How does addition actually work?

First we need to see what these numbers mean. Let's use a different example with more digits

135 + 792

The 135 means **1 hundred, 3 tens, 5 ones**

The 792 means **7 hundred, 9 tens, 2 ones**

When we calculate the result, we write it in the same general shape: hundreds, tens, ones. That's called the **decimal** number system. (Larger numbers can have thousands etc.)

So, if you added these together, you'd get

1 hundred,  3 tens, 5 ones
7 hundred,  9 tens, 2 ones
-------------------------
8 hundred, 12 tens, 7 ones

But 12 has two digits. If we wrote 8127, it would look like we meant 8 thousand, not 8 hundred.

12 tens means 120. And that means we have **1 hundred, 2 tens**. The 10 tens add up to another hundred.

8 hundred, 12 tens, 7 ones
10 tens is 1 hundred
= 9 hundred,  2 tens, 7 ones
= 927

Since smaller values (tens) can add up to larger values (hundreds), we usually start adding from smaller to larger
It's a side effect of our positional number system.  Even before people used a base 10 numbering system they still grouped things together when adding large numbers.

Say you had a bunch of apples and you wanted to count them.   you could count all the apples one at a time:   aaaaaaaaaaaaaaaaa    but that's difficult to recount.    So perhaps you group them in groups of 5 for instance.   aaaaa aaaaa aaaaa aa.   Written that way it's easy to see (and remember) that we have 3 groups of 5 and 2 extra.

Now if we have another bunch of apples:   aaaaaa  and if we group them again and put aaaaa a,  then we can see that we have 1 group of 5 and 1 extra.

The benefit of this grouping is that addition becomes easier.  All together we now have 4 groups of 5 and 3 extra.

24 + 26 = 50 is the same idea.   24 means 2 groups of 10 and 4 extra.  26 means 2 groups of 10 and 6 extra.   Adding them together gives us 4 groups of 10 and 10 extra.   The 10 extra is a new group of 10 so we just have 5 groups of 10, or 50.
Our number system is quite special.  If you look at arithmetic using Roman numerals, for example, it's horrible.  But with the way we write numbers, with different places standing for different powers of 10, and the digit specifying how many we have, makes a lot of things work out nicely.

So for your problem, 24 is 2 tens and 4 ones.  26 is 2 tens and 6 ones.  For the time being, we don't have to know what a one or a ten is.  They could be apples and oranges.  When we put them together, we have 4 tens and ten ones.  However, because we can turn ten ones into one ten, we have 4 tens and then one more ten, for a total of five tens, which we write as 50.

The addition algorithm accomplishes this going one place at a time, without really saying why we are doing what we are doing, or what carrying is.  But carrying is simply what happens when we turn ten ones into one ten, or ten tens into one hundred, or ten hundreds into one thousand.

Multiplication is similar, except we need to use something called the "distributive property" of multiplication.  If you are willing to accept that the area of a rectangle is the base times the height, then the distributive property says that when you take a rectangle and use a vertical line to cut it into two pieces, then the total area is the area of the left rectangle plus the area of the right rectangle.  In symbols, a(b+c)=ab+ac.

So if we wanted to do 24\*26, we could break it into an area problem, but we divide one side into 20 and 4, and the other side into 20 and 6, and this divides  the rectangle into four pieces, each one being not too bad to calculate if we use the fact that 10\*10=100, 20=2\*10, and a few other things that come out of our number system.

The multiplication algorithm takes this idea, but then uses the addition algorithm to add the rows of the rectangle, and then again to add up those results.
You can decompose any number into its parts, for ex 56 = 50 + 6, 164 = 100 + 60 + 4. Now, we can say 56 + 164 = (50 + 6) + (100 + 60 + 4) = 100 + (50 + 60) + (6 +4) by the commutative law. This is why you can add ones and tens individually. Now, in the previous example, 6 + 4, which is in the ones place, equals 10. When you add 10 to the sum, you effectively add 1 to the tens place. Which is why you add one to the tens column.

The basic axioms of addition will be helpful if you want a truly rigorous understanding.
Whatever you tried to draw didn't render at all for me.  But anyway.

Let's say you're adding 24 + 26.  If you're familiar with positional numerals -- that is, where the value of a digit depends on where it's written in the number -- you'll know that 24 is 2 10's + 4 1's, and 26 is 2 10's + 6 1's.  Let's add them: how many 10's?  2 from 24 plus 2 from 26 give you 4 10's.  How many 1's?  4 from 24 and 6 from 26 give you 10 1's.  But 10 1's is the same as 1 10, so we add that to our total of 10's, giving us 5 10's, which we write as 50.

So what did we do?  We added up the 10's, then we added up the 1's, then we combined them into one result.  But when we combined them, we actually had to *change* the number of 10's we had because the number of 1's was too big for one digit.  That's annoying to do, so we should probably add up the 1's first and then the 10's, since the answer to the sum of 1's could affect the number of 10's in the answer, but the sum of 10's can't affect the number of 1's.

Now we have an algorithm, a systematic way to add two numbers by adding their digits.  First we add the 1's, and if there's a number of 10's left over, we add *that* to the sum of the 10's, which, if there's a number of 100's left over, we add that to the sum of the 100's, etc.  But how do we remember the steps?  Easiest way is to write the numbers vertically, lining up the 1's, the 10's, etc.  This way, we can just add down the 1's column and write the answer, remembering to keep track of leftover 10's; then we add the 10's column (plus any leftover 10's) and write that answer, remembering to keep track of leftover 100's, and so on.  You don't *have* to write the numbers that way, and you don't *have* to add the 1's first and work backwards.  But if you don't, you'll have to erase your 10's if your 1's spill over.  It's much easier to keep track of these spillover 10's by writing that number on top of the 10's column, so that you'll easily remember to add it.

Subtraction is exactly the same, except that in subtraction you might need to split a 10 up into 10 1's to make the subtraction work.  If you have 24 and want to subtract 8, well, you can't take 8 1's from 4 1's, so you turn 24 into 1 10 and 14 1's and make your subtraction that way.  You can easily represent this by sticking a little 1 next to the 4 and crossing out the 2 to write in a 1, which we call borrowing.

Multiplication is different.  Here, it's easier to just do algebra.  Let's say we're trying to multiply ABC by DEF (those are all digits).  This is the same as doing ABC · (D·100 + E·10 + F·1).  Multiplying ABC by the single digit F is the same as doing (A·100 + B·10 + C·1) · F = 100·A·F + 10·B·F + 1·C·F.  So you multiply the 1's by F, and if there are any 10's left over, you add it to the total of the 10's times F, etc.  Then you add them all up.  It's all just the distributive property!

Division is completely different in form, but it works on a similar principle: let's say you're dividing 458 by 7.  We'll go digit by digit.  Note that 458 is 4 100's + 5 10's + 8 1's, right?  How many 700's go into 458?  0, so the answer has a 0 in the 100's place.  Subtract 0·700 = 0 from 458 and go to the 10's.  How many 70's go into 458?  6, so the answer has a 6 in the 10's place.  Subtract 6·70 = 420 from 458 to get 38 and go on.  How many 7's go into 38?  5, so the answer has a 5 in the 1's place.  Subtract 5·7 = 35 from 38 to get 3 and go on.  Now, we could go to the 10ths place, 100ths place, etc., and so on forever, or we can declare that 3 is the remainder, since 7 doesn't go into it at all.  The answer is therefore 65 remainder 3.

In all of these cases, it's very helpful in understanding the algorithms to keep track of what the digits actually represent.  A 2 in a number could be 2, or 20, or 20000, or 0.00000002, depending on its position in the number.  These algorithms are optimized in a way that will let you kind of ignore that information, which takes off some of the mental load and makes them really easy to actually execute on paper (other than long division, at least, since you can't do it digit by digit like you can with addition, subtraction, and multiplication).  But when you think about these algorithms *with* the positional information, you can see them for what they are: just breaking down a number and organizing it to make things easier.

I never even questioned this but it was fun to learn about so thank you for asking!
What a solid post and reply.
Do you know how regular addition works, like 2+3 = 5?

How about 20 + 30 = 50?
I just count, starting at the biggest units

&#x200B;

1678  +

7543        I count down the columns starting at the top left. and add the thousands When I get to the bottom, I start on the hundreds. And so on. All you ever two is add two numbers.

one thousand

eight thousand

eight thousand six hundred

nine thousand one hundred

nine thousand one hundred and seventy

nine thousand two hundred and ten

nine thousand two hundred and eighteen

nine thousand two hundred and twenty one   9221   Correct

Two notes

1:   This looks more trouble than it's worth. Of course you don't actually say anything out loud.

It's very fast.

2. You can also add money  very quickly this way.
We start with something called the succesor function which adds 1 to the current number. So, the successor of 0 is 1, the successor of 1 is 2, the successor of 2 is 3, etc.

Addition is repeated succession. So, 0 + 10 is the 10th successor of 0, 5 + 20 is the 20th successor of 5, 30 + 5 is the 5th successor of 30, etc.

Multiplication is repeated addition. So, 4 * 2 is 0 + 4 + 4 which is the 4th successor of the 4th successor of 0. 5 * 3 is 0 + 5 + 5 + 5 which is the 5th succesor of the 5th succesor of the 5th succesor of 0, etc.

When you have a number like 432, you can say that the 4 represents 4 * 100 which is 0 + 100 + 100 + 100 + 100, the 3 represent 3 * 10 which is 0 + 10 + 10 + 10, and the the 2 repesents 2 * 1 which is 0 + 1 + 1. So, you have (4 * 100) + (3 * 10) + (2 * 1) = 0 + 100 + 100 + 100 + 100 + 0 + 10 + 10 + 10 + 0 + 1 + 1 = 0 + 400 + 30 + 2.

In other words, the 400th successor of 0 + the 30th successor of 0 + the 2nd successor of 0 = 432nd of successor of 0.

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