# I never learned what f(a, b) means. . .

1. May 10, 2010

### ripcity4545

1. The problem statement, all variables and given/known data

The problem is :

For the functions from N*N --> N, determine if the following functions are surjective:

f(a, b) = a + b

f(a, b) = ab(b+1)/2

2. Relevant equations

N is all natural numbers

3. The attempt at a solution

My problem is I know the definition of surjective, but I don't know what to do with f(a, b) because I've never seen this before. THanks for the help!

2. May 10, 2010

So, what's the definition of surjective? It's a good idea to start with it.

3. May 10, 2010

### ripcity4545

for a function A-->B, (A is the domain and B is the range), a funtion is surjective if for each b in B, there is at least one x in A such that f(x)=b.

4. May 10, 2010

Okay, so the codomain B of the function equals the image. What about f(a, b) = a + b ? Is there some element of N which can't be expressed in this form?

5. May 10, 2010

### ripcity4545

1? I'm still not clear what f(a, b) means.

6. May 10, 2010

### Tedjn

It is a shorthand convention for f((a,b)), for (a,b) in N×N. The x in your definition of surjective is an ordered pair in N×N.

7. May 10, 2010

Take some ordered pair of natural numbera (a, b). I assume you take 0 to be an element of the naturals. Let a = 0, so you have (0, b), where b can be any natural number. It is obvious that the image of the mapping f is N, since f(0, b) = 0 + b = b.

8. May 10, 2010

### ripcity4545

Sorry i forgot- 0 is not included in naturals in this case. but using your help, I got:

since the codomain is N, we can let f(a,b) = 1.
However, there are no a,b in N that satisfy a+b=1, so the function is not surjective.

Now I cannot find any counterexample of the second equation:

f(a, b) = ab(b+1)/2

but if this function is surjective, I would need to write a short proof, and I still am not sure on the concept.

9. May 10, 2010

### HallsofIvy

f(a,b) means exactly what you are told- it is "shorthand" for "a+ b" in the first formula and ab(b+1)/2 in the second.

Suppose N is a natural number (positive integer). Do there necessarily exist two natural numbers, a and b, such that a+ b= N? (What if N= 1?)

Suppose N is a natural number. Do there necessarily exist two natural numbers, a and b, such that ab(b+1)/2= N?

10. May 10, 2010

### ripcity4545

For N>0, I can't think of any N that cannot be defined by ab(b+1)/2. My problem is proving it. I tried induction but that leaves me with

P(k): ak(k+1)/2= N

and P(k+1): a(k+1)((k+1)+1)/2= N
= ak(k+1)/2 + k+1 =N.

Thanks for your help so far.

11. May 10, 2010

### HallsofIvy

Consider A= N, b= 1. What is ab(b+1)/2?