For every natural number $a ge 2$, every natural number $m$ can be written in base $a$












3















For every natural number $a ge 2$, every natural number $m$ can be written in base $a$, i.e., as sum of powers of $a$: $$m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$$ with $b_1>cdots>b_n$ and $0<k_i<a$ for all $i=overline{1,n}$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We prove by induction on $m$.



For $m=1$, $m$ can be expressed as $a^0cdot 1$ for all natural number $age 2$.



Assume that, for all $k<m$, $k$ can be written in base $a$ for all natural number $age 2$.



Let $b_1=max{bin mathbb{N}mid a^ble m}$. Then there is $k_1$ and $r<a^{b_1}$ such that $m=a^{b_1}cdot k_1+r$. Since $r<a^{b_1}$, $r<m$ and thus $k_1>0$. I claim that $k_1<a$. If not, $k_1ge a$ and thus $k_1=a+Delta$ for some $Deltage 0$. Then $m=a^{b_1}cdot k_1+r=a^{b_1}cdot (a+Delta)+r=a^{b_1+1}+a^{b_1}cdotDelta+r$. This contradicts the maximality of $b_1$. Thus $k_1<a$. Since $r<m$, $r=a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n$ by inductive hypothesis. Since $r<a^{b_1}$, $a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n<a^{b_1}$ and thus $a^{b_2}cdot k_2<a^{b_1}$. It follows that $b_2<b_1$. Hence $m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$ as desired.






elementary-number-theory proof-verification







share|cite|improve this question


















  • 1




    It is correct. $ $
    – Berci
    Nov 20 at 11:29










  • Thank you so much for your confirm @Berci!
    – Le Anh Dung
    Nov 20 at 11:34










  • The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
    – David K
    Nov 20 at 14:42
















3















For every natural number $a ge 2$, every natural number $m$ can be written in base $a$, i.e., as sum of powers of $a$: $$m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$$ with $b_1>cdots>b_n$ and $0<k_i<a$ for all $i=overline{1,n}$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We prove by induction on $m$.



For $m=1$, $m$ can be expressed as $a^0cdot 1$ for all natural number $age 2$.



Assume that, for all $k<m$, $k$ can be written in base $a$ for all natural number $age 2$.



Let $b_1=max{bin mathbb{N}mid a^ble m}$. Then there is $k_1$ and $r<a^{b_1}$ such that $m=a^{b_1}cdot k_1+r$. Since $r<a^{b_1}$, $r<m$ and thus $k_1>0$. I claim that $k_1<a$. If not, $k_1ge a$ and thus $k_1=a+Delta$ for some $Deltage 0$. Then $m=a^{b_1}cdot k_1+r=a^{b_1}cdot (a+Delta)+r=a^{b_1+1}+a^{b_1}cdotDelta+r$. This contradicts the maximality of $b_1$. Thus $k_1<a$. Since $r<m$, $r=a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n$ by inductive hypothesis. Since $r<a^{b_1}$, $a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n<a^{b_1}$ and thus $a^{b_2}cdot k_2<a^{b_1}$. It follows that $b_2<b_1$. Hence $m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$ as desired.






elementary-number-theory proof-verification







share|cite|improve this question


















  • 1




    It is correct. $ $
    – Berci
    Nov 20 at 11:29










  • Thank you so much for your confirm @Berci!
    – Le Anh Dung
    Nov 20 at 11:34










  • The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
    – David K
    Nov 20 at 14:42














3












3








3








For every natural number $a ge 2$, every natural number $m$ can be written in base $a$, i.e., as sum of powers of $a$: $$m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$$ with $b_1>cdots>b_n$ and $0<k_i<a$ for all $i=overline{1,n}$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We prove by induction on $m$.



For $m=1$, $m$ can be expressed as $a^0cdot 1$ for all natural number $age 2$.



Assume that, for all $k<m$, $k$ can be written in base $a$ for all natural number $age 2$.



Let $b_1=max{bin mathbb{N}mid a^ble m}$. Then there is $k_1$ and $r<a^{b_1}$ such that $m=a^{b_1}cdot k_1+r$. Since $r<a^{b_1}$, $r<m$ and thus $k_1>0$. I claim that $k_1<a$. If not, $k_1ge a$ and thus $k_1=a+Delta$ for some $Deltage 0$. Then $m=a^{b_1}cdot k_1+r=a^{b_1}cdot (a+Delta)+r=a^{b_1+1}+a^{b_1}cdotDelta+r$. This contradicts the maximality of $b_1$. Thus $k_1<a$. Since $r<m$, $r=a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n$ by inductive hypothesis. Since $r<a^{b_1}$, $a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n<a^{b_1}$ and thus $a^{b_2}cdot k_2<a^{b_1}$. It follows that $b_2<b_1$. Hence $m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$ as desired.






elementary-number-theory proof-verification







share|cite|improve this question














For every natural number $a ge 2$, every natural number $m$ can be written in base $a$, i.e., as sum of powers of $a$: $$m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$$ with $b_1>cdots>b_n$ and $0<k_i<a$ for all $i=overline{1,n}$.




Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!




My attempt:



We prove by induction on $m$.



For $m=1$, $m$ can be expressed as $a^0cdot 1$ for all natural number $age 2$.



Assume that, for all $k<m$, $k$ can be written in base $a$ for all natural number $age 2$.



Let $b_1=max{bin mathbb{N}mid a^ble m}$. Then there is $k_1$ and $r<a^{b_1}$ such that $m=a^{b_1}cdot k_1+r$. Since $r<a^{b_1}$, $r<m$ and thus $k_1>0$. I claim that $k_1<a$. If not, $k_1ge a$ and thus $k_1=a+Delta$ for some $Deltage 0$. Then $m=a^{b_1}cdot k_1+r=a^{b_1}cdot (a+Delta)+r=a^{b_1+1}+a^{b_1}cdotDelta+r$. This contradicts the maximality of $b_1$. Thus $k_1<a$. Since $r<m$, $r=a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n$ by inductive hypothesis. Since $r<a^{b_1}$, $a^{b_2}cdot k_2+cdots+a^{b_n}cdot k_n<a^{b_1}$ and thus $a^{b_2}cdot k_2<a^{b_1}$. It follows that $b_2<b_1$. Hence $m=a^{b_1}cdot k_1+cdots+a^{b_n}cdot k_n$ as desired.






elementary-number-theory proof-verification




elementary-number-theory proof-verification






share|cite|improve this question













share|cite|improve this question











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asked Nov 20 at 11:14









Le Anh Dung

9491521




9491521








  • 1




    It is correct. $ $
    – Berci
    Nov 20 at 11:29










  • Thank you so much for your confirm @Berci!
    – Le Anh Dung
    Nov 20 at 11:34










  • The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
    – David K
    Nov 20 at 14:42














  • 1




    It is correct. $ $
    – Berci
    Nov 20 at 11:29










  • Thank you so much for your confirm @Berci!
    – Le Anh Dung
    Nov 20 at 11:34










  • The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
    – David K
    Nov 20 at 14:42








1




1




It is correct. $ $
– Berci
Nov 20 at 11:29




It is correct. $ $
– Berci
Nov 20 at 11:29












Thank you so much for your confirm @Berci!
– Le Anh Dung
Nov 20 at 11:34




Thank you so much for your confirm @Berci!
– Le Anh Dung
Nov 20 at 11:34












The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
– David K
Nov 20 at 14:42




The proof of the inductive step, as written, only makes sense if you suppose that you have selected a particular natural number $a$ for the step $b_1=max{bin mathbb{N}mid a^ble m}$. You can accommodate that in the proof, that is, let $a$ be any arbitrary integer with $ageq2,$ prove the statement for that particular $a,$ and since $a$ was arbitrary the statement is true for all $ageq 2.$ A charitable reader might do this part of the reasoning for you, but maybe it's better to be explicit.
– David K
Nov 20 at 14:42















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