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# Number Base Representation

This week, we learn about Number Base Representation.

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How would you convert $$(87654321)_9$$ into base 3? Try and minimize the calculations.

Note by Calvin Lin
3 years, 11 months ago

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

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. Staff · 3 years, 11 months ago

I knew most of this, but what I want to see is how to convert non-integers into different bases. · 3 years, 11 months ago

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. · 3 years, 11 months ago

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. · 3 years, 11 months ago

Is 2221201211100201 in base 3 and 42 374 116 in base 10. · 3 years, 11 months ago

$$87654321$$ 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 = $$2221201211100201$$ · 3 years, 11 months ago

First line itself I did not understand. Please explain. Thanks · 3 years, 11 months ago

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 :) · 3 years, 11 months ago

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. · 3 years, 11 months ago

Ah, it's got to do with the position of the $$7$$, not the $$7$$ itself.

In 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 $$7*9^6$$. · 3 years, 11 months ago

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. · 3 years, 11 months ago

It all makes sense in my head ;)

It really isn't a problem for me to explain myself, so feel free to ask any questions you might have! · 3 years, 11 months ago

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 · 3 years, 11 months ago

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. · 3 years, 7 months ago

I think it has something to do with $$9 = 3^{2}$$ · 3 years, 11 months ago