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Conversion to other basis

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,839
Hello!!! (Wave)

We consider the usual representation of non-negative integers, where the digits correspond to consecutive powers of the basis in a decreasing order.
Show that at such a representation, for the conversion of a number with basis $p$ to a system with basis $q$, where $p=q^n$ and $n$ positive integer, it suffices that each digit of the number is expressed from initial system of basis $p$ to the system of basis $q$, using $n$ digits of the system of basis $q$.
Also the rule should be stated and it should be proved at the reverse case, i.e. when the conversion is done from the system of basis $q$ to the system of basis $p$.

Could you give me a hint how we could show this? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,568
Hey evinda !!

So we have a number like $p_0 + p_1\cdot p + p_2\cdot p^2 +\ldots$ yes?
And we have $p=q^n$ so that number is the same as $p_0 + p_1\cdot q^n + p_2\cdot q^{2n} +\ldots$.
That is a representation in the system with basis $q$ isn't it? 🤔
So if we needed $m$ digits to represent the number in the system with basis $p$, we can represent the number with $m$ digits in the system with basis $q$ as well. 🤔
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,839
Hey evinda !!

So we have a number like $p_0 + p_1\cdot p + p_2\cdot p^2 +\ldots$ yes?
And we have $p=q^n$ so that number is the same as $p_0 + p_1\cdot q^n + p_2\cdot q^{2n} +\ldots$.
That is a representation in the system with basis $q$ isn't it? 🤔
Yes, it is... (Nod) But how do we know that the coefficients $p_0,p_1, \dots$ are also based on the system of basis $q$ ? (Thinking)

So if we needed $m$ digits to represent the number in the system with basis $p$, we can represent the number with $m$ digits in the system with basis $q$ as well. 🤔
Yes, but we need to show that $n$ digits suffice, right? 🧐
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,568
Yes, it is... But how do we know that the coefficients $p_0,p_1, \dots$ are also based on the system of basis $q$ ?
Didn't we just show that? :unsure:
We effectively wrote the same number with respect to the system with basis $q$, didn't we? 🤔

Yes, but we need to show that $n$ digits suffice, right?
Could it be that's a mistake in the problem statement?

Consider for example that $p=100$ and $q=10$. So $n=2$.
We can represent the number $10203$ with respect to $p$ as $123$, and we can represent the same number with respect to $q$ as $10203$.
In this case we need $m=3$ digits to represent the number and $n=2$ digits does not suffice to represent the number. :unsure:
 
Last edited:

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,839
Didn't we just show that? :unsure:
We efectively wrote the same number with respect to the system with basis $q$, didn't we? 🤔
Do we know that $p_0,p_1,p_2, \dots$ are also written in the system with basis $q$, since $p=q^n$ and so $p<q$ ?

Could it be that's a mistake in the problem statement?

Consider for example that $p=100$ and $q=10$. So $n=2$.
We can represent the number $10203$ with respect to $p$ as $123$, and we can represent the same number with respect to $q$ as $10203$.
In this case we need $m=3$ digits to represent the number and $n=2$ digits does not suffice to represent the number. :unsure:
We have that $\frac{10203}{100}=102 \cdot 100+3$ and so $10203=1023$ with respect to $100$ and $\frac{10203}{10}=1020 \cdot 10+3$ and so $10203=10203$ with respect to $10$, right? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,568
Do we know that $p_0,p_1,p_2, \dots$ are also written in the system with basis $q$, since $p=q^n$ and so $p<q$ ?
More or less. It means that the digits with respect to $q$ are the same - except that there are $n-1$ zeroes in between each of the digits. 🤔


We have that $\frac{10203}{100}=102 \cdot 100+3$ and so $10203=1023$ with respect to $100$ and $\frac{10203}{10}=1020 \cdot 10+3$ and so $10203=10203$ with respect to $10$, right? (Thinking)
More accurately, we have $10203=1\cdot 100^2+2\cdot 100^1+3\cdot 100^0$ with respect to $p=100$.
And we have $10203=1\cdot 10^4+0\cdot 10^3 + 2\cdot 10^2+0\cdot 10^1 + 3\cdot 10^0$ with respect to $q=10$. 🤔