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Is $\left\{(x, 4x^2 + x - 3) | x \in \mathbb{N}\right\}$ a subset of $\mathbb{N} \times \mathbb{Z}$ and why?

My guess is that it is a subset because there exist pairs only with natural numbers in the cartesian product like $(1,2), (3,3), (7,4)$.

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2 Answers 2

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Let $(x,y) \in \{(x, 4x^2 + x - 3) | x \in \mathbb{N}\}$, to want to show that $(x,y) \in \mathbb{N} \times \mathbb{Z}$. It suffices to prove that $y=4x^2+x-3$ is in $\mathbb{Z}$

since $x \in \mathbb{N}$, we know that $x \in \mathbb{Z}$

and we can conclude

$$y=4x^2+x-3 \in \mathbb{Z}$$ as the set of integer is closed under multiplication and addition.

Hence, $\left\{(x, 4x^2 + x - 3) | x \in \mathbb{N}\right\}$ is a subset of $\mathbb{N} \times \mathbb{Z}.$

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If your definition of $\mathbb N$ does not include $0$, then $\mathbb N \times \mathbb N$ holds. Because starting with $x=1$, we have $4x^2 +x - 3 = 2\in \mathbb N$, and for any natural number $x \geq 2$, $y\geq 15$.

While we can surely write $$\left\{(x, 4x^2 + x - 3) | x \in \mathbb{N}\right\} \subset \mathbb N \times \mathbb N,$$ since $\mathbb N \subset \mathbb Z$, it is also true that $$\left\{(x, 4x^2 + x - 3) \mid x \in \mathbb{N}\right\} \subset \mathbb N \times \mathbb Z $$


On the other hand, your working with the natural numbers whose minimum element is $0$, then we need to include $-3$ in the codomain, so our codomain must therefor be $\mathbb Z$. Then we must have $$\{\left\{(x, 4x^2 + x - 3) | x \in \mathbb{N}\right\}\subset \mathbb N \times \mathbb Z$$

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