i PLEASE NEED HELP WITH 4010 base five minus 1234 base 5!! please and thank you I know that the numbers add to more than five so they need to be different

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- Apr 11th 2012, 12:47 PMcathyharris4Bases
i PLEASE NEED HELP WITH 4010 base five minus 1234 base 5!! please and thank you I know that the numbers add to more than five so they need to be different

- Apr 11th 2012, 01:00 PMa tutorRe: Bases
I'd just use the usual method taught in schools, 'borrowing', but it's a little difficult to illustrate on here.

Instead you could use 'equal addition' as follows.

4010-1234

adding 1 to both numbers

=4011-1240

adding 10 to both numbers

=4021-1300

adding 200 to both numbers

=4221-2000

=2221

It's rather less efficient though. - Apr 11th 2012, 01:16 PMcathyharris4Re: Bases
is this base five then? and would I use the same method for all bases? thank you so much for your quick response :)

- Apr 11th 2012, 01:32 PMa tutorRe: Bases
Well, you said it was base 5...?

I wouldn't use the method I posted. I'd use 'borrowing'. Yes it works for any base. - Apr 11th 2012, 03:08 PMSorobanRe: Bases
Hello, cathyharris4!

Have you never**ever**done a subtraction in other bases?

Quote:

$\displaystyle 4010_5 - 1234_5$

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 1 & 0 \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$

In C4, we can't subtract 4 from 0, so we "borrow" from C3.

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & ^0\rlap{/}1\; & ^10\; \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$

The "1" we borrowed represents "5", so we have:

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline \end{array}$

Subtract in C4 and we have:

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 4 & 0 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$

In C3, we can't subtract 3 from 0, so we "borrow" from C2.

But C2 has "0"; we can't borrow.

So we must borrow from C1.

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & ^3\rlap{/}4\;\; & ^10\;\; & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array} \quad\Rightarrow\quad \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 5 & 0 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$

Now we can borrow from C2:

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & ^4\rlap{/}5\;\; & ^10\;\; & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array} \quad\Rightarrow\quad \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 4 & 5 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &&&&1 \end{array}$

And we can complete the subtraction:

$\displaystyle \begin{array}{ccccc} & ^{C1} & ^{C2} & ^{C3} & ^{C4} \\ & 3 & 4 & 5 & 5 \\ - & 1 & 2 & 3 & 4 \\ \hline &2&2&2&1 \end{array}$