# Injevtive/surjective Problem - Algebra

1. Jul 19, 2013

### SclayP

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

2. Jul 19, 2013

### hilbert2

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.

3. Jul 19, 2013

### SclayP

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: $\forall$yB,$\exists$xA/y=f(x)

4. Jul 19, 2013

### Number Nine

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 $n+m$?

Please don't do all of the work for the OP, especially when the question is clearly homework.

5. Jul 21, 2013

### Jorriss

To clarify, are you defining N to include 0?