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

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    Injection
  • #1
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Homework Statement


Is it possible to create an injection from {a^2+b^2 l a,b[itex]\in[/itex]Q} to {{a+b l a,b[itex]\in[/itex]Q}

Homework 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?


The Attempt at a Solution

 

Answers and Replies

  • #2
Consider the definition of injection.
 
  • #3
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
Then perhaps it is not possible!
 
  • #5
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
it certainly should be, since the two sets have same cardinality.
Is this necessarily true?

I just need help with coming up with a well defined function that does it.
How about x → x?
 
  • #7
It might simplify the problem somewhat to observe that ##\{a+b \, | \, a,b\in \mathbb{Q}\}## is simply ##\mathbb{Q}##.
 
  • #8
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.
 
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