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Really basic stuff that I am confused on.

  1. Feb 18, 2008 #1
    ok this is really basic stuff that seems obvious I can't do, but I want to know why (I'm an 9th grade alg. 2)

    1=1^(4/4)=(+-1^(2))^(2/4)=(+-1^(2))^(1/2)=+-1^(2/2)=+-1 Why exactly doesn't this work,other then you get -1=1, but other then that in the process.

    I have other questions along similar lines, but I don't have time to post them right now.

    Thanks =]
     
  2. jcsd
  3. Feb 19, 2008 #2
    My best answer for you use is because you choose the wrong root of 1. There are two square roots of one. You equation only is valid for the positive square root.
     
  4. Feb 19, 2008 #3
    Indeed. Suppose:

    a2 = b2

    or, a2 - b2 = 0

    or (a - b)(a + b) = 0

    thus, a = b or (-b)

    In your case since "a" is positive, so is "b".
     
  5. Feb 19, 2008 #4
  6. Feb 19, 2008 #5
    Man that's why I'm all screwed up now, my teacher said you could. =\. I've stopped listening to her now =P Thanks for the reply. So just to clarify if you have x^(4/4), you can't cancel out the 4/4 too 1 or any variation of that, you have to work through it? Thanks, I've got a couple more things I'll put one up at a time.

    is x^(n/m) = to (x^(1/m))^(n)
    thanks for the replys
     
  7. Feb 19, 2008 #6
    Most grade 9 students wouldn't get themselves into trouble by doing as the teacher said. I'm sure the exceptions, which you discovered will not show up in your homework questions. I'm impressed that you are taking an interest in this stuff at grade 9.
     
  8. Feb 19, 2008 #7
    Haha thanks, I'm just curious about maths and science in general, it's something to think about. The problem is that the teacher really isn't very bright, and says wrong stuff all the time, and I took a couple of the things and extended them a bit, and now everything is a bit messed up ='[.

    It's annoying when people make simplifications in stuff like maths and science, because everything you base from there on out is wrong.

    Also I'm a bit annoyed, because last year my teacher wouldn't let me take alg II because of the high schools not liking it, so I'm a bit bored through most of it.

    Also do any of you know what I could do to occupy my time in class, I've already finished reading most of the alg. II text book while not paying attention lol =]

    Also another one is is this right.

    y=x^(1/2)
    x^(2/4)= [(+-x)^(2)]^(1/4)= (+-x)^(1/2)=yi,-yi, y, -y

    Is this right that this has 4 solutions? Or can I not cancel out the exponents. This is one of the extensions I've done from that.
     
    Last edited: Feb 19, 2008
  9. Feb 19, 2008 #8
    Some one who's better at this correct me if I make a mistake.

    If you want to be completely general, I'm not sure that x^(n/m) is well defined unless you put in some restrictions like restricting the roots to the positive roots.

    Rather lets consider the following:

    (x^n)^(1/m)=||x^n||exp(-i*2*k*pi/m)

    (x^(1/m))^n=(||x||exp(-i*2*k*pi/m))^n
    =||x||^n*exp(-i*2*k*pi*n/m)
    =||x^n||*exp(-i*2*k*pi*n/m)

    The first expression has m roots. The second expression will only have m roots if n and m share no common factors.
     
  10. Feb 19, 2008 #9
    Right yea I forgot that. Variables can only be positive real number integers. What exactly does || || mean, I'm not sure if I write it some other way. And what is exp and abbreviation for? thanks, once I know what these stand for I will be able to work out yalls posts a bit better =]
     
  11. Feb 19, 2008 #10
    It seems to work. In general you can't blindly cancel the exponents but when you wrote down the two for the inverse of x^2 you seem to be okay. You just must recognize that 2 of those roots need to be thrown away. So [(x)^(2)]^(1/4) has 4 roots but x^(1/2) has two roots.

    Perhaps I should try to elaborate somewhat on the principle.

    Complex numbers are numbs which in general have both a real and imaginary part writen as:

    z=x+i*y

    where x is the real part and y is the imaginary part.

    Complex numbers can be written in polar form:

    z=||z||exp(-i angle(z))

    where:

    the magnitude of z is given by:

    ||z||=sqrt(x^2+y^2)

    and the angle (argument) of z is given by:

    angle(z)=arctan(x,y)

    Complex numbers in polar form can be converted back to rectangular form by using eular's Identity.

    [tex] exp(-i \theta )=cos( \theta )+i*sin( \theta )[/tex]

    From Eular's identity is that complex numbers which phase (angle) differ by 2 pi are equal.

    see: http://en.wikipedia.org/wiki/Radian

    Returning back to polar form:

    ||z||exp(-i angle(z))=||z||exp(-i angle(z)+2*k*pi)

    for any integer k

    Therefore the Nth root of of z is given by:

    z^(1/n)=||z||^(1/n)exp(-i angle(z)/n+2*k*pi/n)
     
    Last edited: Feb 19, 2008
  12. Feb 19, 2008 #11
    Well, it can have multiple meanings. It is usually used to represent an idea of length. For real numbers length is the absolute value. That is we remove the sign of the number.

    I notice that wikipedia uses |z| instead of ||z||.
    http://en.wikipedia.org/wiki/Comple...forums.com/newreply.php?do=newreply&p=1616959
    Physics Help and Math Help - Physics Forums - Reply to TopicAbsolute_value.2C_conjugation_and_distance

    Anyway, If we think of a complex number as a vector then it it makes sense to use ||z|| for magnitude. That is viewing a the complex number z=x+iy as the vector (x,y) then the magnitude (x,y) is given by:
    ||z||=||(x,y)||=sqrt(x^2+y^2)

    However, I presume that mathematicians wish to distinguish between complex numbers as vectors as you can have vectors of complex numbers so they use |z| instead of ||z|| because |z| is like absolute value, while ||z|| is usually used to represent length (norm).

    So we can think of the absolute value of a complex number as removing the phase of a number just like the absolute value of a real number removes the sign of the number.
     
  13. Feb 19, 2008 #12
    Do you wish to learn more mathematics? I think the easiest subjects for someone of your ability to learn would be Discrete mathematics, or linear Algebra. That advantage of learning these is you don't need to understand Calculus to learn them.

    Linear Algebra is more useful but if your are just playing around with mathematics you might have more fun with discrete mathematics. You seem to have an interest in complex numbers but I don't know if it is advisable to study complex numbers before learning calculus. You might be able to learn set theory but while set theory is valuable for mathematics it is hard to see how to apply it without a certain level of mathematical maturity.
     
  14. Feb 19, 2008 #13
    Ah alright i think I see, so let me see if this is right.

    |x| where x is a real number is absolute value (magnitude)
    ||x|| for a real number is the vector, which is where you get the polar form notation from
    |x| for an imaginary number is the vector
    ||x|| for an imaginary number is the magnitude.

    is that about right? And so where does the exp come in?
     
  15. Feb 19, 2008 #14
    Yea that is what i am really looking for, some mathematics that I can teach myself that I won't be learning through my regular courses in high school. So I just quickly wikipediad both, and the discrete one seems a bit more fun, like you said, as it has more to do with ideas. What maths should I be able to do to do these?

    Thanks for all your help, I really appreciate it =]
     
  16. Feb 19, 2008 #15
    correct. Although we can also used it for absolute value/magnitude of complex numbers.
    Should say ||x|| stands for the magnitude (norm) of x. Usually when we say magnitude (size) we are talking about vectors.
    No, we are still using the symbols for the ideas of absolute value or magnitude. Absolute value is usually used in the context of numbers while magnitude is used in the context of vectors. The reason the terms are somewhat interchangeable is that complex numbers are like vectors in the sense that they obey the same rules with regards to addition.

    Complex numbers add as follows:

    x1+i y1=(x1+x2) + i *(y1 + y2)

    while vectors add as follows:

    (x1,y1)+(x2,y2)=(x1+x2,y1+y2)

    Vector addition could be thought of head to tail addition:

    [​IMG]

    That is you can add vectors by translating one vect towards the other vector without changing the direction of the vector so that the head of one vector connects to the tail to the other. Then the resulting vector is the arrow which points from the tail of one vector to the head of the other vector:
     
    Last edited: Feb 19, 2008
  17. Feb 19, 2008 #16
    I think that the people here are helpful enough that if you are patient you should be able to get though a discrete math book which is geared towards first or second year university students. I think your background is strong enough. You might struggle some with proofs at first as they will be new to you.
     
  18. Feb 19, 2008 #17
    On another note. I bet the answers to your questions aren't as basic as you thought.
     
  19. Feb 19, 2008 #18
    You're going to have to get use to simplifications all though high school and a lot of the way though university.
     
  20. Feb 19, 2008 #19
    Cool, I'll start browsing for online discrete math books!

    No, the answers really weren't lol, I'm going to have to reread everything tomorrow so i get it (it's 11 here, and I'm getting pretty tired, so I am going to have to stop here and do my homework and get some sleep)

    Good, I tend to like proofs =]

    I'll be back tomorrow, probably asking more questions haha, again thanks for your help =D
     
  21. Feb 19, 2008 #20
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