# Is the function N×NN×N to NN surjective given by (x,y)â†¦x(x,y) â†¦ x?

Is the function Nأ—NN أ— N to NN surjective given by (x,y) â†¦ x(x,y) â†¦ x?

And if it is, why?

I know that a function f:S->T is surjective iff for all t that belong to T there exists s that belongs to S with t=fs.

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1

Do you know what it means for a function to be surjective?
– k.stm
2 days ago

1

It means that every element of the target set has to be mapped by at least one element of the source set
– JhonDoe
2 days ago

Let x âˆˆ Nx âˆˆ N be of the target set. Can you think of an element of the source set that maps to xx?
– k.stm
2 days ago

the first element of (x,y), that is x ???
– JhonDoe
2 days ago

Your answer is unclear to me. What is yy? What exactly is your guess for an element of N × NN × N that maps to xx?
– k.stm
2 days ago

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1

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Recall that a function is surjective if every element of the target set is in the range of the function. Your function takes pairs (x,y)(x,y) of elements from NN and maps the pair to the first coordinate. So the question is: In the set of all pairs of elements from NN, does every element of NN appear as the first component of a pair?

I think yes. Isn’t it?
– JhonDoe
2 days ago

1

Yes. If you can explain why, you have your answer. A good way to explain it would be to, given an element x \in Nx \in N, supply a specific pair in N \times NN \times N that is mapped to it.
– Reese
2 days ago

Can I chose like x=j and y=k where k,j are natural numbers, and write:
– JhonDoe
2 days ago

(j,k) -> j ? is this correct to prove it?
– JhonDoe
2 days ago

Sure, that’ll work. It’d be simpler to choose one particular option for yy, but that’s fine. But when you’re writing it up, be precise about what the various letters mean.
– Reese
2 days ago