Functions

skipjack

Forum Staff
Dec 2006
21,479
2,470
You needn't use "$\mathbb N^{*}$", as $\mathbb N$ already means the integers greater than zero.

The original wording was "ascending function", not "strictly ascending function". A constant function is ascending, but not strictly ascending, so it's not excluded by the original wording.

Hence $f(n) = 4\ \forall n \in \mathbb N^{*}$ satisfies all the originally stated conditions.
 
Aug 2018
137
7
România
You needn't use "$\mathbb N^{*}$", as $\mathbb N$ already means the integers greater than zero.

The original wording was "ascending function", not "strictly ascending function". A constant function is ascending, but not strictly ascending, so it's not excluded by the original wording.

Hence $f(n) = 4\ \forall n \in \mathbb N^{*}$ satisfies all the originally stated conditions.
Hello,

I do not understand! I think the function \(\displaystyle f(n)=C\) where \(\displaystyle C\) is a constant is neither ascending and neither descending.
Why can not we consider that the function \(\displaystyle f(n)=C\), where \(\displaystyle C\) is a constant, is descending function?
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Really, if we consider \(\displaystyle f(n)=C\) where \(\displaystyle C\) is a constant, then from the imposed condition it follows that \(\displaystyle C=4\), but first to prove that the function \(\displaystyle f(n)=C\), where \(\displaystyle C\) is a constant, is an ascending function and is not an descending function. Thank you very much!

All the best,

Integrator
 
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skipjack

Forum Staff
Dec 2006
21,479
2,470
The function \(\displaystyle f(n)=C\), where \(\displaystyle C\) is a constant, is both an ascending function and a descending function. The usual terms are "increasing" and "decreasing", rather than "ascending" and "descending".
 
Aug 2018
137
7
România
The function \(\displaystyle f(n)=C\), where \(\displaystyle C\) is a constant, is both an ascending function and a descending function. The usual terms are "increasing" and "decreasing", rather than "ascending" and "descending".
Hello,

I do not understand how the constant function can be increasing and decreasing at the same time. How can we fully solve the proposed problem? Thank you very much!

All the best,

Integrator
 
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skipjack

Forum Staff
Dec 2006
21,479
2,470
This article explains how "increasing" and "decreasing" are used. It also explains how adding the word "strictly" changes the meaning.
 
Aug 2018
137
7
România
This article explains how "increasing" and "decreasing" are used. It also explains how adding the word "strictly" changes the meaning.
Hello,

OK! In conclusion, how do we find all the functions that meet the condition in the problem? Let I understand that \(\displaystyle f(n)=4\) and \(\displaystyle f(n)=n\) are the only functions that meet the condition in the problem? If so, then how do we prove this? Thank you very much!

All the best,

Integrator
 
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