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Mapping a function

  1. Apr 19, 2016 #1
    1. The problem statement, all variables and given/known data

    Hello all, thank you in advance for your help. Please let me know if this is the wrong forum.

    My problem:

    Let x= a + b√2 and let y= the 2 x 2 matrix {a 2b}{b a}.

    Show that x maps to y.

    2. Relevant equations

    As I see it, for multiplication, this is pretty straightforward, a 1 x 2 matrix times a 2 x 2 matrix, will yield a 1 x 2 matrix. But for addition, a required answer, how can we map a + b*sqrt(2) to turn it into the above matrix?


    3. The attempt at a solution

    Aside from teeth gnashing and f-bombs, I really don't have much to offer here. By no means am I asking for an answer, but a hint above what I am finding online would be greatly appreciated.
     
    Last edited: Apr 19, 2016
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  3. Apr 19, 2016 #2

    LCKurtz

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    Could you please post the exact and complete wording of the problem from your source?
     
  4. Apr 19, 2016 #3
    Unfortunately, that is pretty much it as it is stated above.

    Let x= a + b√2 and let y= the 2 x 2 matrix {a 2b}{b a}.

    How does x map to y?
     
  5. Apr 19, 2016 #4

    LCKurtz

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    There must be more context you could give. What are you studying in the section in your book where you got the problem?
     
  6. Apr 19, 2016 #5
    Specifically, a isomorphism. If x is an element of X and y is an element of Y, I am to show that X is isomorphic to Y. I wanted to keep that out of the post initially because I think I have a handle on showing that operations are preserved and that the function would be one-to-one and onto, but just showing that they map to each other is a problem.

    I have to show they are closed under multiplication, which is easy enough -- if I write x as a 1 x 2 matrix and y being a 2 x 2 matrix, the product is simple enough to compute.

    Last, it asks if this isomorphism preserves multiplication as well as addition, which I am totally unsure of, as I think I will need to be able to describe the mapping from x in X to y in Y first.

    Thanks kindly,

    SY
     
  7. Apr 19, 2016 #6

    LCKurtz

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    OK. Now I understand the context. You are dealing with the arithmetic of real numbers of the form ##a + b\sqrt 2## (surds). What you are trying to do is represent those numbers and their arithmetic by matrix operations. So lets let ##\{S,+,\cdot\}## represent the surds. The idea is to identify the surds with special 2x2 matrices of the special form ##\left[\matrix{a & 2b\\b & a}\right ]##. Let's call those matrices with their operations ##\{M,+,\cdot\}##.

    The isomorphism you are supposed to address is the map ##T:S\to M## given by $$T(a+b\sqrt 2) = \left[\matrix{a & 2b\\b & a}\right ]$$Show it preserves operations. Does that help?
     
  8. Apr 19, 2016 #7

    Ray Vickson

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    LCKurtz has explained in Post #6 exactly what the mapping means.

    If ##a,b## are allowed to be real numbers, you will not be able to recover a unique ##(a,b)## pair from a given value of ##a + b\sqrt{2}## so the mapping is not a true function (which would require a single 2x2 matrix for each value of ##a + b \sqrt{2}##). What happens if you restrict ##a,b## to the rationals instead?
     
  9. Apr 19, 2016 #8
    LCKurtz,

    So, all this mapping is literally putting the surds and saying that they equal the given 2 x 2 matrix? That it is as easy as putting an equal sign in between what I have as my x term and y term?

    Great balls of fire, I love/hate mathematics.

    So, the Surds are the x term, (a + b√2), and we have the group {S,+} to represent the set of surds S. The 2 x 2 matrix M is represented as a group by {M,+}. Since our iso is
    T:S→M, and the operation (addition) is the same, so they are preserved.


    Ray Vickson, my given question restricts a,b to the set of rational numbers, I forgot to include that. Since the set of Reals would not give us a 1 to 1 function for the reasons that you stated, restricting the operation to the set of rationals does indeed give a 1 to 1 function, and it is onto, because each unique a and b maps to itself when we go from my x term to my y term.

    I also have them as being closed under the set of rational numbers. because the set of rationals is itself closed under multiplication.

    Last, for the iso being preserved under multiplication, I'm not seeing it. Trying to show this via a counterexample, I have (a+b*√2)^2, which becomes a^2+2ab+2b, and am I correct in seeing that the 2ab term would not map to the 2 x 2 matrix above?

    Thank you both very much for your help, there was quite a bit of weeping and gnashing of teeth over this one.

    SY
     
  10. Apr 19, 2016 #9
    Correction, (a+b√2)^2 would become a^2 + 2ab√2 +2b^2.
     
    Last edited: Apr 19, 2016
  11. Apr 19, 2016 #10

    LCKurtz

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    No. Numbers don't "equal" matrices.

    No. Adding matrices is not adding numbers. And what about multiplying numbers vs multiplying matrices?

    You haven't expanded ##(a+b\sqrt 2)^2## correctly. And you need to use the correct notation using the transformation ##T##. It might help you understand if you used ##+, \cdot## for number operations and ##\oplus,\otimes## for matrix operations. So you could start by showing$$
    T(x + y) = T(x) \oplus T(y)$$You need to write it out and ditto for multiplication.
     
  12. Apr 19, 2016 #11
    Roger, I realized I expanded my term incorrectly after my last post, it becomes a^2 +2√2ab+2b^2, which does not preserve the isomorphism, because the 2√2ab term does not map from x to y for that reason. Is this in the ballpark?

    SY
     
  13. Apr 19, 2016 #12

    LCKurtz

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    No. In fact it is incorrect. Any number that can be put in the form ##c + d\sqrt 2## can be mapped. And, like I said earlier, you need to write the equations using ##T## and show it works for both addition and multiplication.
     
  14. Apr 19, 2016 #13

    Ray Vickson

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    Isn't ##a^2 + b^2 + 2ab \sqrt{2}## a number of the form ##A + B\sqrt{2}## with rational ##A,B##?
     
  15. Apr 19, 2016 #14
    So, T(x+y) maps to T(a)+T(b√2)⊕T{a 2b}{b a}}?

    For multiplication, T(xy) maps to T(a)T(b√2)⊗T{a 2b}{b a}}, and we still have closure by writing x in matrix form and multiplying a 1x2 matrix by a 2x2 matrix?

    As far as the iso not being preserved under multiplication, I am inclined to guess that is so, but I am still not seeing how to prove this to be.

    It's late on the E Coast of the US, thank you both for your time, see you tomorrow.

    Steve
     
    Last edited: Apr 19, 2016
  16. Apr 19, 2016 #15

    LCKurtz

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    No. You are very mixed up. You can't possibly know what T(x+y) maps to without knowing what x, y, and x+y are. And what I have emphasized in red makes no sense at all because T doesn't operate on matrices. It operates on numbers.
    This has nothing to do with 1x2 matrices or writing x in matrix form, whatever that means.

    Maybe you will think more clearly after sleeping. Then go back to post #10 and do what is suggested there.
     
  17. Apr 20, 2016 #16
    OK, as I have my breakfast and coffee, I honestly think that I have gone backwards here.

    The fist part of my question deals solely with addition, so that is the only operation I will look at right now.

    We want to show the iso T:S→M, as defined by T(a+b√2)={{a 2b}{b a}}. The function b is what can make this happen.

    I think the biggest stumbling block I am seeing is one that I have had for several days now. I am not seeing how to map the equation in S to the matrix given in M using function alpha, I have not been able to wrap my mind around that. I am guessing that must be able to happen, and we do need to show what this function is, since it does exist, but apparently is is something that I am able to put together.

    Mathematics. If it were easy, everyone would do it. I think my professor assigned this problem as her "A" level material. If you want an A, this problem must be done to standard. It's the last problem I need to finish in this week's assignment, and while some have required some piecing together, none have absolutely wrapped me around the axle as this one has.
     
  18. Apr 20, 2016 #17

    LCKurtz

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    I will give you another hint. You are trying to show ##T(x+y) = T(x) \oplus T(y)##. Start like this:
    ##\begin{array}{rcl}x& =& a + b\sqrt 2\\
    y &=& ...\\
    x+y& =& ...\\
    T(x+y) & = & ...
    \end{array}##
    When you have that, you will have the left side of the equation you are trying to prove.

    Then calculate ##T(x),~T(y)## and ##T(x)\oplus T(y)##. See if you get the same thing for both sides.

    Then do multiplication the same way.
     
    Last edited: Apr 20, 2016
  19. Apr 20, 2016 #18
    Still work in progress...

    Again, thank you.

    SY
     
  20. Apr 22, 2016 #19
    LCKurtz, I just wanted to say thank you for your time. I spoke to a principal in the district that I work in who has an. MS in mathematics, and he was able to give me a good bit of sorely needed direction.
     
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