Am I working this out correctly? (proof for injective function)

[bushra1175]

Hi everyone. This is the question:


This is how far I got:


I was wondering if the next step is to cross multiply p/q and u/v so that we end up with qu = pv? if so, am I correct in saying that this is not an injective function?

I say this as a pure guess even though I know what it means to have an injective function. I've only answered these types of questions when I'm provided with a set of values for the function but this is the first time I've had to make a proof without one. I appreciate all help with this.

 

[yoscar04]

Hi everyone. This is the question:
View attachment 20554

This is how far I got:
View attachment 20555

I was wondering if the next step is to cross multiply p/q and u/v so that we end up with qu = pv? if so, am I correct in saying that this is not an injective function?

I say this as a pure guess even though I know what it means to have an injective function. I've only answered these types of questions when I'm provided with a set of values for the function but this is the first time I've had to make a proof without one. I appreciate all help with this.

What do you think of the pairs (1,2) and (2,4)? Try to apply the definition injective function to f(p,q)=p/q. What do you get?

 

[Dr.Peterson]

Hi everyone. This is the question:
View attachment 20554

This is how far I got:
View attachment 20555

I was wondering if the next step is to cross multiply p/q and u/v so that we end up with qu = pv? if so, am I correct in saying that this is not an injective function?

I say this as a pure guess even though I know what it means to have an injective function. I've only answered these types of questions when I'm provided with a set of values for the function but this is the first time I've had to make a proof without one. I appreciate all help with this.

Let's start by fixing what you've written. You can't possibly mean it literally, as you are saying that somehow defining a function f causes all fractions to be equal!

I think you mean something more like this: Suppose that [MATH]p, q, r, s\in\mathbb{Q}[/MATH], such that [MATH]f(p,q) = f(r,s)[/MATH]. Then ...

Now, what would this imply if f were injective?

But yoscar04 is actually closer to the right path; think about what it takes to prove something is not true. Does the word counterexample ring a bell?

 

[Steven G]

How do you show that a function is 1-1.

You suppose that f(p,q) = f(r,s) and you conclude that (p,q) = (r,s). If you can't then the function is not 1-1.

Two fractions can be equal and not look the same. yoscar04 supplied you with one in 1/2=2/4. If you really knew the definition of 1-1 you should have seen this happening!

 

[pka]

It occurs to me that this maybe is misunderstanding about notation.
For ordered pairs it is the case that \(\large (1,2)\ne (2,4)\) . But \(\dfrac{1}{2}=\dfrac{2}{4}\).
So if \(F: R\times R\to R\) by \(F: (x,y)\mapsto \dfrac{x}{y}\) then \(F(1,2)=F(2,4)\).