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Math Help - Explain how these proofs work

  1. #16
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    Re: Explain how these proofs work

    Quote Originally Posted by Plato View Post
    I personally do not like the terms domain/codomain. I have used initial set and final set.

    In $f: \mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ the set $\mathbb{Z}\times\mathbb{Z}$ is the initial set and $\mathbb{Z}$ is the finial set.
    Ordered pairs of integers are mapped to an integer: $(m,n)\mapsto k$ so $ f=\{((m,n),k) : (m,n)\in(\mathbb{Z}\times\mathbb{Z})~\&~k\in \mathbb{Z}\}$.
    so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k

    that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?
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  2. #17
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    Re: Explain how these proofs work

    Quote Originally Posted by michaelgg13 View Post
    so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k
    that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?
    I have actually lost the theme of this thread. I even missed - sign on one of your posts.
    Here is what I read as the OP:
    Quote Originally Posted by michaelgg13 View Post
    Let f : Z x Z --> Z state whether the following function is onto or not, then prove or disprove.
    f(m, n) = m2 + n4
    $f: (\mathbb{Z}\times\mathbb{Z})\to\mathbb{Z}$ defined by the rule $(m,n)\mapsto m^2+n^4$.

    If again I have not missed something, it is clearly $f(m,n)\ge0$ so can it be onto?
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  3. #18
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    Re: Explain how these proofs work

    Quote Originally Posted by Plato View Post
    I have actually lost the theme of this thread. I even missed - sign on one of your posts.
    Here is what I read as the OP:

    $f: (\mathbb{Z}\times\mathbb{Z})\to\mathbb{Z}$ defined by the rule $(m,n)\mapsto m^2+n^4$.

    If again I have not missed something, it is clearly $f(m,n)\ge0$ so can it be onto?
    It can't be cause they don't map to anything, correct?



    In post #14 I mentioned
    Quote Originally Posted by michaelgg13 View Post
    Don't give me the answer, but for example, explain how you would incorporate this domain/codomain into proving/disproving f(m,n) = 2m+n
    And then I took a shot at it by stating this
    Quote Originally Posted by michaelgg13 View Post
    so to prove it should I come up with an example set? such that m,n =1 so f(1,1) = 2(1)+1 = 3 = k

    that is, since f(1, 1) = 3 and 3 is an integer, 3 = k (because k is some integer)?
    Sorry for getting you mixed up. When I said OP I wasn't trying to refer to the question as a whole, but to understanding the domain/codomain, which you explained to me very well.
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  4. #19
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    Re: Explain how these proofs work

    Suppose we want to prove $f: \Bbb Z \times \Bbb Z \to \Bbb Z$ given by $f(m,n) = 2m+n$ is onto.

    Then we must show that for ANY integer $k$, there is at least one pair $(m,n)$ with $f(m,n) = k$.

    We can pick $m,n$ any way we want, as long as they are both integers. Suppose we pick: $m = 0, n = k$?

    I can tell you that $f$ is NOT injective. See if you can find two different pairs $(m_1,n_1),(m_2,n_2)$ such that:

    $f(m_1,n_1) = f(m_2,n_2)$. I suggest finding two such pairs with both images of $f$ equal to 3.
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  5. #20
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    Re: Explain how these proofs work

    Quote Originally Posted by michaelgg13 View Post
    Sorry for getting you mixed up. When I said OP I wasn't trying to refer to the question as a whole, but to understanding the domain/codomain, which you explained to me very well.
    @michaelgg13, May I suggest that you should start a new thread if you change the question.
    That is, stick with the PO in any thread..
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  6. #21
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    Re: Explain how these proofs work

    Let's look at the last function:

    $f(n) = n^2 + 1$.

    Can we find an $n$ such that $f(n) = -1$? Suppose that (just for the time being) one existed. Then:

    $n^2 + 1 = -1$
    $n^2 = -2$.

    If $n = 0$, then $n^2 = 0 \neq -2$. So $n$ is not 0.

    If $n > 0$, then $n^2 > 0$ so $n^2 \neq -2$, since $-2 < 0$.

    If $n < 0$ then $n = -k$ for some $k > 0$, and thus $n^2 = (-k)(-k) = (-1)(k)(-1)(k) = (-1)(-1)k^2 = k^2 > 0$, so again $n^2 \neq -2$.

    So there is NO integer $n$ with $f(n) = -2$, so $f$ CANNOT be onto, since the value -2 is not in the range of $f$.

    Can $f$ be injective? Again, no, because $f(-1) = (-1)^2 + 1 = 1 + 1 + 2 = 1^2 + 1 = f(1)$, so $f$ maps -1 and 1 to the same value. That is:

    $f(-1) = f(1)$ but $-1 \neq 1$, and an injective function must satisfy:

    $f(n) = f(m) \implies m = n$ for ALL $m,n$ and this simply isn't true for n= -1, and m = 1.
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