# Create an injection from {a^2+b^2

1. Oct 1, 2013

### aaaa202

1. The problem statement, all variables and given/known data
Is it possible to create an injection from {a^2+b^2 l a,b$\in$Q} to {{a+b l a,b$\in$Q}

2. Relevant equations
I am not sure about this. The problem is that many combinations of a,b can yield the same number, so how do I tackle this obstacle?

3. The attempt at a solution

2. Oct 1, 2013

### UltrafastPED

Consider the definition of injection.

3. Oct 1, 2013

### aaaa202

the definition says:

x1≠x2 => f(x1)≠f(x2)

But the problem is as said that many combinations give rise to the same numbers in the set.

4. Oct 1, 2013

### UltrafastPED

Then perhaps it is not possible!

5. Oct 1, 2013

### aaaa202

it certainly should be, since the two sets have same cardinality. I just need help with coming up with a well defined function that does it. Problem is you can't really say something like f(a^2+b^2)=lal+lbl because of the earlier stated.

6. Oct 1, 2013

### MrAnchovy

Is this necessarily true?

7. Oct 1, 2013

### jbunniii

It might simplify the problem somewhat to observe that $\{a+b \, | \, a,b\in \mathbb{Q}\}$ is simply $\mathbb{Q}$.

8. Oct 1, 2013

### MrAnchovy

I've just looked at a couple of your other posts like this one where again you seem to be confused about what you mean by {a^2+b^2 l a,b∈Q}. Can you see that all elements of this set are in $\mathbb{Q}$, and that a and b are only relevant in selecting which elements of $\mathbb{Q}$ are elements of this set? You can't then define a function f(a,b) and expect it to have any meaning in relation to this set without defining how you get the unique pair (a,b) from $x \in \mathbb{Q}$.

Before you try to think about jections (I hate that word) you need to think more clearly about the membership of sets and the domain, co-domain and range of relations between them.

Last edited: Oct 1, 2013