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Injevtive/surjective Problem - Algebra

  1. Jul 19, 2013 #1
    So, i have to analyze the next function and see if it is injective and surjective
    f: NXN → N
    f(x,y)=x+y

    i know that a function to be injective f(a)=f(b) → a=b

    but here i have that f(a,c)=f(b,d) → (a,c)=(b,d)

    next thing i do: a+c =b+d

    now i dont know what to do...PLEASE HELP ME
     
  2. jcsd
  3. Jul 19, 2013 #2

    hilbert2

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    If the function were injective, it would mean that for every element b in N there's no more than one element a in NXN such that f(a)=b. This is not the case here. For example, f(1,0)=f(0,1).

    Surjectivity means here that for every element b of N there's at least one element a in NXN such that f(a)=b. This function is surjective, because if x is an element of N, then f(x,0)=x+0=x.
     
  4. Jul 19, 2013 #3
    Isn't a way to prove this without a counterexample. The injectivity it's allright if i doi it that way, but to prove that the function it it surjective i have that: [itex]\forall[/itex]yB,[itex]\exists[/itex]xA/y=f(x)
     
  5. Jul 19, 2013 #4
    Write down precisely what you need to show for your function. Are you sure you can't think of a simple way to write any [itex]x \in \mathbb{N}[/itex] as [itex]n+m[/itex]?

    Please don't do all of the work for the OP, especially when the question is clearly homework.
     
  6. Jul 21, 2013 #5
    To clarify, are you defining N to include 0?
     
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