Converting number base of decimals

  • Thread starter RChristenk
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  • #1
RChristenk
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Homework Statement
Transform ##0.7## from base ##8## to base ##5##. Transform ##0.12211## from base ##3## to base ##9##.
Relevant Equations
Conversion between bases
Transforming ##0.7_8## to base ##5## by multiplying ##5##:

##
\begin{array}{r}
0.7 \\
5 \\
\hline
4.3 \\
5 \\
\hline
1.7 \\
\end{array}
##

##0.7_8={0.\dot{4}\dot{1}}_5##. This is the correct answer.

Transforming ##0.12211_3## to base ##9## by multiplying ##9##:

##
\begin{array}{c}
0.&1&2&2&1&1 \\
&&&&&9 \\
\hline
12.&2&1&1&0&0 \\
\end{array}
##

Here I get two digits in the integer portion instead of one. And the correct answer is ##0.573_9##. Clearly ##12 \neq5##. And if I continued to multiply ##12.211## by ##9## to get the next digit position value in base ##9##, I would get ##1221.1## which has four integers. So the method works for me in the first problem, but not the second. I suspect there's nothing wrong with the method of multiplying by the final base, but rather I'm not understanding something or my calculations are wrong. Thanks for the help.
 
Last edited:
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  • #2
12 base 3 is 5.
21 base 3 is ...
10 base 3 is ...

Also, in your base 3 multiply, why not show the 9 in base 3?
 
  • #3
RChristenk said:
Homework Statement: Transform ##0.12211## from base ##3## to base ##9##.
Relevant Equations: Conversion between bases

And the correct answer is ##0.573_9##.
##0.12211_3 = \frac 1 3 + \frac 2 9 + \frac 2 {27} + \frac 1 {81} + \frac 1 {243} = \
\frac 3 9 + \frac 2 9 + \frac 6 {81} + \frac 1 {81} + \frac 3 {729} ##.
Now we're almost there. The trick is to combine the various negative powers of 3 to make negative powers of 9; i.e., ##\frac 1 9, \frac 1 {81}, \frac 1 {729}, \dots ##.
 
  • #4
@Mark44 : I much prefer @RChristenk method. I'm sure that it is the method presented in his class.
The trick they are presenting is operating in one base to directly convert to another base without using base 10 as an intermediate.
What he was doing is fine. He just didn't recognize the 12 base 3 as the first digit (or "nonit"?) of the answer he was looking for.
 
  • #5
.Scott said:
12 base 3 is 5.
21 base 3 is ...
10 base 3 is ...

Also, in your base 3 multiply, why not show the 9 in base 3?
So ##12.211## is in base ##3##, and the integer ##12## would then be ##1\times3^1+2\times3^0=5## in base ##9##? But why wouldn't I multiply by ##9## for base ##9##: ##1\times9^1+2\times9^0=11##?

The same for the next digit. I take the non-integer portion ##0.211## and multiply it again by ##9## and get ##21.1##. So ##21=2\times3^1+1\times3^0=7## which is the correct answer. Why not ##21=2\times9^1+1\times9^0=19## since we are looking for the base ##9## equivalent? Everything looks like it's being done in base ##3## exclusively.

Thanks!
 
  • #6
.Scott said:
I'm sure that it is the method presented in his class.
Well, maybe it is but I was going by how fractional numbers in other bases are defined; i.e., as the sum of negative powers of the base. Perhaps to be clearer I might have written ##0.12211_3 = \frac 1 3 + \frac 2 9 + \frac 2 {27} + \frac 1 {81} + \frac 1 {243}## as ##0.12211_3 = \frac 1 {3^1} + \frac 2 {3^2} + \frac 2 {3^3} + \frac 1 {3^4} + \frac 1 {3^5}##. After all we can expand the decimal fraction ##0.253_{10} = \frac 2{10^1} + \frac 5{10^2} + \frac 3{10^3}## even though 10 is not a digit in the decimal (base-10) system.
 
  • #7
RChristenk said:
But why wouldn't I multiply by ##9## for base ##9##: ##1\times9^1+2\times9^0=11##?

The same for the next digit. I take the non-integer portion ##0.211## and multiply it again by ##9## and get ##21.1##. So ##21=2\times3^1+1\times3^0=7## which is the correct answer. Why not ##21=2\times9^1+1\times9^0=19## since we are looking for the base ##9## equivalent? Everything looks like it's being done in base ##3## exclusively.

Since you are starting in base 3, your multiply and the product are in base 3.

Here are the multiply operations for the first 2 base-9 "digits":

##0.12211_3##
##\times\hspace{0.9em} 100_3## (9, the base used for the solution)
##\hspace{0.5em}\overline{12.211_3}## (yielding ##12_3## as the first digit: equivalent to ##5_9##)

##0.211_3## (remainder from last multiply)
##\times\hspace{0.1em} 100_3## (9, the base used for the solution)
##\hspace{0.5em}\overline{21.1_3}## (yielding ##21_3## as the second digit: equivalent to ##7_9##)
 
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  • #8
@RChristenk :
Since I have spoiled your problem by giving you most of the answer, here is a replacement problem.
Convert ## 2.54321 ## from base 6 (hex digits 0,1,2,3,4,5) to base 12 (dodeca digits 0,1,2,3,4,5,6,7,8,9,a,b)
 
  • #9
.Scott said:
@RChristenk :
Since I have spoiled your problem by giving you most of the answer, here is a replacement problem.
Convert ## 2.54321 ## from base 6 (hex digits 0,1,2,3,4,5) to base 12 (dodeca digits 0,1,2,3,4,5,6,7,8,9,a,b)
The integer ##2## doesn't change because it is ##2## in base ##12## as well. The non-integer ##0.54321## is multiplied by ##12## for each digit, since ##12## is a base. Then the first product is ##11.30420##, and ##11=B## in base ##12##, so the first partial answer is ##2.B##. And then continue multiplying the decimal portion by ##12##. I'm still a bit confused here because the calculations were done in base ##6## (albeit with ##12## considered a single digit), so wouldn't ##11## be in base ##6##? Then ##11_6=1\times6+1=7_{10}## which is also ##7_{12}##. But ##11_6## is taken as ##B## here and not ##7_{12}##. Thanks for your help.
 
  • #10
RChristenk said:
The integer ##2## doesn't change because it is ##2## in base ##12## as well. The non-integer ##0.54321## is multiplied by ##12## for each digit, since ##12## is a base. Then the first product is ##11.30420##, and ##11=B## in base ##12##, so the first partial answer is ##2.B##. And then continue multiplying the decimal portion by ##12##. I'm still a bit confused here because the calculations were done in base ##6## (albeit with ##12## considered a single digit), so wouldn't ##11## be in base ##6##? Then ##11_6=1\times6+1=7_{10}## which is also ##7_{12}##. But ##11_6## is taken as ##B## here and not ##7_{12}##. Thanks for your help.
I think you're confusing yourself because you aren't completing the multiply in base 6.
The first product is ## 15.30420_6 ##. So the digit is ## 15_6 ## or ## b_{12} ##.
After that, there is a "heximal portion" which must be multiplied by ## 20_6 ##.

This seems to be about as far as you got:

Starting at ## 2.54321_6 ##:

The symbol '2' is the same in radix 6 and 12, so it is trivially transferred:
## 2.?????_{12} ##

##\hspace{0.2em}0.54321_6## (remainder after the 2 is transferred.)
##\hspace{0.7em}\times\hspace{1.0em} 20_6## (12, the base used for the solution)
##\hspace{0.1em}\overline{15.3042_6}## (yielding ##15_6## as the second digit: equivalent to ##b_{12}##)
## 2.b????_{12} ##

##\hspace{0.6em}0.3042_6## (remainder from last multiply)
##\hspace{0.7em}\times\hspace{1.0em} 20_6## (12, the base used for the solution)
 
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  • #11
I think I understand a little better now. Generally the entire calculation is done in the original base, and only with a final product do I begin to transform it's value into the desired base. I have another question with transforming ##0.7_8## to base ##5##. How did ##0.7_8\times5 = 4.3_8##? Because I'm thinking ##0.7\times5 = 3.5 = 8\times0.4+0.3##, but then ##0.4## and ##0.3## occupies the same digit position. It works if I think of it in terms of ##0.7\times5=3.5=0.8\times4+0.3##, but it is asking for base ##8## here, and not ##0.8##. Or should I just ignore the decimal initially, calculate ##7\times5=35=4\times8+3=43## then put in decimal with one digit to get ##4.3##? Thanks again.
 
  • #12
When multiplying ## 0.7_8 \times 5 ##, you need to multiply the digits 7 and 5. If you have memorized the base 8 multiplication table, you do that directly. More reasonably, you follow your last choice, if for no other reason than it seems less confusing to you. So you get ## 35_{10} ## which you recognize as ##(4 \times 8) + 3 ##, hence ## 43_8 ##, hence ##4.3_8##.
 
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