. ##\lvert xy\rvert = \sqrt{(xy)^2} = \cdots = \lvert x\rvert\lvert y\rvert##. For each step, you should be able to state a reason to justify that step. For example, you might justify writing ##\lvert xy\rvert = \sqrt{(xy)^2}## as using the definition of the absolute value (I'm assuming that's how it's defined in your class). You can justify the step ##\sqrt{(xy)^2} = \sqrt{x^2 y^2}## based on the property of real numbers, and so on.
 
  • #6
athena810
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So then it is:
|xy|=|xy|
(|xy|)^2 = (|xy|)^2
(|xy|)^2 = |x|^2 * |y|^2
sqrt(|xy|^2) = sqrt(|x|^2) * sqrt(|y|^2)
|xy| = |x| * |y|

Is this correct?
 
  • #7
Avodyne
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No, this is not correct.

First of all, what definition of |x| is given in your text (or by your instructor)? I very much doubt that it is |x|=√x2, which has all sorts of difficulties. The usual definition is |x|=x if x>0, |x|=-x if x<0, and |x|=0 if x=0.

So a valid proof would start out something like this.

If x=0, then |x|=0 by definition, and xy=0 by the multiplication property of zero. If xy=0, then |xy|=0 by definition. Also, if |x|=0, then |x|*|y|=0 by the multiplication property of zero. Thus, if x=0, we have both |xy|=0 and |x|*|y|=0, and so |xy|=|x|*|y| in this case.

If y=0, swap x and y in the proof above.

Now suppose x>0 and y>0. ...

Continue through all possible cases.

Note that each step should be justified.
 
  • #8
epenguin
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I've seen proofs, and I kind of understand that it's basically stating the obvious using theorems and whatever. I've done simpler proofs like proving a^2 > a if a > 1 but I'm not sure what to do with | |.

Start by stating the definition of | | .

If I do it this way, then do I just say something like:
If x, y = 0, then 0*0 = 0*0?


How do I know that it's not -a = b, or does it not matter which one I make negative?
As for that case, in order to disprove it, do I just say that I can't sqrt negatives? Like: "sqrt(-b) is not a Real number."?
Thanks

Be a country girl! You go into relatively complicated stuff you have heard about about squares that is not the point here. (I realise that squaring does something quite reminiscent of | | , namely turning negatives into positives, but here it is not the point.)

Office_Shredder has given you the idea. But what you wrote is just not a proof.

If you said in the case x = 0, y = 0 , what does |x| equal, what does |y| equal, what thence do |x| , |y| equal, hence what does |x|*|y| equal on the one hand, on the other hand what does |x * y| equal, are these two things equal between each other? If so the theorem is true for that case. Then if you could show it was true for the two other cases, then it is true in general and you have proved the theorem.

How do I know that it's not -a = b, or does it not matter which one I make negative?

We have called these two things x and y. Sure we could have called them a and b and the argument is no different.

I think your question meant does it matter whether you make x or y negative, or perhaps you are wondering as well as Office_Shredder's 3 cases, shouldn't there be a 4th: x negative and y positive? This sort of thing often flummoxes beginners and not-so beginners, so get it clear. Algebraic symbols are 'disponible', available, on call. They mean what we decide we want them to mean. So the point here was case 3 is the case where x and y have opposite signs. So if we say let x>0 y<0 or if OTOH we say let x<0 y>0 we are saying exactly the same thing, we have just decided differently what we meant by x and y. So you can use one or the other of those inequality pairs, it doesn't mean anything to use them both. At risk of repetition we are just in case 3 considering the case where one of x, y is a positive number and the other is a negative number and it doesn't matter which we decide to be which.
 
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  • #9
vela
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First of all, what definition of |x| is given in your text (or by your instructor)? I very much doubt that it is |x|=√x2, which has all sorts of difficulties.
What difficulties are those? This is, in fact, how the absolute value is sometimes defined for real numbers.
 
  • #10
vela
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So then it is:
|xy|=|xy|
(|xy|)^2 = (|xy|)^2
(|xy|)^2 = |x|^2 * |y|^2
You run into a problem here because you've used the fact that |xy|=|x||y|, which is what you're supposed to be proving.

Don't square in the beginning. Just keep everything under the square root.

sqrt(|xy|^2) = sqrt(|x|^2) * sqrt(|y|^2)
|xy| = |x| * |y|

Is this correct?
 
  • #11
athena810
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No, this is not correct.

First of all, what definition of |x| is given in your text (or by your instructor)? I very much doubt that it is |x|=√x2, which has all sorts of difficulties. The usual definition is |x|=x if x>0, |x|=-x if x<0, and |x|=0 if x=0.

So a valid proof would start out something like this.

If x=0, then |x|=0 by definition, and xy=0 by the multiplication property of zero. If xy=0, then |xy|=0 by definition. Also, if |x|=0, then |x|*|y|=0 by the multiplication property of zero. Thus, if x=0, we have both |xy|=0 and |x|*|y|=0, and so |xy|=|x|*|y| in this case.

If y=0, swap x and y in the proof above.

Now suppose x>0 and y>0. ...

Continue through all possible cases.

Note that each step should be justified.

It was defined the normal way with the 4 cases. Then it was like: "This theorem can also be proved in a much simpler way. Notice: √x^2 = |x|.

Also, in most of the example proofs, they didn't list the reasons why. It would list all the steps, then at the bottom be like "<this> could happen because of <this property>. Therefore <this> = <that>."

But anyway, do I write the proof in paragraph form like what you did or do I write it like 2 column geometric proofs?

Start by stating the definition of | | .

If you said in the case x = 0, y = 0 , what does |x| equal, what does |y| equal, what thence do |x| , |y| equal, hence what does |x|*|y| equal on the one hand, on the other hand what does |x * y| equal, are these two things equal between each other? If so the theorem is true for that case. Then if you could show it was true for the two other cases, then it is true in general and you have proved the theorem.

So I would say:
(Case 1) x= 0, y= 0...and then be like " |a + b| = a + b= |a| + |b|"?
 
  • #12
athena810
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You run into a problem here because you've used the fact that |xy|=|x||y|, which is what you're supposed to be proving.

Don't square in the beginning. Just keep everything under the square root.

So if I keep it under the radical:
...
sqrt(|xy|) = sqrt(|x| * |y|) ?
 
  • #13
vela
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No, again, you're using what you're trying to prove.

Reread post 5.
 
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  • #14
epenguin
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So I would say:
(Case 1) x= 0, y= 0...and then be like " |a + b| = a + b= |a| + |b|"?

I said, and I don't think anyone here will disagree, that you have to start with the definition of | |.
Actually frame this as a definition of |x| . Because of what I explained about algebraic symbolism, this also covers y, you don't need to state it separately but we will allow you to do that if it helps you.

I cannot help you further until you state a definition and when you have done it make an effort to treat at least one, any one, of the three cases that arise.
 
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  • #15
athena810
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I cannot help you further until you state a definition and when you have done it make an effort to treat at least one, any one, of the three cases that arise.

Will this work?
So definition of | |: The positive value of any real number.
Case 1: If x, y >= 0..then |x| = x, |y| = y, xy >= 0. Thus |xy| =xy = |x||y|?
 
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  • #16
HallsofIvy
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Do you have a specific definition of "The positive value of a real number". Have you ever seen a text book that gives that a the definition of absolute value?

The definition every text book I have ever seen gives is "|x|= x if [itex]x\ge 0[/itex], |x|= -x if x< 0".
 
  • #17
epenguin
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Will this work?
So definition of | |: The positive value of any real number.
Case 1: If x, y >= 0..then |x| = x, |y| = y, xy >= 0. Thus |xy| =xy = |x||y|?

OK you give an informal definition that sounds like what someone said in class explaining it to you. And maybe your memory. H of I is surprised if that is in a book. Use the book.

To be able to use definition you really need the formal one given by HofI.

This bit
Case 1: If x, y >= 0..then |x| = x, |y| = y,

is OK. The rest is all true but it doesn't seem to me a proof argument.

I suggest you do things like this by numbers so it is clear where everything comes from and fits in. I will give my version as model with which you could do the second bit.

1 Definition of |x| as per HofI. (Def 1)

Remark: use of the ≥ has reduced our cases to two not three.

2 Case 1. x ≥ 0, y ≥ 0 (condition A)
By Def 1 (1) |x| = x, |y| = y. Therefore in case 1 |x|*|y| = x * y

3 Still case 1, from condition (A), x * y ≥ 0

4 From Def 1 and 3, |x * y| = x * y

5 Two things that are equal to a third thing are equal to each other.
So the two things we found above each equal to x*y are equal to each other, i.e. |x|*|y| = |x * y|

6 Case 2, |x| ≥ 0, |y| < 0 ...

I don't say this is perfect (it is not my field) but do you see this more like a logical proof, not a handwaving?

It lays bare assumptions, e.g. 3 is I guess an assumption carried over from basic arithmetic and algebra, and line 1 of 5 is an assumed property of " = " , I think you will hear talk of "equivalence relations".
 
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  • #18
athena810
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Do you have a specific definition of "The positive value of a real number". Have you ever seen a text book that gives that a the definition of absolute value?

The definition every text book I have ever seen gives is "|x|= x if [itex]x\ge 0[/itex], |x|= -x if x< 0".
No, it was not from a textbook.

OK you give an informal definition that sounds like what someone said in class explaining it to you. And maybe your memory. H of I is surprised if that is in a book. Use the book.

To be able to use definition you really need the formal one given by HofI.

This bit
Case 1: If x, y >= 0..then |x| = x, |y| = y,

is OK. The rest is all true but it doesn't seem to me a proof argument.

I suggest you do things like this by numbers so it is clear where everything comes from and fits in. I will give my version as model with which you could do the second bit.

1 Definition of |x| as per HofI. (Def 1)

Remark: use of the ≥ has reduced our cases to two not three.

2 Case 1. x ≥ 0, y ≥ 0 (condition A)
By Def 1 (1) |x| = x, |y| = y. Therefore in case 1 |x|*|y| = x * y

3 Still case 1, from condition (A), x * y ≥ 0

4 From Def 1 and 3, |x * y| = x * y

5 Two things that are equal to a third thing are equal to each other.
So the two things we found above each equal to x*y are equal to each other, i.e. |x|*|y| = |x * y|

6 Case 2, |x| ≥ 0, |y| < 0 ...

I don't say this is perfect (it is not my field) but do you see this more like a logical proof, not a handwaving?

It lays bare assumptions, e.g. 3 is I guess an assumption carried over from basic arithmetic and algebra, and line 1 of 5 is an assumed property of " = " , I think you will hear talk of "equivalence relations".

Oh, ok, I kind of understand.
Is it required that i write a reason for doing why I did what I did or are the reasons implied?
 
  • #19
arildno
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" The rest is all true but it doesn't seem to me a proof argument."
Why not?
In cases x,y positive, we have
|x*y|=x*y=|x|*|y|, i.e identity verified.
 
  • #20
Avodyne
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Is it required that i write a reason for doing why I did what I did or are the reasons implied?
This is really a question for your instructor, but to be unambiguous the reason should be given for each step.

Also, the definition |x|=√x2 is OK provided that the symbol "√" is defined to mean the positive square root. Then, to use this definition in a proof, you would need to assume rules about square roots such as: if x≥0 and y≥0, then √(xy) = √x * √y. (If you had to prove this rule first, then the proof would probably be longer than one based on the simpler definition of |x| as x if x≥0 and -x if x<0.)
 
  • #21
epenguin
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Oh, ok, I kind of understand.
Is it required that i write a reason for doing why I did what I did or are the reasons implied?

To me that argument of mine seems to provide is own reasoning, I don't think you need to say more.
Except it helps to say what you are trying to do e.g. 'I try to prove that each side of the original equation equals x*y .
Then if you don't get to a result a teacher or examiner credits you for a thought process.

I think we need to hear from the algebra and didactic specialists, also for #19.
 
  • #22
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A different approach. Take x and y as complex numbers in exponential form then try proving it. A proof for complex numbers will be proof for all the numbers(real, non-real, rational..etc)
 
  • #23
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First of all, what definition of |x| is given in your text (or by your instructor)? I very much doubt that it is |x|=√x2, which has all sorts of difficulties. The usual definition is |x|=x if x>0, |x|=-x if x<0, and |x|=0 if x=0.

Errmm... what? There are no difficulties to that definition and it is far superior to defining it conditionally.

In fact, the generalized norm |x| for some x = (x1, x2, .. , xn) in R^n is precisely defined as:

sqrt(x1² + x2² + ... + xn²)

For R, this is exactly what the OP has shown. That is the definition of |x|, what you have is a description of its behavior.
 
  • #24
epenguin
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Errmm... what? There are no difficulties to that definition and it is far superior to defining it conditionally.

In fact, the generalized norm |x| for some x = (x1, x2, .. , xn) in R^n is precisely defined as:

sqrt(x1² + x2² + ... + xn²)

For R, this is exactly what the OP has shown. That is the definition of |x|, what you have is a description of its behavior.

Isn't that contrary to the spirit of abstract algebra? Which is not about introducing extra unnecessary ideas but about minimising the number of axioms and assumptions. Thus laying bare what they really are, which is sometimes or initially surprising. Anyway I suspect your superior idea has the one you're trying to get away from hidden in it.
 
  • #25
vela
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Is it required that i write a reason for doing why I did what I did or are the reasons implied?
When you write a proof, you are writing. Just like in any other writing, you have to know your intended audience. You have to decide what details are important to explain and what you can you can assume the reader already knows. For classwork, you should realize your goal is not only to find a valid proof but to show that you know what you're doing, so perhaps to demonstrate to the grader you know your stuff, you might include more detail than normal.

In any case, even if you don't explicitly write down a reason for each step, you should be able to justify each step in your head.

Also, the definition |x|=√x2 is OK provided that the symbol "√" is defined to mean the positive square root.
That's exactly what the radical symbol means for real numbers. It denotes the principal square root. When you write ##\sqrt{4}##, it unambiguously means +2.

Then, to use this definition in a proof, you would need to assume rules about square roots such as: if x≥0 and y≥0, then √(xy) = √x * √y. (If you had to prove this rule first, then the proof would probably be longer than one based on the simpler definition of |x| as x if x≥0 and -x if x<0.)
By the same token, one could argue that you have to prove, for instance, that the product of two positive numbers is positive. There's almost always some level of knowledge about the properties of real numbers and operations that you're going to have to take as established.

Isn't that contrary to the spirit of abstract algebra? Which is not about introducing extra unnecessary ideas but about minimising the number of axioms and assumptions.
From what athena810 posted earlier, it sounds like using ##|x|=\sqrt{x^2}## was suggested as another way of proving |xy|=|x||y|.
 
  • #26
1MileCrash
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Isn't that contrary to the spirit of abstract algebra? Which is not about introducing extra unnecessary ideas but about minimising the number of axioms and assumptions. Thus laying bare what they really are, which is sometimes or initially surprising. Anyway I suspect your superior idea has the one you're trying to get away from hidden in it.

The idea of generalized norms doesn't need to be introduced, I was just showing why it is true, unambiguously and completely, that |x| = sqrt(x^2)

Furthermore, this proof is literally half of a line when that definition is used, rather than involving cases.

And, of course this definition has the conditional definition hidden in it.. they are equivalent.

But, in no way, shape, or form, is the conditional definition "simpler," nor does it require fewer assumptions to apply. You do not need to prove that (xy)^2 = (x^2)(y^2), this can be taken as an elementary fact. If you demand that elementary facts be proven when using the modern definition, but then let them exist freely when using the archaic definition, and then call that reason for the proof "being longer," you're just being silly.
 
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  • #27
epenguin
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This is really a question for your instructor, but to be unambiguous the reason should be given for each step.

Also, the definition |x|=√x2 is OK provided that the symbol "√" is defined to mean the positive square root. Then, to use this definition in a proof, you would need to assume rules about square roots such as: if x≥0 and y≥0, then √(xy) = √x * √y. (If you had to prove this rule first, then the proof would probably be longer than one based on the simpler definition of |x| as x if x≥0 and -x if x<0.)

Errmm... what? There are no difficulties to that definition and it is far superior to defining it conditionally.

The √(x2) definition is conditional if you spell it out.

If you just say 'the positive square root' this is informal, similar to and no better than, indeed not even as good as the OP's informal
So definition of | |: The positive value of any real number.


If you do it formally you get a conditional definition the same as the 'conditional' definition you were trying to get away from, but with a totally unnecessary addition.

Any number of functions, f, with multiple inverses could be used any you could say |x| is f-1f(x) then adding some condition about which inverse you take, a simple one could be the smallest positive value of tan-1(tan x) - ridiculous.
 
  • #28
1MileCrash
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The √(x2) definition is conditional if you spell it out.

If you just say 'the positive square root' this is informal, similar to and no better than, indeed not even as good as the OP's informal

I don't have to say anything, the √ symbol doesn't mean anything other than the "positive square root" for real numbers. There is no ambiguity.

If you do it formally you get a conditional definition the same as the 'conditional' definition you were trying to get away from, but with a totally unnecessary addition.

Please, show this formal conditional definition.

Show me what conditions need to be considered for defining |x| = √(x²). Show me how this definition isn't equivalent to the conditional definition unless I add a condition to consider. I'll wait.

Any number of functions, f, with multiple inverses could be used any you could say |x| is f-1f(x) then adding some condition about which inverse you take, a simple one could be the smallest positive value of tan-1(tan x) - ridiculous.

I'm not sure what any of this has to do with anything.
 
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  • #29
epenguin
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I don't have to say anything, the √ symbol doesn't mean anything other than the "positive square root" for real numbers. There is no ambiguity.

Surely not true. √9 = ± 3. √-9 = ± 3 i

Please, show this formal conditional definition.

It would be just like the one already given for |x|. Instead of
|x|=x if x>0, |x|=-x if x<0, and |x|=0 if x=0.
we'd have something like |x|= √(x²) if √(x²)> 0, |x|= -√(x²) if √(x²)<0, and |x|=0 if x=0, introducing and then inverting an operation to no purpose.


Show me what conditions need to be considered for defining |x| = √(x²). Show me how this definition isn't equivalent to the conditional definition unless I add a condition to consider. I'll wait.

I'm not sure what any of this has to do with anything.

Your √(x2) is just one example out of a myriad we could find of this, f in this example is squaring, √ is finding an inverse of f. Your f maps one point to one other, the inverse maps one point to two in your example, or to an infinite number of points in mine, unless in both cases you spell out a condition.

......
 
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  • #30
D H
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Surely not true. √9 = ± 3. √-9 = ± 3 i
1MileCrash is correct. This is not. √9 is 3, never -3. √x is the function that uniquely maps the non-negative reals to the non-negative reals such that (√x)2=x. This function can be extended to the negative reals and to all of the complex numbers by defining a a branch cut. There are some problems with this extension (e.g., (ab)c = abc is no longer universally true), but √z still a function, not a multifunction.
 
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  • #31
D H
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This massive off-topic misunderstanding over the meaning of √x was a bit unwarranted.

As an aside, proving the conjecture with |x| defined as √(x2) is rather trivial.
 
  • #32
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Surely not true. √9 = ± 3. √-9 = ± 3 i
D H already dealt with this, but I'll say a bit more. It's true that 9 has two square roots, but the symbol √9 is generally recognized to mean the positive square root, or 3.
It would be just like the one already given for |x|. Instead of we'd have something like |x|= √(x²) if √(x²)> 0, |x|= -√(x²) if √(x²)<0, and |x|=0 if x=0, introducing and then inverting an operation to no purpose.
For any real number x, √(x2) is always nonnegative.
 
  • #33
epenguin
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1MileCrash is correct. This is not. √9 is 3, never -3. √x is the function that uniquely maps the non-negative reals to the non-negative reals such that (√x)2=x. This function can be extended to the negative reals and to all of the complex numbers by defining a a branch cut. There are some problems with this extension (e.g., (ab)c = abc is no longer universally true), but √z still a function, not a multifunction.

OK I guess you're right, now I think of it otherwise we wouldn't write ± in the quadratic solution formula.

However I maintain the appearance of greater simplicity is merely verbal or notational and apparent - you happen to have a convention for something a bit elaborate, you have this convention because you do these operations often. But it is not simpler than Avodyne's definition. It represents an operation on x which is then inverted with a condition. Two unnecessary operations which no one in their senses would ever do. Of course anyone presented with calculating √(x2), say √(-237124.6292) can do it in their heads, write the answer immediately and what do they actually do? They write |x| as calculated according to the Avodyne's definition! :tongue2:
 
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  • #34
1MileCrash
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OK I guess you're right, now I think of it otherwise we wouldn't write ± in the quadratic solution formula.

However I maintain the appearance of greater simplicity is merely verbal or notational and apparent - you happen to have a convention for something a bit elaborate, you have this convention because you do these operations often. But it is not simpler than Avodyne's definition. It represents an operation on x which is then inverted with a condition. Two unnecessary operations which no one in their senses would ever do. Of course anyone presented with calculating √(x2), say √(-237124.6292) can do it in their heads, write the answer immediately and what do they actually do? They write |x| as calculated according to the Avodyne's definition! :tongue2:

It's not just an appearance of greater simplicity. It is simpler in practice. This proof is one example, just as any proof involving the absolute value of a real number. Other examples aren't hard to find. What is the derivative of |x|? That any person would use Avodyne's definition for a quick computation means it is a useful way to quickly compute something, not that it is a better working definition in proofs or studying qualities of the function.

And another thing I want to mention is that √(x2) isn't just chosen because it's a "function and inverse" that works out like we want it to. Consider that the absolute value for a complex number (the set of which is a superset of the reals, so this MUST hold in the reals too) is |a + bi| = √(a2 + b2). That's what it is, there is no way to define it with another function and its inverse. |x| = √(x2) is the exact same thing, the imaginary component is 0. That's why we say that |x| = √(x2) and not some other "function and inverse." It does not require any conditional at all (I still don't see what you mean by "inverted with a condition"), and is guaranteed to work every time.
 
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