Subtracting Whole Numbers Using the Standard Algorithm

Hello. In the previous lesson, you used place value to add whole numbers: digits were aligned by column, you began at the ones place, and regrouped when ten smaller units made one larger unit. Subtraction uses the same place-value structure, but the exchange goes in the opposite direction.

In this lesson, you will subtract whole numbers with the standard written method. You will learn when regrouping is needed, why exchanging one ten for ten ones keeps a number’s value unchanged, how to regroup through larger places, and how to check an answer using addition.

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Set up subtraction by place value

In a subtraction expression,

$$\text{minuend}-\text{subtrahend}=\text{difference}$$

the minuend is the starting amount, the subtrahend is the amount removed, and the difference is the answer.

For this course, we will place the larger whole number on top so the difference is a whole number.

Just as in written addition, align matching places in vertical columns:

• ones below ones,
• tens below tens,
• hundreds below hundreds,
• and so on.

Then subtract beginning with the ones column and move left.

For example:

$$\begin{array}{r}546\\ -425\\ 121\end{array}$$

Work one place at a time:

• Ones: $6-5=1$
• Tens: $4-2=2$
• Hundreds: $5-4=1$

So,

$$546-425=121$$

Subtraction works smoothly here because each top digit is at least as great as the digit beneath it in the same column. But often, a column will not have enough units to subtract. Then we regroup.

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Regrouping means exchanging one larger unit for ten smaller units

Consider:

$$95-28$$

Start in the ones column. We need to subtract 8 ones from 5 ones, but 5 ones are not enough. The solution is not to change the value of 95; it is to rewrite that value in a more useful form.

The number 95 means 9 tens and 5 ones. Exchange one of its tens for 10 ones:

$$9\text{ tens}+5\text{ ones}=8\text{ tens}+15\text{ ones}$$

The total is still 95. We have only changed how it is grouped.

Now subtract:

1. Ones: $15-8=7$.
2. Tens: after giving away one ten, there are 8 tens left. So $8-2=6$.

$$\begin{array}{rr}& \cancel{9}\cancel{5}\\ & 815\\ & 95\\ -& 28\\ & 67\end{array}$$

A less crowded way to show the same written work is:

$$\begin{array}{rr}8& 15\\ 9& 5\\ -2& 8\\ 6& 7\end{array}$$

The numbers on the first row show the regrouped values: 8 tens and 15 ones.

The key idea is:

When a top digit is too small, take 1 unit from the place to its left and exchange it for 10 units in the current place.

So:

• 1 ten becomes 10 ones;
• 1 hundred becomes 10 tens;
• 1 thousand becomes 10 hundreds.

Calling this process “borrowing” is common, but regrouping describes the mathematics more accurately: nothing is borrowed from another number. You are exchanging units within the same number.

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Regrouping in more than one column

The same reasoning works with hundreds.

Calculate:

$$532-276$$

First align the places:

$$\begin{array}{r}532\\ -276\end{array}$$

Ones column

We cannot subtract 6 ones from 2 ones. Regroup 1 ten from the 3 tens.

• The 3 tens become 2 tens.
• The 2 ones become 12 ones.

Now:

$$12-6=6$$

Tens column

We now have 2 tens, but need to subtract 7 tens. Regroup 1 hundred from the 5 hundreds.

• The 5 hundreds become 4 hundreds.
• The 2 tens become 12 tens.

Now:

$$12-7=5$$

Hundreds column

Finally:

$$4-2=2$$

So the completed calculation is:

$$\begin{array}{rrr}4& 12& 12\\ 5& 3& 2\\ -2& 7& 6\\ 2& 5& 6\end{array}$$

Therefore,

$$532-276=256$$

Notice what changes each time you regroup:

Column where you need more	Unit exchanged from the left	New amount in current column
Ones	1 ten	10 more ones
Tens	1 hundred	10 more tens
Hundreds	1 thousand	10 more hundreds

Do not subtract using the original top digit after regrouping. For example, in the tens column above, you must calculate $12-7$, not $3-7$, because one hundred has been exchanged for ten tens.

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What if the place to the left is zero?

A zero represents no units in its place, so it cannot directly give one unit to the column beside it. You must first regroup from the nearest nonzero digit farther to the left.

Consider:

$$5103-2046$$

Start at the ones:

$$3-6$$

This requires regrouping, but the tens digit is 0. There are no tens available to exchange. Move left again: the hundreds digit is also 1, so it can help.

1. Exchange 1 hundred for 10 tens. The hundreds digit changes from 1 to 0, and the tens place becomes 10 tens.
2. Exchange 1 of those 10 tens for 10 ones. The tens place becomes 9 tens, and the ones place becomes 13 ones.
3. Subtract column by column.

$$\begin{array}{rrrr}5& 0& 9& 13\\ 5& 1& 0& 3\\ -2& 0& 4& 6\\ 3& 0& 5& 7\end{array}$$

Now calculate:

• Ones: $13-6=7$
• Tens: $9-4=5$
• Hundreds: $0-0=0$
• Thousands: $5-2=3$

Thus,

$$5103-2046=3057$$

The zero in the difference must be written. It holds the hundreds place. Without it, writing 357 would incorrectly mean 3 hundreds, 5 tens, and 7 ones.

A reliable way to handle zeros is to say the exchanges aloud:

“I exchange one hundred for ten tens. Then I exchange one of those tens for ten ones.”

This makes clear why the tens digit becomes 9, rather than 10: one of the newly created tens was immediately exchanged.

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A dependable subtraction routine

For any whole-number subtraction problem, follow this routine:

1. Check the order. Put the larger number on top when finding a whole-number difference.
2. Align the digits by place value. Begin at the right edge, so the ones line up.
3. Begin in the ones column.
4. Subtract each column. If the top digit is large enough, subtract normally.
5. Regroup when necessary. Exchange one unit from the place to the left for ten units in the current place.
6. Continue leftward, using the changed digits after regrouping.
7. Keep zeros wherever they belong in the answer.
8. Check with addition.

The check uses the inverse relationship between addition and subtraction:

$$\text{difference}+\text{subtrahend}=\text{minuend}$$

For the earlier problem,

$$532-276=256$$

check it by adding:

$$256+276=532$$

Because the sum returns to the starting number, the subtraction is consistent.

You can also make a quick estimate. Since $532$ is close to $500$ and $276$ is close to $300$,

$$500-300=200$$

The exact difference, $256$, is reasonably close to 200. This does not prove the answer is correct, but it can reveal major place-value mistakes, such as writing 2,056.

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Key takeaways

The standard written method for subtraction depends on place value.

• Align digits by ones, tens, hundreds, and larger places.
• Subtract from right to left, beginning with ones.
• When a top digit is too small, regroup from the place to its left.
• One larger unit always exchanges for ten of the next smaller unit.
• When regrouping crosses a zero, first obtain units from the nearest nonzero place to the left.
• Keep zeros in their proper columns.
• Check subtraction with addition: difference plus subtrahend should equal the minuend.

You have now completed the whole-number addition and subtraction part of this module. Next, the course moves to multiplication, where equal groups and place value will again support a standard written method.