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

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@Siniša Bubonja 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. Calvin Lin Staff · 3 years, 11 months ago

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

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@Justin Wong 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. Michael Tong · 3 years, 11 months ago

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@Justin Wong 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. Hariram K · 3 years, 11 months ago

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

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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 = \(2221201211100201\) Ton De Moree · 3 years, 11 months ago

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@Ton De Moree First line itself I did not understand. Please explain. Thanks Snehal Shekatkar · 3 years, 11 months ago

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@Snehal Shekatkar 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 :) Ton De Moree · 3 years, 11 months ago

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@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. Snehal Shekatkar · 3 years, 11 months ago

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@Snehal Shekatkar 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 \(7*9^6\). Ton De Moree · 3 years, 11 months ago

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@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. Snehal Shekatkar · 3 years, 11 months ago

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@Snehal Shekatkar 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! Ton De Moree · 3 years, 11 months ago

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@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 Snehal Shekatkar · 3 years, 11 months ago

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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. Tianbo Chen · 3 years, 7 months ago

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I think it has something to do with \( 9 = 3^{2} \) Tan Li Xuan · 3 years, 11 months ago

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