Transforming a formal set into a informal description.

[Dima_11]

Hi,

I have been stuck on this problem which I am sure is easy, but I am just not seeing what the informal description of the following set may be:
S = {...,-16, -8, -4, -2, 0, 2, 4, 8, 16,...}.
At first I thought it would be: "The set of all even integers", but then I realized I don't see the 6 in there, so it can't be all. Any assistance will be appreciated.

 

[Deleted member 4993]

Hi,

I have been stuck on this problem which I am sure is easy, but I am just not seeing what the informal description of the following set may be:
S = {...,-16, -8, -4, -2, 0, 2, 4, 8, 16,...}.
At first I thought it would be: "The set of all even integers", but then I realized I don't see the 6 in there, so it can't be all. Any assistance will be appreciated.

You were almost on the right track - work with '2' a bit more......

 

[Dima_11]

The other way I see it, is "the set of even integers multiplied by 2" or power of 2? If it is power of 2, what about the zero?

 

[pka]

I have been stuck on this problem which I am sure is easy, but I am just not seeing what the informal description of the following set may be:
S = {...,-16, -8, -4, -2, 0, 2, 4, 8, 16,...}.

[imath]\mathcal{S}=\{0\}\cup\left\{\pm 2^n\right\}[/imath] where [imath]n\in\mathbb{N}^+[/imath]
[imath][/imath][imath][/imath]

 

[Dima_11]

[imath]\mathcal{S}=\{0\}\cup\left\{\pm 2^n\right\}[/imath] where [imath]n\in\mathbb{N}^+[/imath]
[imath][/imath][imath][/imath]

Would the informal descriptions be: "The set containing a zero or power of 2 where n is all positive natural numbers"?

 

[pka]

Would the informal descriptions be: "The set containing a zero or power of 2 where n is all positive natural numbers"?

Well your aet [imath]\mathcal{S}[/imath] contains both [imath]16~\&~\large -16[/imath]!
Thus [imath]\mathcal{S}=\{0\}\cup\{\pm 2^n|n\in\mathbb{N}^+\}[/imath]

 

[Dima_11]

Well your aet [imath]\mathcal{S}[/imath] contains both [imath]16~\&~\large -16[/imath]!
Thus [imath]\mathcal{S}=\{0\}\cup\{\pm 2^n|n\in\mathbb{N}^+\}[/imath]

Oh yeah, I see. Thank you!