1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 19, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove:
    |xy| = |x| * |y|?

    2. Relevant equations
    |x| = sqrt(x^2)


    3. 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.
     
  2. jcsd
  3. Sep 19, 2013 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
     
  4. Sep 19, 2013 #3

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  5. Sep 19, 2013 #4
    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

    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 | |.
     
  6. Sep 19, 2013 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  7. Sep 19, 2013 #6
    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?
     
  8. Sep 19, 2013 #7

    Avodyne

    User Avatar
    Science Advisor

    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.
     
  9. Sep 19, 2013 #8

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    Start by stating the definition of | | .

    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.

    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.
     
    Last edited: Sep 20, 2013
  10. Sep 19, 2013 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What difficulties are those? This is, in fact, how the absolute value is sometimes defined for real numbers.
     
  11. Sep 19, 2013 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.

     
  12. Sep 19, 2013 #11
    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?

    So I would say:
    (Case 1) x= 0, y= 0...and then be like " |a + b| = a + b= |a| + |b|"?
     
  13. Sep 19, 2013 #12
    So if I keep it under the radical:
    ...
    sqrt(|xy|) = sqrt(|x| * |y|) ?
     
  14. Sep 19, 2013 #13

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, again, you're using what you're trying to prove.

    Reread post 5.
     
    Last edited: Sep 20, 2013
  15. Sep 20, 2013 #14

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    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.
     
    Last edited: Sep 20, 2013
  16. Sep 20, 2013 #15
    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|?
     
    Last edited: Sep 20, 2013
  17. Sep 20, 2013 #16

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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".
     
  18. Sep 20, 2013 #17

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    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".
     
    Last edited: Sep 20, 2013
  19. Sep 20, 2013 #18
    No, it was not from a textbook.

    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?
     
  20. Sep 20, 2013 #19

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    " 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.
     
  21. Sep 20, 2013 #20

    Avodyne

    User Avatar
    Science Advisor

    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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How do I prove |xy| = |x| * |y|?
Loading...