20. Give an example of a function from N to N that is a) one-to-one but not onto. b) onto but not one-to-one. c) both onto and one-to-one (b

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20. Give an example of a function from N to N that is a) one-to-one but not onto. b) onto but not one-to-one. c) both onto and one-to-one (but different from the identity function). d) neither one-to-one nor onto.

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Thiên Thanh 4 years 2021-08-16T10:44:14+00:00 1 Answers 1122 views 0

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  1. Dulcie
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    2021-08-16T10:46:09+00:00
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    Answer:

    Step-by-step explanation:

    a) To provide an example of a function N → N that is one-to-one but not onto.

    Suppose f:N\to N  to be f(n)=n^2

    Then; \text{a function } \ f: A \to B\ \text{is one-to-one if and only if } f(a) = f(b) \implies a = b \ for \ a, b \ \epsilon \ A.

    \text{a function } \ f: A \to B\ \text{is onto if and only if for every element } b \ \epsilon \ B \\ \text{there exist an element a} \ \epsilon\ A \ such \ that f(a) = b}

    Now, assuming a \ \Big {\varepsilon} \ N \& \ b \ \epsilon \ N;

    Then f(a) = f(b)

    a^2 = b^2 \\ \\ a = b

    The above function is said to be one-to-one

    \text{it is equally understandable that not every natural number is the square of a natural number}e.g

    2 is not a perfect square, hence, it is not regarded as the image of any natural no.

    As such, f is not onto.

    We can thereby conclude that the function  f(n) = n^2 is one-to-one but not onto

    b)

    Suppose f: N \to N be

    f(n) = [n/2] \\ \\ For \ n =1, f(1) = [1/2] = [0.5] = 1 \\ \\ For \ n=2 , f(2) = [2/2] = [1] = 1

    It implies that the function is not one-to-one since there exist different natural no. having the same image.

    So, for n \epsilon N , there exists an image of 2n in N

    i.e.

    f(2n) = [2n/2] = [n] = n

    Hence, the function is onto

    We thereby conclude that the function f(n) = [n/2] \text{ is onto but not one-to-one}

    c)

    let f: N\to N be  f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right.

    So, if n, m is odd:

    Then:

    f(n) = f(m) \\ \\ n-1 = m-1 \\ \\ n = m

    Likewise, if n, m is even:

    Then;

    f(n) = f(m) \\ \\ n+1 = m+ 1 \\ \\ n = m

    The function is then said to be one-to-one.

    However, For n \epsilon N and is odd, there exists an image of n - 1that is even;

    f(n - 1) = n -1 + 1 =n

    For n \epsilon N and is even, there exists an image of n + 1that is odd;

    f(n - 1) = n +1 - 1 = n

    where(; implies such that)

    Hence, this function is said to be onto.

    We can therefore conclude that the function f(n) = \left \{ {{n+1, \ if \ n \ is \ even } \atop n-1 , \ if \ n \ is \ odd} \right. is both onto and one-to-one.

    d)

    Here, to provide an example where the f:N \to N is neither one-to-one nor onto.

    SO;

    Let f : N \to N is defined to be f(n)=0

    Then, since every integer has the same image as zero(0), the function is not one-to-one.

    Similarly, the function is not onto since every positive integer is not an image of any natural number.

    We, therefore conclude that, the function f(n)=0 is neither one-to-one nor onto.

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