Injevtive/surjective Problem - Algebra

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The function f: NXN → N defined by f(x,y) = x + y is not injective because multiple pairs (a,c) can yield the same sum, such as f(1,0) = f(0,1). However, it is surjective since for every element b in N, there exists a pair (x,0) such that f(x,0) = x = b. The discussion emphasizes the need for clarity in definitions, particularly whether N includes 0. Participants suggest that proving surjectivity can be approached without counterexamples, focusing on expressing any natural number as a sum of two non-negative integers. The conversation revolves around understanding the properties of injective and surjective functions in the context of this specific function.
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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 don't know what to do...PLEASE HELP ME
 
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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.
 
hilbert2 said:
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.

Isn't a way to prove this without a counterexample. The injectivity it's all right if i doi it that way, but to prove that the function it it surjective i have that: \forallyB,\existsxA/y=f(x)
 
SclayP said:
Isn't a way to prove this without a counterexample. The injectivity it's all right if i doi it that way, but to prove that the function it it surjective i have that: \forallyB,\existsxA/y=f(x)

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 x \in \mathbb{N} as n+m?

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.

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