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Note that this works only because 9 is a power of 3. In most other cases, it is advisable to convert to base 10, unless you are at ease with calculation in different bases.

$87654321$ becomes $807060504030201$ by only using $9=3^2$. Furthermore $3n+a$ in base $9$ is written as $na$ in base $3$ and luckily $n$ is not greater than $2$ because we have a number in base $9$.

Applying this to $807060504030201$ = 8 07 06 05 04 03 0201 we get 22 21 20 12 11 10 0201 = $2221201211100201$

We look at the $7$ for example. In base $9$ it represents $7*9^6$. Using $9=3^2$ we get $7*9^6=7*(3^2)^6=7*3^{12}$. With this we see that every power in base $9$ is doubled when moving to base $3$. Also, we see that this transformation only gives even powers of $3$, hence all the zeroes on the places for odd powers of $3$.

@Ton de Moree
–
Sorry but I still don't understand. $7*9^6$ is very big number. How come 7 is equal to 7*9^6 in base 9? I have studied different bases and all but this is very confusing to me.

@Ton de Moree
–
Thank you very much. I really did not see all these genius intricacies. :)
But Still I am trying to understand your whole argument. It seems that you have taken too many short-cuts and for people like me it is very hard to understand your argument.

@Ton de Moree
–
No no.. I completely understand this.. For example.. as a chess player, I see 5-6 moves deep combination easily many times but my amateur friends often don't understand it without me explaining them all the elements. I am amateur in maths so I am just saying that you please describe your answer in little more detail. Thank you

If you want only an approximation upto a few digits, one way to do this by multiplying with appropriate power of base. Suppose you want the base 3 representation of 100.150 upto 5 trinary places. Then start by multipling 100.150 with 3^5. This is 24336.45. Now get 24336 in trinary. This will be 1020101100.
So we can say $(100.15)_{10} = (10201.01100..)_3$
The .. in the end means there are more digits in the expansion.

It's the same way, but it's much more complex / less applicable. Since $510.63_{10}$ implies $5(10^2) + 1(10^1) + 0(10^0) + 6(10^{-1}) + 3(10^{-2})$, then in say base 3, this number would equal $2(3^5) + 2(3^2) + 2(3) + 1(3^{-1}) + 2(3^{-2})... = 200220.12..._3$. You don't see non-integers in other bases often because it's just so awkward.

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## Comments

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TopNewest$\mathit{9=3^2}\\ \\ 8=(2)\cdot 3+(2)\\ 7=(2)\cdot 3+(1)\\ 6=(2)\cdot 3+(0)\\ 5=(1)\cdot 3+(2)\\ 4=(1)\cdot 3+(1)\\ 3=(1)\cdot 3+(0)\\ 2=(0)\cdot 3+(2)\\ 1=(0)\cdot 3+(1)\\ \\ (87654321)_9=(2)\cdot 3^{15}+(2)\cdot 3^{14}+(2)\cdot 3^{13}+(1)\cdot 3^{12}+(2)\cdot 3^{11}+(0)\cdot 3^{10}+(1)\cdot 3^9+(2)\cdot 3^8\\+(2)\cdot 3^7+(1)\cdot 3^6+(1)\cdot 3^5+(0)\cdot 3^4+(0)\cdot 3^3+(2)\cdot 3^2+(0)\cdot 3^1+(1)\cdot 3^0=\\(2221201211100201)_3$

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Good explanation of the step that you are using.

Note that this works only because 9 is a power of 3. In most other cases, it is advisable to convert to base 10, unless you are at ease with calculation in different bases.

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$87654321$ becomes $807060504030201$ by only using $9=3^2$. Furthermore $3n+a$ in base $9$ is written as $na$ in base $3$ and luckily $n$ is not greater than $2$ because we have a number in base $9$.

Applying this to $807060504030201$ = 8 07 06 05 04 03 0201 we get 22 21 20 12 11 10 0201 = $2221201211100201$

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First line itself I did not understand. Please explain. Thanks

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No problem :)

We look at the $7$ for example. In base $9$ it represents $7*9^6$. Using $9=3^2$ we get $7*9^6=7*(3^2)^6=7*3^{12}$. With this we see that every power in base $9$ is doubled when moving to base $3$. Also, we see that this transformation only gives even powers of $3$, hence all the zeroes on the places for odd powers of $3$.

I hope this cleared things up :)

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$7*9^6$ is very big number. How come 7 is equal to 7*9^6 in base 9? I have studied different bases and all but this is very confusing to me.

Sorry but I still don't understand.Log in to reply

$7$, not the $7$ itself.

Ah, it's got to do with the position of theIn base $10$ the number 321 means $3*100 + 2*10 + 1*1$ (multiples of powers of $10$.

In base $9$ we use powers of $9$, so 321 then means $3*81 + 2*9 + 1*1$.

Now, in the number $87654321$ in base $9$ means $8*9^7 + 7*9^6 + 6*9^5 + 5*9^4 + 4*9^3 + 3*9^2 + 2*9^1 + 1*9^0$.

So the $7$ in $(87654321)_9$ stands for $7*9^6$.

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It really isn't a problem for me to explain myself, so feel free to ask any questions you might have!

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Is 2221201211100201 in base 3 and 42 374 116 in base 10.

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I knew most of this, but what I want to see is how to convert non-integers into different bases.

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If you want only an approximation upto a few digits, one way to do this by multiplying with appropriate power of base. Suppose you want the base 3 representation of 100.150 upto 5 trinary places. Then start by multipling 100.150 with 3^5. This is 24336.45. Now get 24336 in trinary. This will be 1020101100. So we can say $(100.15)_{10} = (10201.01100..)_3$ The .. in the end means there are more digits in the expansion.

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It's the same way, but it's much more complex / less applicable. Since $510.63_{10}$ implies $5(10^2) + 1(10^1) + 0(10^0) + 6(10^{-1}) + 3(10^{-2})$, then in say base 3, this number would equal $2(3^5) + 2(3^2) + 2(3) + 1(3^{-1}) + 2(3^{-2})... = 200220.12..._3$. You don't see non-integers in other bases often because it's just so awkward.

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I think it has something to do with $9 = 3^{2}$

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I believe using this trick: https://brilliant.org/discussions/thread/investigation-faster-conversion-through-bases/?ref_id=54083

would also work out nicely. We would just have to work backwards. $(87654321)_9 = |22|21|20|12|11|10|02|01|_3$

$8 = |22|_3$

$7 = |21|_3$ and etc.

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