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Let f be a function such that f: N*N -> N such that:If (x,y) !=(0,0),then f(x,y) = gcd(x,y).if (x,y) = (0,0),then f(x,y) = 0.Is f surjective only,injective only,bijective,or none of these options?

Question

Let f be a function such that f: N*N -> N such that:If (x,y) !=(0,0),then f(x,y) = gcd(x,y).if (x,y) = (0,0),then f(x,y) = 0.Is f surjective only,injective only,bijective,or none of these options?

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Solution

The function f is surjective only.

Here's why:

  1. Surjectivity: A function is surjective (or onto) if for every element in the codomain, there is at least one element in the domain that maps to it. In this case, the codomain is the set of natural numbers N. For every natural number n, we can find a pair (n,n) in NN such that f(n,n) = gcd(n,n) = n. Therefore, every natural number has a preimage in NN, so the function is surjective.

  2. Injectivity: A function is injective (or one-to-one) if every element in the codomain is mapped to by exactly one element in the domain. In this case, the function f is not injective. For example, consider the pairs (2,4) and (3,6) in NN. We have f(2,4) = gcd(2,4) = 2 and f(3,6) = gcd(3,6) = 3. So, the same natural number 2 is the image of two different pairs in NN. Therefore, the function is not injective.

  3. Bijectivity: A function is bijective if it is both surjective and injective. Since the function f is not injective, it is not bijective.

So, the function f is surjective only.

This problem has been solved

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