Injevtive/surjective Problem - Algebra

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Homework Help Overview

The discussion revolves around analyzing the function f: NXN → N defined by f(x,y) = x + y, specifically examining its injectivity and surjectivity properties.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the definitions of injective and surjective functions and is exploring how to apply these definitions to the given function. Some participants provide examples to illustrate the function's properties, while others question the need for counterexamples in proving surjectivity.

Discussion Status

Participants are actively discussing the injectivity and surjectivity of the function, with some providing examples and counterexamples. There is a mix of interpretations regarding the definitions and proof strategies, and guidance has been offered to help clarify the original poster's understanding.

Contextual Notes

There is a question about whether the set N includes 0, which may affect the analysis of the function's properties.

SclayP
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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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