1. ## Bijection and Functions

Hi, I need some help with understanding bijective functions.

Is the function f(a,b) = (a+1,b) for all (a,b) bijective? I think not because x is not equal to y. Can some please help me with this?

Also similar to this,

f(a,b) = a-1 for all (a,b). Is this function onto or bijective?

Thnaks alot.

2. Originally Posted by kurac
Hi, I need some help with understanding bijective functions.

Is the function f(a,b) = (a+1,b) for all (a,b) bijective? I think not because x is not equal to y. Can some please help me with this?

Also similar to this,

f(a,b) = a-1 for all (a,b). Is this function onto or bijective?

Thnaks alot.
please state the domain and ranges for the functions you have. where are a and b from, where are x and y from?

a function is bijective if it is both one to one and onto, so just check those properties. without more info, can't help you beyond that.

Well, lets just consider Z^2, the set of all pairs (a,b), where a, b elements (E) of Z. (Z is domain, collection of all integers)

Hope that is clearer, its all i have been supplied with in my textbook.

4. ignore x and y, i meant to refer to a and b. I dont want to confuse anyone or myself know, im confused enough

5. ## Bijection

Hello kurac
Originally Posted by kurac
Is the function f(a,b) = (a+1,b) for all (a,b) bijective?
Yes, if the domain is the whole of $\mathbb{Z}^2$. You can think of $\mathbb{Z}^2$ as a lattice of points in the $(x, y)$ plane with integer coordinates. The function $f(a, b) = (a+1, b)$, then, maps a point onto its immediate neighbour 1 unit to the right. So it's both one-to-one and onto.

f(a,b) = a-1 for all (a,b). Is this function onto or bijective?
Presumably, this function is from $\mathbb{Z}^2$ to $\mathbb{Z}$?

If so, it will certainly be onto, because $\forall n \in \mathbb{Z}, \exists a \in \mathbb{Z}, a-1=n$.

But it's not one-to-one, because for a given value of $a$, there are infinitely many possible values of $b$ for which $f(a,b) = a-1$. So it is not bijective.