Visual walkthrough — Addition and subtraction — carrying, borrowing, word problems

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Step 1 — A number is stacks of ten-blocks

WHAT. Before any arithmetic, we need to see what a digit means. Take the number $472$. Picture three bins standing in a row: a ones bin, a tens bin, a hundreds bin. The ones bin holds single cubes. The tens bin holds rods — each rod is exactly $10$ cubes glued together. The hundreds bin holds flats — each flat is $10$ rods, i.e. $100$ cubes.

WHY. Every rule below is just "you may swap $10$ small blocks for $1$ next-size block, and back again." To trust that, you must first see that a rod really is ten cubes. Nothing is created when we swap — the total number of cubes never changes. This is the meaning of the expanded form

$472=\underset{\text{how many flats}}{\underbrace{4}}\times \underset{\text{cubes per flat}}{\underbrace{100}}+\underset{\text{how many rods}}{\underbrace{7}}\times \underset{\text{cubes per rod}}{\underbrace{10}}+\underset{\text{loose cubes}}{\underbrace{2}}\times \underbrace{1}.$

PICTURE. Look at the three bins: the amber flats on the left, cyan rods in the middle, white cubes on the right.

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Step 2 — Adding = pour both piles together, column by column

WHAT. To compute $487+356$, pour the two collections into the same bins. Loose cubes join loose cubes, rods join rods, flats join flats. In the ones bin we now have $7+6=13$ cubes.

WHY. We add within a column because only same-size blocks can be counted together — you cannot add a rod to a cube and call the result a single digit. This is exactly why we align numbers by place value (right-edge on the ones): mis-alignment would mean pouring cubes in with rods.

$\underset{\text{ones of }487}{\underbrace{7}}+\underset{\text{ones of }356}{\underbrace{6}}=\underset{\text{too many for one bin}}{\underbrace{13}}.$

PICTURE. The ones bin overflows: it is holding $13$ cubes, but a bin is only allowed to show a single digit $0$–$9$.

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Step 3 — The overflow forces a carry (this is the Division Algorithm)

WHAT. A bin may show at most $9$. We have $13$ cubes. So we grab $10$ of them, glue them into $1$ rod, and move that rod to the tens bin. Left behind: $3$ loose cubes.

WHY this exact split? Ask: "How many complete rods hide inside $13$ cubes, and how many cubes are left over?" That is the Division Algorithm question — divide by the base $10$:

$13=\underset{\text{base}}{\underbrace{10}}\times \underset{q=\text{carry}}{\underbrace{1}}+\underset{r=\text{digit we write}}{\underbrace{3}},0\le r\le 9.$

• $q$ (the quotient) is how many groups of ten, so it is the carry — the rod we send up.
• $r$ (the remainder) is what cannot form a full rod, so it is the digit we keep.

WHY the carry is always $0$ or $1$. Each digit is at most $9$, and a carry-in is at most $1$, so the biggest a column can ever hold is $9+9+1=19$. And $\lfloor 19/10\rfloor =1$. You can never need to carry a $2$.

PICTURE. Ten of the thirteen cubes fuse into one cyan rod (the amber arrow shows it climbing to the tens bin); $3$ cubes stay.

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Step 4 — The carry rides into the next column and repeats

WHAT. Now the tens bin. It had $8+5=13$ rods, plus the $1$ rod that just climbed up: $8+5+1=14$ rods. Same overflow, same fix — bundle $10$ rods into $1$ flat, send it up, keep $4$.

WHY. The rule is identical at every column because every column is ten times the one to its right. This is the payoff of place value: one rule, reused forever. Formally, in column $i$ with carry-in $c_i$:

$s_i=(\underset{\text{top digit}}{\underbrace{a_i}}+\underset{\text{bottom digit}}{\underbrace{b_i}}+\underset{\text{carry-in}}{\underbrace{c_i}})\mathrm{mod}10,c_{i+1}=\lfloor \frac{a_i+b_i+c_i}{10}\rfloor .$

The $\mathrm{mod}10$ means "the leftover after removing whole tens" ( $=r$ ); the $\lfloor \rfloor$ (floor, "round down") counts the whole tens ($=q$). In the hundreds bin: $4+3+1=8$, which is under $10$, so no overflow, no carry — we simply write $8$.

$487+356=843.$

PICTURE. The tens bin overflowing to $14$, ten rods becoming one flat, the flat climbing to hundreds.

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Step 5 — Subtracting = the swap run backwards

WHAT. Now $605-278$. Start at the ones bin: we must remove $8$ cubes, but only $5$ are there. We are short.

WHY. You cannot take $8$ from $5$ and stay in the whole numbers here (that would need Negative Numbers and Integers, which this column is not allowed to use). So we do the reverse of a carry: walk to the tens bin, take one rod, and break it into $10$ cubes. Now the ones bin holds $5+10=15$ cubes — plenty.

$\underset{\text{too few}}{\underbrace{5}}\stackrel{\text{break 1 rod}}{\to }\underset{=15}{\underbrace{5+10}}-\underbrace{8}=7.$

This is exact because we add $10$ here and subtract that same $10$ next door — nothing changes overall:

$t-u=(t+10)-u-10,\text{the extra }-10\text{ becomes }-1\text{ in the tens column.}$

PICTURE. A cyan rod from the tens bin cracks into $10$ white cubes pouring into the ones bin; the amber arrow points down/right (borrow bows down).

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Step 6 — Borrowing across a zero (the tricky case)

WHAT. After Step 5 the ones bin took a rod from the tens bin. But the tens bin held $0$ rods! So the tens bin is now $0-1=-1$ — itself short. It must borrow from the hundreds bin: break one flat into $10$ rods. The tens bin momentarily holds $10$ rods, immediately lends $1$ down to the ones (already gone), and is left with $10-1=9$ rods before its own subtraction.

WHY every middle zero becomes a $9$. A $0$ that must lend has nothing, so it borrows and lends in the same breath: it receives $10$, gives away $1$, keeps $9$. That is the whole "chain." Now finish the tens: $9-7=2$. And the hundreds, having lent one flat, is $6-1=5$, then $5-2=3$.

\text{ones: } 15-8=7 \\ \text{tens: } \overbrace{0}^{\text{was }0}\to 10 \to 10-1=9,\quad 9-7=2 \\ \text{hundreds: } 6\to 6-1=5,\quad 5-2=3 \end{array}\qquad\Rightarrow\qquad 605-278=327.$$ **PICTURE.** The hundreds flat cracks into $10$ rods; the middle bin shows $10\to9$; each amber label marks a step of the chain. ![[deepdives/dd-maths-1.1.03-d2-s06.png]] > [!mistake] The zero-borrow trap > Do **not** think "there's nothing in the $0$ to borrow, so skip it." The $0$ is a *relay*: it borrows from the left, lends to the right, and settles at $9$. Skipping it makes your answer wrong by a full $100$. --- ## Step 7 — The check: addition undoes subtraction **WHAT.** Every borrow you did was a break-a-block move; every carry was a glue-a-block move. They are opposites. So if $605-278=327$ is correct, then rebuilding must return the start: $$\underbrace{327}_{\text{difference}}+\underbrace{278}_{\text{subtrahend}}=\underbrace{605}_{\text{minuend}}.\qquad (a-b)+b=a.$$ **WHY it always works.** Adding $327+278$ *carries* exactly in the places where subtracting *borrowed* — the same tens moving back the other way. If a single borrow was mishandled, the rebuilt total will not match. This is why the parent note insists: **always add the difference back.** **PICTURE.** Two bin-rows side by side: the borrow (left, arrows down) and its mirror carry (right, arrows up), meeting at the identity $(a-b)+b=a$. ![[deepdives/dd-maths-1.1.03-d2-s07.png]] --- ## The one-picture summary **WHAT.** One figure holds the whole story: a carry *climbs* (glue $10\to1$, move up) and a borrow *bows* (break $1\to10$, move down). Same door, opposite directions. The digit you write is always the *leftover* ($\bmod 10$); what moves is always the *whole tens* ($\lfloor\ /10\rfloor$). ![[deepdives/dd-maths-1.1.03-d2-s08.png]] > [!recall]- Feynman retelling — the whole walkthrough in plain words > Every number is just blocks: loose cubes, rods of ten cubes, flats of ten rods. **Adding** means pouring two block-piles together and tidying up: any time a bin holds ten or more, glue ten into one bigger block and send it one bin to the left — that climbing block is the **carry**, and because the most a bin can ever hold is $9+9+1=19$, you never send more than one. **Subtracting** is the exact reverse tidy-up: if a bin doesn't have enough to give away, walk left, grab one bigger block, and smash it into ten smaller ones — that's a **borrow**. If the bin you walk to is empty (a $0$), it grabs from *its* left, keeps nine, and passes one along — that's why every middle zero turns into a nine. Since gluing and smashing are opposites, you can always check your answer by rebuilding: add the difference back to the subtrahend and you must land on the number you started with. No cube is ever created or destroyed — you only ever *repackage*. > [!mnemonic] One line to keep > **Carry Climbs (glue ten, go up); Borrow Bows (break one, come down).** The digit is the leftover; the tens do the travelling. --- ## Connections - [[Addition and Subtraction — Carrying, Borrowing, Word Problems]] — the parent this page derives. - [[Place Value and Base-Ten System]] — why bins are worth $1,10,100$. - [[Division Algorithm]] — the $13=10\cdot1+3$ split *is* carrying. - [[Negative Numbers and Integers]] — where "$5-8$" can be answered without borrowing. - [[Estimation and Rounding]] — a fast pre-check before you trust the columns.