# Find the image: F: N * N -> R , F(x) = m^2 + 2n

#### KOO

##### New member
F:N*N -> R, F(x) = m^2 + 2n

I think the answer is N. Am I right?

#### Chris L T521

##### Well-known member
Staff member
I think the answer is N. Am I right?
Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$).

Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$.

If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.

(In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.)

I hope this makes sense!

#### KOO

##### New member
Based on your title, we have that $F:\mathbb{N}\times\mathbb{N}\rightarrow \mathbb{R}$ ($\mathbb{N}\ast\mathbb{N}$ makes no sense at all) where $F(x) = m^2+2n$ (I assume here that $x=(m,n)$).

Now, depending on who you talk to, $\mathbb{N}$ may or may not include zero (the general consensus from what I've seen is that $0\notin\mathbb{N}$, but there are some professors/authors that include zero in their definition of the natural numbers; hence why I think it's best to clarify this right from the get go); that is, either $\mathbb{N}=\{x\in\mathbb{Z} : x\geq 0\}$ or $\mathbb{N}=\{x\in\mathbb{Z}: x\geq 1\}$.

If $0\in\mathbb{N}$, then $\mathrm{Im}(F)=\mathbb{N}$. If $0\notin\mathbb{N}$, then $\mathrm{Im}(F) = \mathbb{N}\backslash\{1,2,4\}$ since there aren't pairs $(m,n)\in\mathbb{N}\times\mathbb{N}$ such that $m^2+2n=1$, $m^2+2n=2$, or $m^2+2n=4$.

(In $\mathrm{Im}(F)$, note that all the positive odd numbers greater than or equal to 3 are generated by pairs of the form $(1,n)$ for $n\in\mathbb{N}$ since $1^2+2n=2n+1$, and all positive even numbers greater than or equal to 6 are generated by pairs of the form $(2,m)$ for $m\in\mathbb{N}$ since $2^2+2m = 2(m+2)$; this is why I can claim that $F(\mathbb{N}\times\mathbb{N}) = \mathbb{N}\backslash\{1,2,4\}$ for $0\notin\mathbb{N}$.)

I hope this makes sense!
Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.

Anyways what does \ mean?

#### Chris L T521

##### Well-known member
Staff member
Actually I had this question on a test and you're right x=(m,n) and we're told N does not include 0.

Anyways what does \ mean?
Ah, that's one notation for set difference. I could have also written it as $\mathbb{N}-\{1,2,4\}$.