Discussion between gurinderhans and Jonas Gomes - 2014-11-01

last day (14 days later) »

3:07 AM

1

Q: is f: Z x Z → Z onto, where f(m,n)=m^2+n

I have a good understanding of onto functions, one-to-one functions. However the problem here as I am having is the how would one graph this function. For ex. if f(x)=x then we know its onto since for every pre-image there is an image. But this function I cannot seem to create a visual of how thi...

functions discrete-mathematics

What happens when $m=0$?

I'm sorry but I do not know what you mean

Your function associates a pair of integers to another integer. So, $(1,2)$ is associated with $3$, for example. I said that $(0,n)$ is associated with $n$ and therefore $f$ is onto.

@JonasGomes I understand all of what you have said. Just a last quick question. Why do we only pick (0,n)? Why not (m,0)?

If we'd pick $(m,0)$ the function would be $m^2$, which is not onto as you probably knows.

3:07 AM

Yes that's true and you do get onto but why pick (0,n)? Isn't a function onto if it is onto for anything?

Hi! So, you want to prove that the function is onto. All you have to do is pick someone in the codomain and prove it is the image of $f$ under some element of the domain

So just one set of pairs for this function?

But then what if I just picked two integers (1,1)?

Let us think about something. If I ask you to find m n such that f(m,n) = 18, can you find them?

yes

And what would they be?

3:12 AM

m = 3 and n = 9

Good, try to give me all such pairs.

(You'll understand soon)

(0,18) , (1,17), (2,14), (3,9), (4,3)

Still not sure where you're trying to take me

last one should be (4,2)

Ok, very good.

Now, if I ask you to give me a pair for 810, it'll be probably easier to just pick (0,810)

Yes

and can garantee that f(0,810) = 810, or, rephrasing, 810 is an element of the image of f.

3:20 AM

I do understand that (0,n) creates a linear graph of the image values which is onto.

For every element n of Z, you can say that n = f(0,n) and then n is an element of the image.

But not sure why just (0,n)

It's not just (0,n), as our example shows

but its enough to say it's onto.

Hmmm, not getting it yet

For a function to be onto

every element of its codomain

needs to be in its image

3:23 AM

yes

but the case (m,0)?

So, given a n in Z, you proved that f is in the image, because n = f(0,n)

*you proved that n is in the image of f

And therefore f is onto.

You don't need to find all the pairs which go to n, one is enough

I see

so for any single pair it is enough?

The pairs are in the domain

First, you look at the codomain, gets an arbitrary elements and then you go to the domain and find an element (a pair in this case)

which goes to the element you picked in the codomain

If you can always do that, then f is onto

So, you should probably say "for any n it is enough"

I see. For any number in the codomain we can find an element in domain which like you previously said would simply be (0,n)

I think I understand

Thank you so much!!

You are welcome :)

last day (14 days later) »

Transcript for

Nov '141

$Mathematics$ Discussion between gurinderhans and J…

Imported from a comment discussion on math.stackexchange.com/q...

discrete-mathematics functions

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile