Converting Bases with Decimals

[geekily]

I'm reviewing all my worksheets in preparation for a test on Tuesday, and I came across a problem involving a decimal that I hadn't done. We've gone over converting bases and I understand them just fine (Questions like, "What is the base 10 equivalent of 4012 base 5?") We never went over how to do this when the number has a decimal, but I found this one problem and she has a habit of making our tests a lot harder than the problems we're used to, so I figured I'd better be prepared.

The questions is, "what is the base 10 equivalent of 21.12 base 3?" The way I've been doing this with whole numbers is by making a chart: For 4012 base 5 to base 10, I would put the 4 under 5^3, 0 under 5^2, 1 under 5^1, and 2 under 5^0, add up those equations, and I'd have my answer. When it gets into decimals, though, I don't get it. I tried continuing the chart once I got to 3^0 with 3^-1 and 3^-2 because that's how we figure out expanded notation, but since there's no such thing as a negative square root, that doesn't work. I also tried making them 1 and 2, because in expanded notation we just carry over whatever's after the decimal, it seems. That gives me 10, but according to an online base converter, the answer is 7. Does anyone know how to get to this answer?

Thank you so much!

 

[arthur ohlsten]

decimal numbers are are for example: 4021.234
4x10^3+0x10^2+2x10^1+1x10^0+1x10^(-1)+3x10^(-2) + 4x10^(-3)
4 thousands + 0 hundreds + 2 tens +1 one + 1 tenth +3 one hundreths+4 one thousanths
4,000+20+1+2/10+3/100+4/1000

==============================================
ok now then 21.12 base 3 is
2[3]+1[1]+1/3+2/9
7+ 5/9
7 5/9 answer

Arthur

 

[soroban]

Hello, geekily!

There is a simple misunderstanding . . .

The question is: "What is the base 10 equivalent of \(\displaystyle 21.12_3\)?"

When it gets into decimals, though, I don't get it.
I tried continuing the chart once I got to \(\displaystyle 3^0\) with \(\displaystyle 3^{-1}\) and \(\displaystyle 3^{-2}\)
because that's how we figure out expanded notation,
but since there's no such thing as a negative square root, that doesn't work.

Be careful . . . \(\displaystyle 3^{-2}\) is neither a negative nor a square root.
. . \(\displaystyle 3^{-2}\) means: \(\displaystyle \frac{1}{3^2}\) . . . remember?


Your chart should look like this:

\(\displaystyle \;\;2\;\;|\;\;1\;\;|\;\;1\;\;|\;\;2\;\;|\)
\(\displaystyle \,- - - - - - - - - - - - - -\)
\(\displaystyle \;\;3^1\;|\;\;3^0\;|\;\,3^{-1}\;|\;3^{-2}\:|\)


So we have: \(\displaystyle \:\left(2 \times 3^1\right) \,+\,\left(1 \times 3^0\right)\,+\,\left(1 \times \frac{1}{3}\right) \,+\,\left(2 \times \frac{1}{9}\right)\;=\;6\,+\,1\,+\,\frac{1}{3}\,+\,\frac{2}{9}\;=\;\L\frac{68}{9}\)

 

[geekily]

Ok, that makes sense. Thank you very much!