About a function over natural numbers which does not exist.

eraldcoil · **Posted:** Mon, 28 Nov 2011 05:33:53 UTC

Show that is not there a function $f:\mathbb {N}\rightarrow \mathbb {N}$ such that $f^2(n)=n^+ +3, \ (n^+=n+1)$

Proof: (For

)

It easy view:

f(n+1)=f(f(fn)))=f(n)+

1 for all

.

By induction it is proved that for all

,

For

It is trivial, If this is true for n-1

then:

n+1=f(f(n))=f(n)-1+f(1)=n-1+f(1)-1+f(1)=n-2+2f(1)

, $\therefore f(1)=3/2$ contradiction.

As would be for f^2(n)=n^+ +3

?

Thanks in advance.

Shadow · **Posted:** Mon, 28 Nov 2011 05:38:15 UTC

eraldcoil wrote:

Show that is not there a function $f:\mathbb {N}\rightarrow \mathbb {N}$ such that $f^2(n)=n^+ +3, \ (n^+=n+1)$

Proof: (For

)

It easy view:

f(n+1)=f(f(fn)))=f(n)+

1 for all

.

By induction it is proved that for all

,

For

It is trivial, If this is true for n-1

then:

, $\therefore f(1)=3/2$ contradiction.

As would be for f^2(n)=n^+ +3

?

Thanks in advance.

What? the way you've written it you want f^2(n)=n+4

, why not just write that? And presumably you want f(f(n))

and not the square of f(n)

, otherwise it's trivial. In any case, clarify your function definition, it seems based on what you've written that you want just f(f(n))=n+4

, which I sincerely doubt is the case, since you can just let

be the (functional) square of the successor function if so.

qleak · **Posted:** Mon, 28 Nov 2011 13:36:43 UTC

Shadow wrote:

eraldcoil wrote:

Show that is not there a function $f:\mathbb {N}\rightarrow \mathbb {N}$ such that $f^2(n)=n^+ +3, \ (n^+=n+1)$

Proof: (For

)

It easy view:

f(n+1)=f(f(fn)))=f(n)+

1 for all

.

By induction it is proved that for all

,

For

It is trivial, If this is true for n-1

then:

, $\therefore f(1)=3/2$ contradiction.

As would be for f^2(n)=n^+ +3

?

Thanks in advance.

What? the way you've written it you want f^2(n)=n+4

, why not just write that? And presumably you want f(f(n))

and not the square of f(n)

, otherwise it's trivial. In any case, clarify your function definition, it seems based on what you've written that you want just f(f(n))=n+4

, which I sincerely doubt is the case, since you can just let

be the (functional) square of the successor function if so.

functional square of the successor function is a fancy way of saying
f(n)=n+2 which has
f(f(n))=f(n+2)=(n+2)+2=n+4

Shadow · **Posted:** Mon, 28 Nov 2011 18:15:38 UTC

qleak wrote:

Shadow wrote:

eraldcoil wrote:

Show that is not there a function $f:\mathbb {N}\rightarrow \mathbb {N}$ such that $f^2(n)=n^+ +3, \ (n^+=n+1)$

Proof: (For

)

It easy view:

f(n+1)=f(f(fn)))=f(n)+

1 for all

.

By induction it is proved that for all

,

For

It is trivial, If this is true for n-1

then:

, $\therefore f(1)=3/2$ contradiction.

As would be for f^2(n)=n^+ +3

?

Thanks in advance.

What? the way you've written it you want f^2(n)=n+4

, why not just write that? And presumably you want f(f(n))

and not the square of f(n)

, otherwise it's trivial. In any case, clarify your function definition, it seems based on what you've written that you want just f(f(n))=n+4

, which I sincerely doubt is the case, since you can just let

be the (functional) square of the successor function if so.

functional square of the successor function is a fancy way of saying
f(n)=n+2 which has
f(f(n))=f(n+2)=(n+2)+2=n+4

Yes, but that isn't what he wants, at least not apparently, he wants to show that there is NO such function.

eraldcoil · **Posted:** Mon, 28 Nov 2011 18:22:47 UTC

eraldcoil wrote:

Show that is not there a function $f:\mathbb {N}\rightarrow \mathbb {N}$ such that $f^2(n)=n^+ +3, \ (n^+=n+1)$

Proof: (For

)

It easy view:

f(n+1)=f(f(fn)))=f(n)+

1 for all

.

By induction it is proved that for all

,

For

It is trivial, If this is true for n-1

then:

, $\therefore f(1)=3/2$ contradiction.

As would be for f^2(n)=n^+ +3

?

Thanks in advance.

Shadow · **Posted:** Mon, 28 Nov 2011 23:43:17 UTC

Why did you just quote your own post? Do you have something new to say?

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About a function over natural numbers which does not exist.

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