Question

Let f : Z × Z → Z × Z be defined as f(m, n) = (3m + 7n, 2m + 5n). Is f a bijection, i.e.,one- to-one and onto? If yes then give a formal proof, based on the definitions of one-to-oneand onto, and derive a formula for f−1. If no then explain why not.

I did:

assume f(m1,n1) = f(m2,n2)
and I just equated both side of the equation making a sys. of equations and finding that m1 = m2 and n1=n2 making it one to one.

However I can't prove how it is onto, any help?

Answer 1

It seems more complicating cuz its a function that is dependent on 2 variables m and n
Lemme just think how we would go abt this

Answer 2

I know I can replace a and b in the equation, but seems kinda hard to evaluate

Answer 3

Ok i think i got smth

Answer 4

I am getting thrown off cuz its a multivariate function ...

Let set up our function as follows:
3m+7n=a
2m+5n=b

$$\frac{a-3m}{7}=n$$
So $$2m+5n=b \to 2m+5\left( \frac{a-3m}{7} \right )=b$$

$$2m+\frac{5}{7}a-\frac{3}{7}m=b$$
$${11}{7}m=b-\frac{5}{7}a$$
$$m=\frac{7b-5a}{11}$$

Answer 5

but, how does proving it in a system of equation prove that it is onto

Answer 6

Hey gimme a sec ... lemme think abt this

Answer 7

HI!@!

Answer 8

Ok well the way im thinking abt it is as follows:
We basically have 2 equations for each coordinate f(m,n)=(a,b)
Now f(m,n)=3m+7n=a for the x-coordinate
and f(m,n)=2m+5n=b for the y-coordinate

Now obv we can rewrite one of the equations in terms of n
Once we have one of the equations in terms of n we can replace the n in the other equation since thats what n equals
And then we can solve the newly revised equation in terms of m

Answer 9

pick an arbitrary element of \(\mathbb{Z}\times \mathbb{Z}\) say \((p,q)\) and show that there is something that gets mapped to it
it is the same as finding the inverse exactly

Answer 10

Ya but its difficult finding the inverse when its a multivariate function

Answer 11

Therefore the way to go about this is setting the coordinates as a and b
And then use one equation to eliminate one variable

Answer 12

Once we have done that then its easy to find its inverse

Answer 13

Wait, so once you eliminated m ( or n) what would you do afterwards? plug it back in in the function to find its inverse?

Answer 14

Reread My original post ...
I first solved for n in my first equation
Then I plugged that in to my second equation and eliminated the n
And then I found my inverse
And my inverse's domain was \(Y \in R\)

Answer 15

hence it is onto

Answer 16

completely sorry, my browser is screwing up some part of openstudy. I just saw, but your inverse is in terms of a,b so it is for some arbitrary value. Does that change anything? I know proving there is an inverse ( so finding a function in term of m,n) would prove its bijection.

Answer 17

ya thats no problem
We are choosing an arbitrary term for f(m,n)
$$(a,b) \in f(m,n)$$

Answer 18

Basically (a,b) is an element of the range

Answer 19

ah so a,b is a range for which there exist a function to describe m. but if you can prove it for a smaller set, It also holds true for the whole domain?

Answer 20

hmmmm
I wldnt say a,b is the range but rather (a,b) is an arbitrary point in the range or in other words f(m,n)
So basically are dependent values are m and n
We give m and n and then according to what we set m and n as that's what our range , a and b , will be
So basically view f(m,n) as our "y"
Now when we are proving that a function is onto we are proving that every y in our codomain has an x-value associated
So we find our inverse and show that if we plug in ANY SINGLE VALUE for a or b - which is our range - we will definitely find our m and n

For example \(y=x^2-1\) is not onto
So lets find our inverse which is :
$$x=\sqrt{y+1}$$
Now for y<1 this function does not exist and therefore the function is not onto

So we can basically bring the same principle to this multivariate example

Answer 21

thanks a lot , you were super helpful!!!