Question

Determine which of the function F:ZxZ -> Z is not onto:

A) F (m,n) = 2m - n
B) F (m,n) = m^2 - n^2
C) F (m,n) = m + n + 1
D) F (m,n) = |m| - |n|

Answer 1

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Answer 2

Thanks, hope you can help me in my question :D

Answer 3

They all look onto to me. I don't see any function that cannot produce a specific integer (assuming any m,n values can be used). Then again I could be wrong and maybe B could have some gaps in it?

Answer 4

oh wait

m^2 - n^2 = (m-n)(m+n)

so that means that if we want F(m,n) to be some prime number P, then

m-n = 1
m+n = P

or

m-n = P
m+n = 1

If P is not equal to 2, then add up the equations

2m = 1+P

m = (1+P)/2

so m is definitely an integer since P is odd, 1+P is even. That forces n to be an integer. So we could get any prime number we want. I also think we could get any other integer as well.

Answer 5

also, I've graphed each equation and I'm noticing that it is possible to hit every single number in the range of integers Z

Answer 6

@jim_thompson5910, thanks for your reply but I am really stuck at Discrete Mathematics, What's the final answer?
is it B) F (m,n) = m^2 - n^2?

Answer 7

I don't know honestly. That seems like the best choice, but I'm not sure which integers are left out.

Answer 8

It's ok, thanks for your reply and effort.

Answer 9

Let k=-n^2-1. k is integer since n is an integer. Can we show there is a pair of integers m and n such that f(m,n)=-n^2-1
so we have m^2-n^2=-n^2-1

Answer 10

Try solving for m^2 and see what happens

Answer 11

All I did was think about the graph f(m)=m^2-4
This is a parabola with y-intercept -4 and it is open upwards.
There are no integers below -4 that get hit.
So f(m)=m^2-4 is not surjective over the integers.
So I thought of f(m,n)=m^2-n^2.
So I just thought well -n^2-1 will not get hit and -n^2-k where k>0 will also not get hit
but we just need one example to show it is not surjective

Answer 12

I found this page

http://www.cse.msu.edu/~luciwmat/hw5.pdf

Look on page 4. On that page, it says f(m,n) = m^2 - n^2 is not onto since 2 is not in the range. Basically there are no solutions to m^2 - n^2 = 2 where m,n are integers

Answer 13

that works too

Answer 14

and I wasn't thinking earlier

let's say we want f(m,n) = 2 and we want to find the m,n values that are the solution

f(m,n) = 2

m^2 - n^2 = 2

(m-n)(m+n) = 2*1

m-n and m+n are integers since m,n are integers. So we equate factors to get

m-n = 2
m+n = 1


Add up the equations: 2m+0n = 3 ---> m = 3/2

but m is an integer, so that's a contradiction which means there are no solutions for f(m,n) = 2 where m,n are integers

Answer 15

@jim_thompson5910 are you sure that there is no solution? as I can only choose one of the four choices, it's university homework :/

Answer 16

you're misreading

Answer 17

the fact that f(m,n) = 2 has no solutions when m,n are integers means that f(m,n) = m^2 - n^2 is not onto

Answer 18

f(m,n) = m^2 - n^2 is onto only if you are able to generate ANY integer

this is because you're going from ZxZ to Z

if you made restrictions on Z, then perhaps it could be onto

Answer 19

also, look at that link I posted

Answer 20

Sorry for misreading and thanks alot for your effort!