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Weird?about inequalities

  1. Sep 27, 2012 #1
    Hi,
    I am trying to solve x^2 - 10x + 26 < 0
    Here are my steps:
    1.Find the roots of x by using quadratic formula:
    x = [10 ± √(100-104)]/2
    = 5±i
    2.Rewrite x^2 - 10x + 26 into [x-(5+i)][x-(5-i)]
    3.Now we have:
    [x-(5+i)][x-(5-i)]<0
    x-(5+i) < 0 and x-(5-i) > 0 or x-(5+i) > 0 and x-(5-i) < 0
    x < 5+i and x > 5-i or x>5+i and x<5-i (rejected)
    So we get the answer 5-i<x<5+i

    Here is the bit that I found weird:
    When x = 5,
    5-i<5+0i<5+i
    ∴x^2 - 10x + 26 = 25 - 50 + 26
    = 1
    1>0?What's wrong with the above steps?
    Thx a lot!
     
  2. jcsd
  3. Sep 27, 2012 #2

    Mark44

    Staff: Mentor

    Homework and textbook-type problems should be posted in the Homework & Coursework section, not in the math technical section. I am moving your thread there, under Precalc Mathematics.
    Since the roots are complex and not real, that means that the expression x2 - 10x + 26 is never zero. That means that x2 - 10x + 26 is always either positive or negative.
     
  4. Sep 27, 2012 #3

    pwsnafu

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    x is real not complex.
     
  5. Sep 27, 2012 #4
    Oops,I don't know it is a textbook-type question coz I thought it by myself.Anyway thx for your replies.
    Do you mean we cannot solve [x-(5+i)][x-(5-i)] <0 by using the above algebraic method as we cannot compare the size between real no and complex no?
    Any other methods to solve this question? Thx a lot
     
  6. Sep 27, 2012 #5

    Mark44

    Staff: Mentor

    Nonreal complex numbers can't be compared using > or < because the complex numbers are not an ordered field. The real numbers are an ordered field, so can be compared this way.
    See what I said in my previous post.
     
  7. Sep 27, 2012 #6

    Ray Vickson

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    Homework Helper

    Besides what Mark44 has told you, you should also realize that in complex numbers there is no concept of > or <, so statements like 5-i < 5 < 5+i have absolutely NO MEANING at all.

    RGV
     
  8. Sep 30, 2012 #7
    sorry for my foolishness.
    Consider 5+i and 6-2i,you cannot compare their size as 5 is less than 6 but i is greater than -2i
    However,consider 5+i and 5-i
    Logically,i think 5-i is less than 5+i because 5-i has less i than 5+i and their real part are equal.
    Same case in 5+i and 6+2i.I always think that 5+i<6+2i...I can't understand
     
    Last edited: Sep 30, 2012
  9. Sep 30, 2012 #8

    Ray Vickson

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    What you are describing is called "lexicographic ordering", but it does NOT have all the desirable properties of a true ordering. See, eg.,
    http://www.cut-the-knot.org/do_you_know/complex_compare.shtml ,
    which illustrates the problems that arise from complex multiplication. In other words, we can have z1 < z2 and z3 > 0, but z1*z3 > z2*z3, which is not what we want.

    RGV
     
  10. Sep 30, 2012 #9
    Why if a<c then (a + ib) < (c + id)?
    e.g. (-1 + i) < (2 - 6i)
    -1<2,but i>-6i,how can you compare their size?
     
  11. Sep 30, 2012 #10

    Ray Vickson

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    Lexicographic ordering does NOT compare sizes. If you want to compare sizes, you can compare "magnitudes" and say z1 ##\prec## z2 if |z1| < |z2|, where |a + i*b| = √(a2+b2). That has SOME of the properties you want, but not all. For example, you can have z1 ##\prec## z2 but z2 + a ##\prec## z1 + a.

    The fact is: there is NO ordering on the complex numbers that preserves all the properties you want or need. Every proposed ordering fails for some of the desired properties, but different orderings fail in different ways.

    It is good that you are thinking about these issues, but don't allow those issues to side-track your approach to the original problem. Did you actually grasp the points explained to you before? The equation f(x) = 0 has NO real roots, so f(x) never changes sign for REAL values of x. To see which sign applies, just compute f(x) at some convenient point, such as x = 0. That has nothing at all to do with whether or not complex numbers can be ordered.

    RGV
     
    Last edited: Sep 30, 2012
  12. Oct 1, 2012 #11

    HallsofIvy

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    You don't. The complex numbers is not an ordered field under any definition of "order".

    A "field" is a a set with "addition" and "multiplication" such that all the "usual" arithmetic properties (commutative and associative, distributive, etc.) hold and that all members have an additive inverse (negative) and every member except 0 has a multiplicative inverse.

    An "ordered field" is a field together with an order (<) such that
    i) If a< b and c is any member of the field then a+ c< b+ c
    ii) If a< b and 0< c then ac< bc
    iii) For any a, b in the field one and only one must hold:
    1) a< b
    2) b< a
    3) a= b

    There is NO way to define an order on the complex number such that they form an ordered field.

    Suppose there were. Clearly i is not equal to 0 because 1+i is not 1. So, by (iii), we must have either i>0 or i< 0.

    If i> 0 then i(i)> i(0) or -1> 0. That, by itself, is not a contradiction since this is not necessarily and extension of the ordering of the real numbers. But if -1> 0 then i(-1)> i(0) or -i> 0. By (iii) we cannot have both i> 0 and -i> 0. Contradiction.

    If i< 0 then -i> 0. Then (-i)(-i)> (-i)(0) so [itex]-i^2= -1> 0[/itex]. And now (-i)(-1)= i> 0. Again that contradicts (iii).
     
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