A University-Level Understanding of Number with an eye towards elementary education

Subsection 4.2.1 Standard Subtraction Algorithm

1. Write one number below the other so that each place value is in its own column. Now begin at the rightmost column.

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2. 1. If the top digit (or digit plus regrouping) is larger than or equal to the bottom digit, subtract and put the difference below in the same column.

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   2. If the top digit is smaller than the bottom digit, determine the first digit in the top number to the left of the current place that is non-zero. Decrease that place by 1, (if there are any) change the 0 digits between that digit and our current place to 9s (or whatever the largest digit in your base is, if you are subtracting in a non-ten base), and add 10 to the current top digit.

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3. If there are no non-zero digits in any column to the left, you are done. Otherwise move to the next column to the left and go to Step 2.

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Example 4.2.1.

Let’s subtract \(4052-2457\text{.}\) We first write the numbers aligned vertically, and view each number in terms of place value blocks:

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Now let’s subtract the rightmost place. Note that the top 2 is smaller than the bottom digit, so we must borrow. The the place directly to the left has a non-zero top digit, so we regroup 1 of this place (tens) to make 10 of our current place (ones) and thus decrease the digit in that place by 1.

Now we have \(2+10=12\) ones and we do the subtraction \(12-7=5\text{.}\)

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Now we move to the next place to the left; the tens place. Note that the top digit is now a 4, so it is smaller than the bottom digit 5. So again, we must borrow.

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We have a zero immediately to the left of the (what was a 5 but is now a) 4, so we move to the left one more time (to the thousands place) since there is a non-zero digit in the top number there. We borrow one thousand, thus decreasing the thousands digit by 1 and increasing the hundreds by 10.

Then we ungroup that into 9 hundreds and 10 tens. In our calculation, this means changing the zero in the hundreds place in the top number to a 9, and the 4 in the tens place to a 14.

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Now we have 14 tens and we do the subtraction in the tens column \(14-5=9\) . Similarly, the top number in the hundreds and thousands place is smaller so we do those subtractions as well: \(9-4=5\) and \(3-2 = 1\) respectively.

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As we have no non-zero digits to the left of the thousands place, we are finished and we know \(4052-2457=1595.\)

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Example 4.2.2.

Let’s calculate \(2201_{four} - 1332_{four}\) and since we are still getting used to arithmetic in other bases, let’s draw some base four blocks to represent our calculation.

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We start from the ones digit on the right. Since the top digit is smaller than the bottom digit, we must borrow. The first non-zero digit to the left is the 2 (in the squares, or sixteens, place). We ungroup \(1_{four}\) sixteen (square) to \(3_{four}\) fours (longs) and \(10_{four}\) ones (remember that \(10_{four}\) represents the number "four"). We can see this since \(16 = 3 \times 4 + 4 \times 1\text{,}\) similar to our situation in the previous example in base ten. Thus we decrease the squares digit in the top number by 1 and replace the 0 in the longs place by a 3, and thus we have \(11_{four}\) ones (which, in base four, is five ones).

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Now, we can do the subtraction in the ones place \(11_{four} - 2_{four} = 3_{four} \text{,}\) since we can see that \(2_{four}+3_{four} = 11_{four}\text{.}\) We can subtract the longs digit as well: \(3_{four} - 3_{four} = 0 _{four}\text{.}\) We write these down in our calculation.

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Now, in the squares place, the digit 1 in the top number is smaller than the 3 in the bottom number, so we need to borrow again. We ungroup \(1_{four}\) cube (or sixty-four) to \(10_{four}\) squares, decreasing our 2 cubes to 1 cube.

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Thus we now have \(10_{four}+1_{four} = 11_{four}\) squares and can now do our subtraction \(11_{four}-3_{four} = 2_{four}\text{.}\) We can also do the subtraction in the cubes place: \(1_{four}-1_{four} = 0_{four}\text{.}\) We now have our final difference: \(2201_{four} - 1332_{four} = 203_{four}.\text{.}\)

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Subsection 4.2.2 Partial Differences

Example 4.2.4.

Let’s subtract \(4052-2457\) again. Like before, we line up the numbers vertically by place value. In terms of blocks, we can view each place as a separate subtraction calculation with blocks of the same place value.

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Order isn’t important in this algorithm, so let’s start from left and go right. Subtracting the thousands place, we know that \(4000-2000 = 2000\text{.}\) We write this partial difference in our calculation.

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Now, for the hundreds place, we have \(000-400=-400.\) (note that writing 0 as \(000\) is to remind you that we have \(0\) hundreds.) So we mark down \(-400\) in our calculation. Similarly, we have \(50-50=0\) and \(2-7=-5\text{.}\) We write these down in our calculation as well.

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So we must combine our partial differences to obtain the difference we are looking for. We have \(2000-400+0-5\text{.}\) Working from left to right we obtain.

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\begin{equation*} 1600+0-5 \end{equation*}

\begin{equation*} =1595. \end{equation*}

And we have our final answer.

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Subsection 4.2.3 Same Change Algorithm

1. Addition is easier than subtraction

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2. For any number \(a, a-0=a\)

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3. For any numbers \(a,b,c, a-b = (a+c) - (b+c)\)

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1. Write the subtraction vertically (as usual) with place values aligned in columns. Start at the rightmost column.

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2. If the top digit in this place is less than the bottom digit, add the opposite digit number of places of the bottom digit to both numbers to make a new equivalent subtraction problem (this makes the bottom digit a 0, for easy subtraction). If the top digit in this place is not less than the bottom digit, do nothing.

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3. If there are non-zero digits in any place to the left of this place, move to the left one place and go to step 2. Otherwise, go to the next step.

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4. Each top digit should be greater than or equal to the bottom digit in its respective place. Subtract the digits in each place to obtain your final answer.

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Example 4.2.5.

Let’s subtract \(4052-2457\) using the same change algorithm. We line up the two numbers as usual (and we’ll keep block representations along with our calculation to see the process in a pictorial way)

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Starting at the ones place, we notice that the top digit is smaller than the bottom digit, we add 3 to both the top and bottom numbers (\((4052+3)-(2457+3))\)) to obtain the new subtraction problem \(4055-2460\text{.}\) Notice that the bottom digit is now 0, so this will eventually make subtraction easy (at the cost of doing a bit of addition)

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Now we move left to the tens place. As before, we notice that the top digit is less than the bottom digit, so add the opposite digit number of tens to both numbers. Since the opposite of \(6\) is \(4\) we add \(40\) to both numbers (\((4055+40)-(2460+40)\)) to make the new subtraction problem \(4095-2500.\)

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Moving to the hundreds place, since the top digit is less than the bottom digit, we add the opposite number of hundreds, that is \(500\text{,}\) to both numbers to obtain the new subtraction problem \(4595-3000\text{.}\)

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Moving left to the thousands place, we notice that the top digit is greater than or equal to (in this case, greater than) the bottom digit, so we do nothing. Also, there are no non-zero digits in any places to our left (thousands is the largest place in our numbers) so we can now easily subtract place by place to obtain our final answer \(4595-3000=1595.\)

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Subsection 4.2.4 Equal Additions Algorithm

1. Write the subtraction vertically (as usual) with place values aligned in columns. Start at the rightmost column.

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2. If the digit in the top number is less than the digit in the bottom number (including additional ones if one is there), add ten of that place to the top digit (signified by a 1) while adding an additional 1 to the place to the left in the bottom number. Then subtract the bottom number from the top number in that place. If the digit in the top number is greater than the digit in the bottom number, do the place subtraction without any borrowing.

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3. If there are any non-zero digits in any places to the left of the current place, move one place to the left and go to step 2. Otherwise you are finished.

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Example 4.2.6.

Let’s subtract \(4052-2457\) using the equal addition algorithm. As usual we line up our numbers so digits of the same place value are in the same column, but this time we will have just a little bit of space between the bottom number and our difference (for any borrowed 1’s that will appear). And, as usual, we will include blocks for added pictorial understanding.

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Starting with the ones place, we see that the digit \(2\) in the top number is less than the digit \(7\) in the bottom. So we "borrow" 10 ones, but unlike the standard subtraction algorithm we are "in debt" 1 ten to the bottom number. Thus, we have \((10+2)-7=12-7=5\) ones in our difference. We mark this, and our additional debt of 1 ten, in our calculation:

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Now we move to the tens place. We see that the \(5\) in the top number is less than the \(5+1\) in the bottom number, so we "borrow" 10 tens from 1 hundred in the same way as the previous step to have \((10+5)-(5+1)=15-6=9\) tens in our difference. We mark this, and our additional debt of 1 hundred, in our calculation:

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Now we move to the hundreds place. We see that the \(0\) in the top number is less than the \(4+1\) in the bottom number, so we "borrow" 10 hundreds from 1 thousand in the same way as our previous two steps to have \((10+0)-(4+1)=10-5=5\) hundreds in our difference. We mark this, and our additional debt of 1 thousand, in our calculation:

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Finally we move to the thousands place. We see that \(4\) in the top number is greater than the \(2+1\) in the bottom number, so we subtract without borrowing: \(4-(2+1)=4-3=1\) and write this in our answer. As there are no more non-zero digits to the left (that is; no larger places than thousands in our numbers,) we are finished, and we have calculated \(4052-2457=1595.\)

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