# Prove That The Function Is Onto

• Aug 14th 2012, 03:07 PM
ravenkaw
Prove That The Function Is Onto
I have already done the proof, but I just want to make sure it is correct!

f: ZxZ -> Z f(m,n)=m+n

Prove that f is onto using the definition of onto.

Proof
Suppose that f(m,n)=m+n.
Let a∈Z.
Then, f(m,n)=a iff m+n=a iff m=a-n which is in Z.
Therefore, f(m+n)=m+n is onto.

Is this correct? Or proper?

Thanks.
• Aug 14th 2012, 03:21 PM
Plato
Re: Prove That The Function Is Onto
Quote:

Originally Posted by ravenkaw
I have already done the proof, but I just want to make sure it is correct!
f: ZxZ -> Z f(m,n)=m+n
Prove that f is onto using the definition of onto.
[/COLOR]

Suppose that $\displaystyle n\in\mathbb{Z}$ what is $\displaystyle f(n,0)~?$
• Aug 14th 2012, 03:57 PM
ravenkaw
Re: Prove That The Function Is Onto
Would that not be f(n,0)=n+0=n? And we already know n∈Z.
I'm not sure where you're going with that or how 0 can be arbitrarily assigned...
• Aug 14th 2012, 04:05 PM
Plato
Re: Prove That The Function Is Onto
Quote:

Originally Posted by ravenkaw
Would that not be f(n,0)=n+0=n? And we already know n∈Z.
I'm not sure where you're going with that or how 0 can be arbitrarily assigned...

I showed that $\displaystyle f : (m,n) \mapsto m+n$ is onto!
You see $\displaystyle \forall n\in\mathbb{Z}$ we know that $\displaystyle (n,0)\in\mathbb{Z}\times\mathbb{Z}$
• Aug 14th 2012, 04:17 PM
ravenkaw
Re: Prove That The Function Is Onto
Oh, okay! So, we can arbitrarily set the original n to 0 in order to prove that (n,0)∈ZxZ?
• Aug 14th 2012, 04:57 PM
Plato
Re: Prove That The Function Is Onto
Quote:

Originally Posted by ravenkaw
Oh, okay! So, we can arbitrarily set the original n to 0 in order to prove that (n,0)∈ZxZ?

It is a bit more than that.
Given any $\displaystyle k\in\mathbb{Z}$ then $\displaystyle f(k,0)=k$ where clearly $\displaystyle (k,0)\in\mathbb{Z}\times\mathbb{Z}.$
• Aug 14th 2012, 08:54 PM
Deveno
Re: Prove That The Function Is Onto
you can think of a function as something which starts at a source "a", and "hits a target b":

f:a→b

the only restriction is that each source can only pick "one target".

normally, the unique target of a, is called f(a), so we can write: f:a→f(a).

a function is onto if: "every target gets hit". what that means is: given any target b, we have to find at least one source a with f:a→b, that is at least one a with f(a) = b, for every b.

in YOUR function, the targets live in the set of integers. so to prove that f is onto, we need to find a pair (ANY pair) that adds to a given integer k, and we have to do this for EACH integer k.

well, one such pair is (k,0)...another is (0,k). for both of these pairs, it is straightforward to verify that:

f(k,0) = k+0 = k
f(0,k) = 0+k = k.

but, pick your favorite integer m (it doesn't matter which one you pick, just pick one, and stick with it).

then the pair (m,k-m) will work just as well:

f(m,k-m) = m+(k-m) = (m+k)-m = (k+m)-m = k+(m-m) = k+0 = k. the thing to be clear about, is that while we are doing this for each k, it is conceptually clearer, if we use the same "m", over and over. Plato chose m = 0, from which i deduce that 0 is his favorite integer.