Base 8

rchristian

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Joined
Sep 14, 2011
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I'm having trouble understanding how to add and subtract using different bases.

If I have 742 base 8
- 364 base 8
_______

I think I need to borrow from the 4 to over to the 2 because 2 is smaller than the 4 I'm trying to subtract. When I borrow how much am I borrowing? Do I end up with 8 plus 2 = 10? And then do I subtract 4 from 10 to get 6?

When I count in base 8 what happens after 8?
 
Hello, rchristian!

Your reasoning is correct.


I'm having trouble understanding how to add and subtract using different bases.

If I have: .\(\displaystyle \begin{array}{cccccccc}&7&4&2& \text{(base 8)} \\ - & 3 & 6 & 4 & \text{(base 8)}\\ \hline \end{array}\)

I think I need to borrow from the 4 to over to the 2,
. . because 2 is smaller than the 4 I'm trying to subtract. . Right!
When I borrow, how much am I borrowing? . Eight
Do I end up with 8 plus 2 = 10? . Yes!
And then do I subtract 4 from 10 to get 6? . Correct!

We "borrow one" from the 4 and "add it" to the 2.

It looks like this: .\(\displaystyle \begin{array}{ccccc}& 7 & \rlap{/}4^3 & ^12 \\ - & 3 & 6 & 4 \\ \hline \end{array}\)

Since \(\displaystyle 12_8 = 10_{10}\), the right column becomes: .\(\displaystyle 10 - 4 \,=\,6\)

And the problem looks like this: .\(\displaystyle \begin{array}{cccccc}& 7 & 3 & 2 \\ - & 3 & 6 & 4 \\ \hline &&&6 \end{array}\)


To subtract 6 from 3, we must "borrow 1" from 7 and "add it" to the 3.

So we have: .\(\displaystyle \begin{array}{cccccc}& \rlap{/}7^6 & ^13 & 2 \\ - & 3 & 6 & 4 \\ \hline &&&6 \end{array}\)

Since \(\displaystyle 13_8 = 11_{10}\), the middle column becomes: .\(\displaystyle 11 - 6 \,=\,5\)

And the problem looks like this: .\(\displaystyle \begin{array}{cccccc}& 6 & 3 & 2 \\ - & 3 & 6 & 4 \\ \hline &&5 & 6 \end{array}\)


Finally, \(\displaystyle 6 - 3 \,=\,3\)
. . and the problem looks like this: .\(\displaystyle \begin{array}{cccccc} & 6 & 3 & 2 \\ - & 3 & 6 & 4 \\ \hline & 3 & 5 & 6 \end{array}\)

The final answer is: .\(\displaystyle \begin{array}{cccccc}& 7 & 4 & 2 \\ - & 3 & 6 & 4 \\ \hline & 3 & 5 & 6 \end{array}\)




When I count in base 8, what happens after 8?

You mean "after 7".
In base-eight, there are eight digits: 0, 1, 2, 3, 4, 5, 6, 7.


When we reach the quantity "eight", it is written \(\displaystyle 10.\)
 
counting in base 8

If when I'm counting in base 8 it goes: 0,1,2,3,4,5,6,7, do I then have 10? What does 11 look like and so on?
 
I'm having trouble understanding how to add and subtract using different bases.

If I have 742 base 8
- 364 base 8
_______

I think I need to borrow from the 4 to over to the 2 because 2 is smaller than the 4 I'm trying to subtract. When I borrow how much am I borrowing? Do I end up with 8 plus 2 = 10? And then do I subtract 4 from 10 to get 6?

When I count in base 8 what happens after 8?
Remember : For base " n " you always borrow from the left " n "units.
 
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