Bound a natural by two consecutive powers

Question

I'm working in the following problems: Given two naturals $m$ and $n$, there exist a natural $d$ such that

$$m^{d}\leq n \leq m^{d+1}.$$

Afterwards I need to show that: If one chooses an arbitrary $b$, one can find a $c$ such that

$$m^{c}\leq n^b \leq m^{c+1}.$$

However I'm not able to solve even the first problem, I'm puzzled on how to show the existence of that $d$, and proving by contracition is leading me nowhere.

Answer 1

As long as $1 \lt m \le n$ you can just compute it. Take the base $m$ log of the inequality, giving $d \le \log_m n \le d+1$. You can just take $d=\lfloor \log_m n\rfloor$

Answer 2

You can easily prove the first one using Archimedean Property for real numbers $\log m$ and $\log n$

And for the second one again use the Archimedean Property for real number $b*\log_{m} n$

Knowing that Archimedean Property states thas for every real number $a$ and $b$ there exists only one natural number $n$ such that :

$(n-1)*a\leq b<n*a$

Proving the second one you should take number $a=1$

Answer 3

The statement is not true if $m = 0,1; n > m$. Nor if $n = 0$ and $m \ge 1$. So let's assume $m \ge 2; n \ge 1$

Now I'm assuming you do know that for any real value $M$ you can find $k \in \mathbb N$ so that $k > M$.

And if $m > 1> 0$ we know that $m^2 = m*m > 1*m =m$ so by induction, $1 < m < m^2 < m^3 < ....$ is an infinitely increasing sequence.

Now a pretty trivial result is that if $m > 1$ and $n\in \mathbb N$ then $m^n > n$. $m^2 = m*m = (m-1)*m + m \ge m+m > 1 + 1= 2$ and by induction $m^{n+1} = m^n*m \ge 2*m^n > 2n = n + n \ge n + 1$.

Okay, that was silly.

But that means of all the intervals $[1,m]$ and $[m,m^2]$ and all the $[m^i, m^{i+1}]$ that $n$ must lie in one of them (because $m^n > n$ so $n \in [m^i, m^{i+1}]$ for some $i < n$.)

And that's the first part.

The only difference with the second part is that if $b$ is not natural then $n^b$ need not be rational.

But you do know that $n^b > 0$

If $m > 1$ then we know that $.... < m^{-3}< m^{-2} < m^{-1} < 1 < m < m^2 < ....$. Of all the intervals $[m^c, m{c+1}]$ we know that $n^b$ must lie in one of them.

!EXCEPT! we do not know if it is possible for $n^b$ to be smaller than all $m^{-k}$. and we don't know if it is possible for $n^b$ to be bigger than all $m^k$.

Okay. $n^b$ is a value so we can find an natural number, $w$ so that $w > n^b$ and so $m^w > w > n^b$ so $n^b$ must be in one of the in one of those intervals $[m^i, m^{i+1}]$ . (More importantly we have proven that $m^i$ is unbounded.)

And we know that $\frac 1{n^b} > 0$ so we can fid a natural number so that $\frac 1{n^b} < w < m^w$. So $m^{-w} < \frac 1w < n^b$. So it is not possible for $n^b$ to the smaller than all $m^{-k}$.

Answer 4

(1).Some restrictions apply. Assume $m>1$ and allow the possibility $d=0.$

(2). Part of the axiomatic definition of $\Bbb N$ is the Principle of Induction: Let $P(n)$ mean that $n$ has property $P,$ which can be any property stated in the formal language of arithmetic. If $P(n)\implies P(n+1)$ for all $n\in \Bbb N$, and if $P(1)$, then $P(n)$ is true for all $n\in \Bbb N.$

From this we can prove that if $P(n)$ for some $n\in \Bbb N$ then there is a least $n\in \Bbb N$ for which $P(n)$ is true.

(3).The Principle of Induction implies that if $m,n\in \Bbb N$ and $m>1$ then there exists $e\in \Bbb N$ for which $n< m^e.$

For if we suppose not, we obtain a contradiction as follows: Let $P(n)$ be "$m^e\leq n$ for all $e \in \Bbb N$". Let $n_0$ be the least $n\in \Bbb N$ with property $P.$

Then $n_0-1\in \Bbb N$ because $n_0\geq m^1=m>1.$

For any $e\in \Bbb N$ we have $e'=e+1\in \Bbb N$ so $m^{e'}\leq n_0,$ so $$m^e=m^{e'}/m\leq n_0/m\leq n_0/2\leq n_0-1.$$ Thus $m^e\leq n_0-1$ for all $e\in \Bbb N, $ and since $n_0-1\in \Bbb N$ we have $P(n_0-1).$ But this contradicts the requirement that no member of $\Bbb N$ which is less than $n_0$ has property $P.$

(4). So let $e_0$ be the $least $ $ e\in \Bbb N$ such that $n<m^e.$ Let $d=e_0-1.$

If $e_0=1$ then $d=0$ and $1=m^0=m^d=m^0=1\leq n <m^{e_0}=m^{d+1}.$

If $e_0>1$ then $d\in \Bbb N$ so $n\geq m^d$ (otherwise $m^{e_0}$ would not be the least positive power of $m$ that is greater than $n.$) So again we have $m^d\leq n<m^{e_0}=m^{d+1}.$

....................Remark:Arithmetic on $\Bbb N$ can be extended to arihmetic on a larger collection $\Bbb N^*$ which has members that are larger than any member of $\Bbb N,$ but the Principle of Induction does not apply to $\Bbb N^*$. So we cannot avoid this principle and still prove that $d$ exists. (Otherwise we could be talking about $m,n\in \Bbb N^*$ and $d\in \{0\}\cup \Bbb N,$ where in some cases, $d$ will not exist.)