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Constructible numbers!

  1. Apr 7, 2006 #1
    i'm suppose to show how to construct each of the following:

    a+b,ab,a/b,a-b, a^(1/2) .... a,b >0

    This has to do with constructible numbers using the ruler and compass.

    So i know that a number is constructible if it the distance (or negative distance) between two constructible points, but I suppose I don't really need such a fact. hmm... SO how do i start this? Can I just look at points a, and b as vectors. I'm sure if I get help with a+b, I can start working on the rest. thanks..
     
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  3. Apr 7, 2006 #2

    HallsofIvy

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    First, don't talk of a and b as "points"- they are line segments.

    a+ b is easy. You have line segment a (actually a line segment of length a) and line segment b. Draw a line, L, and use your compasses to mark off a segment of L congruent to a (i.e. having the same length as a). Now from one end of that copy of a, use your compasses to mark off a segment having the same length as b. You now have a segment corresponding to a+ b.
     
  4. Apr 7, 2006 #3
    thanks for your help. Similarly, a-b follows... take a segments of length a and b... construct a segment L that is congruent to a, similarly use the compass to produce another segment congruent to b, but in the opposite direction of a. This gives a-b. [describing such actions is odd to me for some reason]

    The product ab then, I would assume, could either be a box or a line (if one straightened out that box)- depending on how you want to construct it (right?).

    a/b, and a^(1/2) still leave me clueless... I'm still trying...thanks again.
     
  5. Apr 7, 2006 #4

    NateTG

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    You need to get a line segment. "Straighten out a box" doesn't make sense.
     
  6. Apr 7, 2006 #5
    It's enough to show ab is constructible (for all constructible a,b). Since then ab^{-1} is certainly constructible.
    Note that the perimeter of that box is 2a+2b, not ab, so that does not give a proof for ab.
     
  7. Apr 7, 2006 #6

    Hurkyl

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    You can't just assume it would be so -- it still requires a proof.

    Maybe he'll figure out how to invert his construction, or maybe constructiong b^{-1}, or even some other method. Maybe one of these will be trivial. But it can't just be assumed!
     
  8. Apr 7, 2006 #7
    You're right, I was thinking we had already known that b^{-1} is constructible if b is, but I see that isn't the case.
     
  9. Apr 8, 2006 #8

    AKG

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    Suppose you have a triangle with an angle of 60 in there somewhere. Call the side opposite to this angle c. Then we get the formula:

    c² = a² + b² - ab
    ab = a² + b² - c².

    Given a and b, it's easy to construct a triangle such that the angle between a and b is 60. You know how to add and subtract, so you just need to know how to square lengths. However, I can't think of how to do this.

    If you have vectors a and b, then (a.b)a/|a| is constructible: it is projection of the vector b onto the line generated by a. So if you vectors a and b with their tails touching, then the vector above is obtained by extending a line through b, and finding the point P on this line such that the line perpendicular to this line through a which passes through the point P meets the tip of a. Note (a.b)a/|a| = |b|cosXaa, where X is the angle between a and b. Let's say the vector b has coordinates (b,0,0) hence has length b. Then the above becomes bcosXa. With the appropriate choice of X, can you use the above construction of the vector projection to construct the products ba1, ba2, ba3? I have a feeling this won't work, I can't think of what to do...
     
  10. Apr 8, 2006 #9
    I'm still lost, regarding products and roots...I appreciate all of your help. I will post again 24hrs or less from now, after I have studied the topic a little more. Just thought I'd let you all know that I've taken the time to read your responses.


    [ To let you know also, I haven't really been taught anything on the subject, but my prof intends to put a question on constructions on the final (optional of course), and I just intend to answer it. He said I'd have to construct the given segments, define what it means for a real number to be constructible, and solve a certain classical problem (of the ancient greeks)... FYI)
     
  11. Apr 8, 2006 #10

    Curious3141

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    This is a most interesting question. I pondered on it for awhile, and here's what I could come up with :

    Here are the basic stipulations. One is given two unequal line segments, length a and b (without loss of generality, a > b). With this alone, and with the aid of unmarked ruler and compasses, what line segments can be constructed ?

    Immediately, one can see how to work out (a + b) and (a - b) : just superimpose the lines and extend or demarcate as needed. In fact, any linear combination of length (ma + nb) can be worked out for integral m and n.

    Now, one can also bisect ad nauseam to give all lengths of the form [tex]\frac{ma}{2^p} + \frac{nb}{2^q}[/tex] where m and n are integers and p and q are naturals.

    Further, using this ingenious method of trisecting a line segment, we can extend the list of possibilities to : [tex]\frac{ma}{(2^p)(3^r)} + \frac{nb}{(2^q)(3^s)}[/tex] I don't know if extension to further prime divisors is possible, if I'm not mistaken, Gauss came up with some results in this area.

    Now, let's go to roots. Interestingly, [tex]\sqrt{ab}[/tex] can be fairly easily constructed, like so :

    [tex](a+b)^2 - (a-b)^2 = 4ab[/tex] so what we need to do is to construct line segment of length (a-b), then use one end to draw a circle of radius (a+b). From the other end, we need to construct a perpendicular to this line segment, and connect the intersection of the perpendicular and the circle to the first point. This is the hypotenuse of a right triangle and it has length [tex]2\sqrt{ab}[/tex]. Bisect that to get [tex]\sqrt{ab}[/tex]. So now, we can get both the arithmetic mean and the geometric mean of two line segments.

    From this we can get various combos of the form : [tex]\sqrt{ma^{p/2}b^{q/2} + na^{r/2}b^{s/2}[/tex] and work up to higher binary exponents. The problem here is that neither the addition nor either of the a or b terms can be eliminated. Only composites involving sums of products of various binary roots of a and b can be gotten.

    If we are given an additional piece of critical info, namely, the length of the unit line segment (in the form of a line segment which is defined to be of unit length), then we can do a lot more very easily.

    Given a, b and 1 (unity), immediately, a/b and b/a can be seen as gradients on a Cartesian plane where the a and b segments are marked out perpendicular to each other. To get line segments corresponding to those just mark out unity on either the vertical or horizontal axis, produce and voila !

    Reciprocals (1/a, 1/b, 1/ab, and other combos) are also easy. The harmonic mean of a and b can now also be constructed.

    Similarly, products like (ab), and all integral powers of a and b individually and combos of the form [tex]a^mb^n[/tex] with m and n integral (negatives too) can be gotten. Using the previous results, combos of binary roots and so forth can also be derived.
     
    Last edited: Apr 8, 2006
  12. Apr 8, 2006 #11

    Hurkyl

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    By the way, I think it might be a good exercise to prove that you need to use a segment of length 1 in order to multiply two line segments. The proof is very short once you figure out what's going on.
     
  13. Apr 8, 2006 #12

    Curious3141

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    I would be interested to see a formal proof of that. I mean, I know how to "see" it (the product or quotient needs a similar triangle construction with a unit segment as reference), but how do you go about proving it rigorously ?
     
  14. Apr 8, 2006 #13

    Hurkyl

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    Progressive hints in white:

    ---------------

    Consider what happens when you change your mind about which segment should count as length 1.

    ---------------

    In particular, choose two segments that aren't congruent. Construct the product ab when you treat the first one as the unit segment. Then, redo the construction when you treat the other one as the unit segment. Compare the two products you constructed.

    ---------------

    So different unit segments lead to different products. So if you never use the unit segment in your construction of ab, how do you know which one is right?

    ---------------
     
  15. Apr 8, 2006 #14

    Curious3141

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    That's a good idea. To make things a little clearer, I would pose the problem like so : Given two unequal line segments and that one (not told which) is of unit length, construct a line segment with a length equal to the product of the two.

    Obviously, not knowing which segment is of length unity would mean two distinct answers are possible, violating commutativity.

    This doesn't of course apply to finding the geometric mean because, conceptually, that would lie between the two line segments so that a single definitive construction is possible.
     
  16. Apr 8, 2006 #15

    Hurkyl

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    Here's a second approach! I was thinking along these lines originally, but when trying to make it rigorous, I came up with the other argument.


    Euclidean geometry is invariant under dilations -- if you double the size of everything, any construction you make will still work.

    But the product ab is not invariant under dilations! No construction based on the line segments of length a and b can be guaranteed to produce a line segment of length ab. Why? If you double the size of everything, then we would have 2ab being the product of 2a and 2b!

    So it's clear we need to divide: we really want to construct ab/1, not ab.
     
  17. Apr 8, 2006 #16

    Curious3141

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    NEAT !! :approve:

    It's so cool how everything works out. [tex]\sqrt{ab}[/tex] is OK without a unit segment because [tex]\sqrt{(2a)(2b)} = 2\sqrt{ab}[/tex] :cool:
     
  18. Apr 9, 2006 #17

    AKG

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    How are you going to multiply 2a by 2b, and how are you going to take the square root of its product?
     
  19. Apr 9, 2006 #18

    Curious3141

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    You don't have to : quoting from my previous post (the long one).

     
    Last edited: Apr 9, 2006
  20. Apr 9, 2006 #19
    To trisect a segment (and to divide it into equal parts of any multiple) cant we just place [itex]n[/itex] segments end-on-end perpendicular to our original segment, join a straight a line from the end of the n segments to the end of our first segment (thus creating a right angled triangle, with 2 perpendicular lengths [itex]a[/itex] and [itex]an[/itex]), and then draw straight parallel lines to this hypotenuse from the points where the n segments join together.
     
  21. Apr 9, 2006 #20

    Curious3141

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    Yes you can. That's a similar method to one given here : http://www.jimloy.com/geometry/trisect0.htm

    So arbitrary division is possible,
     
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