Question

Determine whether f : Z x Z --> Z is onto if

f (m, n) = m + n + 1.

Two parts: first, prove or disprove the function is
one-to-one AND second, prove or disprove the function is onto.

Answer 1

Hi Haichi

Answer 2

Injective
f(a,b)=f(c,b)
see if you have a=c and b=d
If you do then it is injective.

Answer 3

Hello, sorry I was away from the computer

Answer 4

how do I prove if it is one to one function?

Answer 5

basically we want to suppose f(a,b)=f(c,d)
and if we show (a,b)=(c,d) then we have shown it is injection .
Or we can try to think of an example that shows it isn't injective.
Can you think of (a,b) not equal to (c,d)
such that f(a,b) equal f(c,d)

Answer 6

f (m, n) = m + n + 1? how do I use (a,b) = (c,d) equation?

Answer 7

f(a,b)=f(c,d)
a+b+1=c+d+1
a+b=c+d

Does this imply a=c and b=d?
Well it could actually imply the other way around couldn't it?
a=d and b=c
So maybe we can try to think of a counterexample:
think of choosing one where a=d and b=c

Answer 8

(a,b) is what we are plugging in
(c,d) is what we are plugging in

Answer 9

oh ok so any integer will match?

Answer 10

I'm saying you can provide a counterexample to the injective part

Answer 11

And I gave you a hint as to what to pick

Answer 12

that will only work if I pick one integer for all a,b,c,d?

Answer 13

I will copy what I pasted above:
"So maybe we can try to think of a counterexample:
think of choosing one where a=d and b=c"

Answer 14

I am sorry. I dont know what counterexample to use for this question

Answer 15

I'm asking you to think of one where a=d and b=c

Answer 16

to prove that it is incorrect?

Answer 17

We are proving this it is not injective.

Answer 18

Because we can show that (a,b) not equal to (c,d) implies f(a,b)=f(c,d) for some values a,b,c, and d.

Answer 19

so if i want a=d and b=c
I will randomly select two points such that that happens
(4,3) and (3,4)

Answer 20

now what is f(4,3) and what is f(3,4)?

Answer 21

right so it is not injective?

Answer 22

im sorry im not sure what you're asking

Answer 23

remember that f(m,n)=m+n+1?

Answer 24

ok so that is not one to one function?

Answer 25

Right because we were about to find (a,b) not equal (c,d) such that f(a,b)=f(c,d)

Answer 26

(3,4) doesn't equal (4,3)
but f(3,4)=f(4,3)=8

Answer 27

That is called a counterexample

Answer 28

ok what about onto function?

Answer 29

You want to suppose b is an element of the codomain which happens to be the set of integers in this case.
You want to prove there exists (m,n) in the ZxZ for which f(m,n)=b

Answer 30

f(m,n)=m+n+1=b
We want to chose an m and n such that does =b

Answer 31

I will let you choose m and n
such that when you simplify m+n+1 you get b

Answer 32

b would be in co domain? does it have to match domain numbers?

Answer 33

what does that mean exactly?

Answer 34

if (m,n) is in domain and m + n +1 is in codomain, these wont match?

Answer 35

sorry im just confused lol

Answer 36

I kinda don't understand what you say saying exactly
I'm just asking you to choose an m and n such that
when simplify m+n+1 you get b

Answer 37

if I chose (3,2) and b would be 6?

Answer 38

well the problem with that is you aren't showing you get all the integers

Answer 39

that would be 5 = 6 in form of (m,n) = m + n + 1?

Answer 40

m+n+1=b
solving for either m or n
m=b-n-1
So I will choose m=b-n-1 and n=n
so f(b-n-1,n)=(b-n-1)+n+1=b
so this shows that every element of the codomain does get hit
Therefore the function is surjective

Answer 41

f(b-n-1,n)=(b-n-1)+n+1=b ?

Answer 42

-n+n=0
-1+1=0
only thing left is b

Answer 43

how do you know the difference between injective and surjective?

Answer 44

I understand them by the diagrams but not in the equations

Answer 45

f:A->B
injective:
For every x,y in A..if x does not equal y then f(x) does not equal f(y)
--as you seen we found a counterexample for this above. we found x does not equal y such that we had f(x) equals f(y)

surjective:
If for every b in B there is a in A with f(a)=b
---This is an exisential proof. All we have to is find an element a in the domain such that we get f(a)=b where b represents every element of the codomain B.
m and n were both integers
b was an integer
choosing m to b-n-1 was fine because b-n-1 is also an integer for any integer b
b is an integer
the set of integers is closed under addition/subtraction

Answer 46

Ok, i think I get it. Thank you for your help!

Answer 47

So in others words for this problem that wasn't the only way to choose m and n.

Answer 48

we could have chose to left m as is and to select n as b-m-1

Answer 49

of we could have even chose m as 0 and n as b-1

Answer 50

or chose m as b-1 and n as 0

Answer 51

We just needed to show we get every element of the codomain

Answer 52

for the surjective part

Answer 53

Ok I think I understand it now. Thank you!