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

[KOO]

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

I think the answer is N. Am I right?

 

[Chris L T521]

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]

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]

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 $\mathbb{N}-\{1,2,4\}$.

 

[MarkFL]

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

I think the answer is N. Am I right?

I just wanted to let you know that I moved this topic from Calculus to Pre-Calculus (this is a better fit) and copied the problem from the title into the body of the first post. It's okay to put the problem in the title when it is short, but we ask that it also be included in the post as well for clarity. :D