Question

Determine whether or not the function f:ZxZ->Z is onto, if f((m,n))=mn. Determine whether or not the function f:ZxZ->Z is onto, if f((m,n))=mn. @Mathematics

Answer 1

And what does ZxZ->Z mean?

Answer 2

a function from the 2-dim lattice of integers to the integers.

Answer 3

Z = the integers = {0, +1, -1, +2, -2, +3, -3, ....}

Answer 4

and ZxZ is the cartesian product of Z with itself, which you can think of as a lattice of points on the usual 2-dimensional plane.

Answer 5

Now, if you understand that and you know the definition of onto, it should be easy for you to determine whether this function f is onto or not.

Answer 6

so you're saying it's a function that can point to more than one thing?

Answer 7

What is the formal definition of this statement
a function f: X -> Y is ONTO if ....what?

Answer 8

I actually don't know haven't learnt onto yet

Answer 9

I'll look it up give me a sec

Answer 10

It means that for every \( y \in Y \) there exists an \( x \in X \) such that \(f(x) = y\).

On the real numbers, the function f(x) = x is onto (and also 1:1). The function f(x) = x^3 is also onto. The function f(x) = x^3 - x is also onto, but not 1:1.

The function f(x) = sin(x) is not onto because for y = 2 for instance, there is no x such that
sin(x) = 2.

Answer 11

The function \( f : R^3 \rightarrow R^2 \) given by
\[ f(x,y,z) = (x,y) \]
is also onto but not 1:1

Answer 12

Oh I see I still don't get what zxz->z means so all integers including 0?

Answer 13

\( \mathbb{Z} \) is the standard symbol in mathematics for the integers, including negatives and zero.

Answer 14

okay but why does it do zxz it's saying it's 2 dimensional?

Answer 15

Is it onto because for every number if we make m=k and n=1 we'll get k?

Answer 16

\( A \times B \) is the Cartesian product of sets,
\[ A \times B = \{ (a,b) \ | \ a \in A, b \in B \} \]
This is why we write \( \mathbb{R}^2 \) for the two-dimensional Cartesian plane:
\[ \mathbb{R}^2 = \mathbb{R} \times \mathbb{R} \]
Similarly,
\[ \mathbb{R}^3 = \mathbb{R} \times \mathbb{R} \times \mathbb{R} \]

Answer 17

Thus
\[ \mathbb{Z} \times \mathbb{Z} = \{ (p,q) \ | \ p, q \in \mathbb{Z} \} \]

Answer 18

Now,
"Is it onto because for every number if we make m=k and n=1 we'll get k?"
yes.

Just formalize that argument so it makes exactly the definition.

So the argument starts like this.

"Given an arbitrary integer \( p \in Z \) let (m,n) = ( ...), then f(m,n) = p. Hence, by definition, f is onto Z"

Answer 19

Thank you!! That makes a lot more sense!