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Discrete Mathematics : Functions and Relations : Question 2c

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    c) Is 'g' a surjective function (onto) ? Justify your answer.

    2. Relevant equations

    Let 'f' be a relation on ℤ (the set of integers) , defined by the entrance requirement :

    (x;y) ∊ ƒ iff y = x + 15

    and let 'g' be the function on ℤ defined by the entrance requirement :

    (x;y) ∊ g iff y = 5x(to the power of 2) + 7

    3. The attempt at a solution

    The steps I follow is the following :
    1. Clarify to myself what surjective is.
    2. Confirm the correct function or relation to use, and substitute the x and y with real values.
    3. Write out the formula with the proof in the required format.

    1 : Surjective :

    When we have a function , with a set A and a set B, as example, and all the elements of set B is mapped to an element in A. f: A→B (function f on set A to B), is a surjective function if the range of f is equal to the codomain of f, ie, f[A]=B.

    2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

    g(1) = 5(1)2 + 7 = 32
    g(2) = 5(2)2 + 7 = 107
    g(3) = 5(3)2 + 7 = 232

    Rewriting the (x;y) in each of the above examples :
    (1;32)
    (2;107)
    (3;232)

    * 3.

    This is how far i have gotten. My set 'A' can be defined as {1;2;3} used in my example, but i am not sure what my set B is, and if i am not sure what my set B is , then i cannot really say if every element of my set B can be mapped to set A. Not sure if this makes sense :(
     
  2. jcsd
  3. Jul 31, 2012 #2

    Mark44

    Staff: Mentor

    The question is about g, so the information above about f seems to be extraneous.
    You aren't doing these correctly. To evaluate 5x2, replace x by whatever number you're working with, square that number, and then multiply by 5. There's a difference between 5x2 and (5x)2, which seems to be what you are doing.

    For example, g(1) = 5(1)2 + 7 = 5*1 + 7 = 12, not 32.
    Isn't the domain of g all of the integers? If so, that would be your set A, assuming that's what you mean by A. To be able to say whether g is surjective, you need to get some more (correct) function values. When you do that, you will probably be able to answer the question.
     
  4. Jul 31, 2012 #3
    Sorry, i see now, i am fixing this immediately.
     
  5. Jul 31, 2012 #4
    Ok, so i have miscalculated this by far :

    Incorrect values :

    -------


    2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

    g(1) = 5(1)2 + 7 = 32
    g(2) = 5(2)2 + 7 = 107
    g(3) = 5(3)2 + 7 = 232

    Rewriting the (x;y) in each of the above examples :
    (1;32)
    (2;107)
    (3;232)

    -------

    Correct Values :


    2. The function to use is : (x;y) ∊ g iff y = 5x2 + 7

    g(1) = 5(1)2 + 7 = 12
    g(2) = 5(2)2 + 7 = 27
    g(3) = 5(3)2 + 7 = 52

    Rewriting the (x;y) in each of the above examples :
    (1;12)
    (2;27)
    (3;52)

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

    And the question is, is 'g' a surjective function.

    Isn't the domain of g all of the integers? If so, that would be your set A, assuming that's what you mean by A. To be able to say whether g is surjective, you need to get some more (correct) function values. When you do that, you will probably be able to answer the question.

    The domain would be ℤ , yes, all integers. I cannot find a limitation, and no limitation was specified. I have added the correct values for the range.

    My problem is this :

    a) I have 'domain' (x value) examples, and know it consist of ℤ. The range cannot possibly b
    e ℤ, however if it is g: ℤ→ℤ , then ℤ includes negative elements as well? To me if the x is multiplied with itself, it will definitely be positive , and cannot be negative. (Correct me if i am wrong, looking at the function 'g')

    Can you confirm what the range values will be ? After that, I should be able to see if it is surjective or not.

    Thanks!!
     
  6. Jul 31, 2012 #5

    Mark44

    Staff: Mentor

    The numbers you calculated are in the range (or codomain), but they aren't the range.
    Z includes all of the integers: positive, negative, and zero.
    I think you're on the right track, so I'll give you a push. All you need to do to establish that g is not surjective is to find one value in Z that is not in the range of g.

    It might be helpful to sketch a graph of g. Note that the graph is not a smooth curve -- it's a bunch of disconnected points that lie on a curve.
     
  7. Jul 31, 2012 #6
    Thanks for your quick reply, I appreciate your assistance. i am quickly reading your reply. And thank you once again for the assistance, you are seriously helping me an insane amount here.
     
  8. Jul 31, 2012 #7
    Sheez, i am still struggling a bit

    I have attached the graph, however , if i go into a minus, it still gives positive values ( this graph was done on excel).

    Further to this i did some more reading, and found something else which confuses me a bit more :)

    Let me share this with you :


    ran(g) = {g(x) | x ∊ ℤ}
    = {5x2 + 7 | x ∊ ℤ }
    = {y | for some x ∊ ℤ, y = 5x2 + 7 }
    = {y | 5x2 = -7 } [this cannot be possible because it must be positive ]
    = { y | dont know what else.. but it is confusing :D }

    it cannot be equal to the codomain
     

    Attached Files:

  9. Jul 31, 2012 #8

    Mark44

    Staff: Mentor

    OK to here, but incorrect in the next step. You are assuming that y = 0 for some x. Your graph should show you that this is not true. IOW, 0 is not in the range of g.
    Who is "it"?
     
  10. Jul 31, 2012 #9
    I see ( i think i see.. :)


    So i took away the y, which i cannot really do. but after a cup of coffee I saw.

    so let me do the next step again :

    y = 5x2 + 7
    y - 7 = 5x2

    y-7
    ---- = x2
    5

    So y cannot be a couple of things :

    a) it cannot be less than 7
    b) it must be a multiple of 5, and if it is, it must be possible to calculate the square root, because x is a ℤ, and thus the left hand side of the formula must amount to the sqaure root of any ℤ,

    Not sure if i am on the right track here :)
     
  11. Jul 31, 2012 #10
    I double checked the graph values - and i think that the scale is just right, however y will be 7 when x = 0 :


    X Y
    -10 507
    -9 412
    -8 327
    -7 252
    -6 187
    -5 132
    -4 87
    -3 52
    -2 27
    -1 12
    0 7
    1 12
    2 27
    3 52
    4 87
    5 132
    6 187
    7 252
    8 327
    9 412
    10 507
     
  12. Jul 31, 2012 #11

    Mark44

    Staff: Mentor

    a) Yes, y cannot be less than 7 (equivalently, y >= 7).
    b) No, y is not a multiple of 5, but y - 7 must be a multiple of 5. Further, y - 7 must be a multiple of 5 such that (y - 7)/5 is a perfect square. For example, y = 12 satisfies these criteria, because 12 - 7 = 5, and 5/5 = 1, which is a perfect square.

    However, I think you're getting a bit lost in the weeds. All you need to do is find a number in Z (the domain of g) that is not also in the range of g. If you can find such a number, which should be easy now, you can say whether g is surjective.


     
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