Prove that there does not exist a strictly increasing function satisfying $f(2)=3$ and $f(acdot b)=f(a)cdot...
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I stumbled upon the following problem, but I can't seem to find a way to prove it.
Show that there does not exist a strictly increasing function that maps the set of natural numbers to natural numbers, satisfying $f(2)=3$ and $f(acdot b)=f(a)cdot f(b)$ for any $a,b$ in the naturals.
The only information I was able to obtain from the probably are trivialities, such as $f(2)=3,f(1)=1,f(4)=9, 9<f(5)<27, f(6)=3f(3),$ if it is assumed that the function is actually increasing. Altough, I can't find a pattern nor disprove by contraddiction. How would I go about this?
real-analysis
asked Nov 20 at 0:05
DMH16
539217
add a comment |
up vote
3
down vote
favorite
I stumbled upon the following problem, but I can't seem to find a way to prove it.
Show that there does not exist a strictly increasing function that maps the set of natural numbers to natural numbers, satisfying $f(2)=3$ and $f(acdot b)=f(a)cdot f(b)$ for any $a,b$ in the naturals.
The only information I was able to obtain from the probably are trivialities, such as $f(2)=3,f(1)=1,f(4)=9, 9<f(5)<27, f(6)=3f(3),$ if it is assumed that the function is actually increasing. Altough, I can't find a pattern nor disprove by contraddiction. How would I go about this?
real-analysis
asked Nov 20 at 0:05
DMH16
539217
Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I stumbled upon the following problem, but I can't seem to find a way to prove it.
Show that there does not exist a strictly increasing function that maps the set of natural numbers to natural numbers, satisfying $f(2)=3$ and $f(acdot b)=f(a)cdot f(b)$ for any $a,b$ in the naturals.
The only information I was able to obtain from the probably are trivialities, such as $f(2)=3,f(1)=1,f(4)=9, 9<f(5)<27, f(6)=3f(3),$ if it is assumed that the function is actually increasing. Altough, I can't find a pattern nor disprove by contraddiction. How would I go about this?
real-analysis
asked Nov 20 at 0:05
DMH16
539217
I stumbled upon the following problem, but I can't seem to find a way to prove it.
Show that there does not exist a strictly increasing function that maps the set of natural numbers to natural numbers, satisfying $f(2)=3$ and $f(acdot b)=f(a)cdot f(b)$ for any $a,b$ in the naturals.
The only information I was able to obtain from the probably are trivialities, such as $f(2)=3,f(1)=1,f(4)=9, 9<f(5)<27, f(6)=3f(3),$ if it is assumed that the function is actually increasing. Altough, I can't find a pattern nor disprove by contraddiction. How would I go about this?
real-analysis
real-analysis
asked Nov 20 at 0:05
DMH16
539217
asked Nov 20 at 0:05
DMH16
539217
asked Nov 20 at 0:05
DMH16
539217
asked Nov 20 at 0:05
DMH16
539217
asked Nov 20 at 0:05
DMH16
539217
539217
Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18
add a comment |
Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18
Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18
Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18
add a comment |
1 Answer
1
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oldest
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up vote
3
down vote
accepted
Hint: Let $f(3)=k$. Then
$$f(2^m)=3^m$$
and
$$f(3^n)=k^n.$$
Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint: Let $f(3)=k$. Then
$$f(2^m)=3^m$$
and
$$f(3^n)=k^n.$$
Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
add a comment |
up vote
3
down vote
accepted
Hint: Let $f(3)=k$. Then
$$f(2^m)=3^m$$
and
$$f(3^n)=k^n.$$
Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint: Let $f(3)=k$. Then
$$f(2^m)=3^m$$
and
$$f(3^n)=k^n.$$
Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
Hint: Let $f(3)=k$. Then
$$f(2^m)=3^m$$
and
$$f(3^n)=k^n.$$
Can you find a pair $(m,n)$ so $2^m<3^n$ but $3^m>k^n$ or vice versa?
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
answered Nov 20 at 0:21
Carl Schildkraut
10.5k11438
10.5k11438
add a comment |
add a comment |
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Probably a noteworthy clarification: do the naturals in this context include $0$ or no? Depending on the context or professor/author, people seem to differ on whether $0 in mathbb{N}$. If it is allowed, then it makes finding such a counterexample fairly easy, I think, but best to ask first.
– Eevee Trainer
Nov 20 at 0:18