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Math Help - Doubts in functions.

  1. #1
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    Doubts in functions.

    I had a few doubts:

    a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
    Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

    b) Prove that this relation need not hold, if f: A-->B and X ,Y are subsets of A.
    f(X ^ Y) = f(X) ^ f(Y)
    So is it enough if I just give a counter example or should I prove it?
    and is this example right?

    Say f:A---->B given by f(x) = a for all x in A,
    Let X = {1,2,3}
    Y = {4,5}
    so here X ^ Y is null,
    whereas f(X) =f(Y) = {1}
    so f(X^Y) need not be equal to f(X)^ f(Y)

    c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

    I just wanted to know if my answer is correct.
    # There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

    # From what I could see there are (2^n - 2) functions.
    That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
    I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
    ..
    Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

    So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

    So are these right?
    Have I gone wrong anywhere?
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by poorna View Post
    c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

    I just wanted to know if my answer is correct.
    # There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

    # From what I could see there are (2^n - 2) functions.
    That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
    I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
    ..
    Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

    So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

    So are these right?
    Have I gone wrong anywhere?
    The (c) part seems right.
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  3. #3
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    hey guys, please help me with the questions!
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by poorna View Post

    a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
    Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

    I think its just a whim...but look at the fallacy below.

    Your question reminded me the following from my textbook:
    \sqrt{-1}.\sqrt{-1}=\sqrt{(-1)^2}=1
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    ah, yes but that's (-1)^(1/2 + 1/2) = -1 right?
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    Quote Originally Posted by poorna View Post
    I had a few doubts:

    a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
    Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

    b) Prove that this relation need not hold, if f: A-->B and X ,Y are subsets of A.
    f(X ^ Y) = f(X) ^ f(Y)
    So is it enough if I just give a counter example or should I prove it?
    and is this example right?

    Say f:A---->B given by f(x) = a for all x in A,
    Let X = {1,2,3}
    Y = {4,5}
    so here X ^ Y is null,
    whereas f(X) =f(Y) = {1}
    so f(X^Y) need not be equal to f(X)^ f(Y)

    c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

    I just wanted to know if my answer is correct.
    # There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

    # From what I could see there are (2^n - 2) functions.
    That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
    I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
    ..
    Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

    So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

    So are these right?
    Have I gone wrong anywhere?

    For the (b) part ,since  X\cap Y = \emptyset,you must prove 1st that :

    f(\emptyset) = \emptyset so you can show:

     f(X\cap Y)\neq f(X)\cap f(Y).

    Also how do you prove that:

    f({1,2,3}) = {1} ??

    It would be easier if you put :

    X ={1} and Y ={2}


    FOR the (c) part,according to your formula for n=2 we have that:

     2^n - 2 = 2^2 -2 = 4 -2 =2

    Here is a counter example:

    A = {1,2} , B={ a,b},then we have the following functions:


    1) f(1) = a , f(2) =a

    2) f(1) = b ,f(2 ) = b

    3) f(1) = a ,f(2) = b

    4) f(1) = b , f(2) = a
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  7. #7
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    Quote Originally Posted by xalk View Post
    FOR the (c) part,according to your formula for n=2 we have that:
     2^n - 2 = 2^2 -2 = 4 -2 =2

    Here is a counter example:
    A = {1,2} , B={ a,b},then we have the following functions:
    1) f(1) = a , f(2) =a
    2) f(1) = b ,f(2 ) = b
    3) f(1) = a ,f(2) = b
    4) f(1) = b , f(2) = a
    His/her part c) is correct.
    The question is "how many onto (surjections) are there?"
    There two in blue above.
    The other two are not onto.
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    Quote Originally Posted by Plato View Post
    His/her part c) is correct.
    The question is "how many onto (surjections) are there?"
    There two in blue above.
    The other two are not onto.
    her formula for n=2 ,gives two functions from A ={1,2} to B={a,b},while my counter example gives 4 functions
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  9. #9
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    Quote Originally Posted by poorna View Post
    I had a few doubts:

    a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
    Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?
    Hi poorna. The analytic definition of exponential with basis a is

    a^x=e^(xlog a)

    Hence, it is only well defined for a>0.

    That is also the reason underlying because some mathematical programs only give the graphic representation of (x)^(1/3) for x positive, because it makes the calculation fot e^((1/3)(log x)). A lot of programs avoid the problem because this function can be easily defined for negative numbers
    as -e^((1/3)log(-x)
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  10. #10
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    Quote Originally Posted by xalk View Post
    her formula for n=2 ,gives two functions from A ={1,2} to B={a,b},while my counter example gives 4 functions
    zalk,
    Please learn the vocabulary of set theory before correcting a posting.
    The question asked for the number of onto functions (the number of surjections).
    Two is the correct answer. I told you that above.

    From the set \left\{ {1,2,3, \cdots ,n} \right\} to the set \left\{ {0,1} \right\} there 2^n -2 surjecions.

    Please be sure of the vocabulary before posting a correction to questions that you may not understand.
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  11. #11
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    Quote Originally Posted by poorna View Post



    c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?


    # From what I could see there are (2^n - 2) functions.
    HERE she asks for how many functions from A into B FIRSTLY AND then for the onto functions.



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