How addition and multiplication works

Question

Lets say i am doing 12 + 13 by using the addition method that we know. i mean first we write 13 below 12, then we do 2+3 and then 1+1. The result can be validated as 25 (or true) by doing the counting manually. But for larger numbers, what is the guarantee (or proof) that the addition method that we use is indeed right?

Edit:

1. Could you please tell me how multiplication also works using the same logic used in the accepted answer(by Adriano) ?

2. In the accepted answer, the proof starts with the assumption that a number, lets call it 'abc' can be expressed in the form a*100 + b*10 + c. How can we prove this assumption for large numbers?

Answer 1

Suppose we add $372$ and $594$. Using "the addition method that we know", we would first add $2$ and $4$ to get $6$. Then we would add $7$ and $9$ to get $16$, so we write down a $6$ then carry the $1$. Then we add the carried $1$ with the $3$ and $5$ to get $9$, for a final answer of $966$. Why does this work?

Well what we're really doing is taking advantage of how numbers are written in base $10$: \begin{align*} 372 + 594 &= (300 + 70 + 2) + (500 + 90 + 4) \\ &= (300 + 500) + (70 + 90) + (2 + 4) \\ &= (300 + 500) + (70 + 90) + (6) \\ &= (300 + 500) + (160) + (6) \\ &= (300 + 500) + (100 + 60) + (6) \\ &= (300 + 500 + 100) + (60) + (6) \\ &= (900) + (60) + (6) \\ &= 966 \end{align*} This idea indeed generalizes, even to numbers written in other bases.

Answer 2

This is in answer to the edits (I'm going to start with your second question):

#2: How we know that we can always represent a number by adding factors of base-10 numbers

There is no proof needed. Decimal notation is simply a shorthand for expressing numbers in terms of base-10 numbers. $342$ is DEFINED to mean $(3×10^2)+(4×10^1)+(2×10^0)$, where we take the numbers $0 - 10$ and the operations $\times, +$, and ^ (exponentiation) as "primitives" (things we don't NEED to define because we all hopefully have the same idea in mind as to what they mean). Because of this definition it doesn't matter how big the number is, you can ALWAYS reduce it to this form (okay, not ALWAYS, but if we assume that a number HAS a decimal representation, then it can be reduced to this form). For instance $ 12345678909876543210.54321 := 1 \times 10^{19} + 2 \times 10^{18} + 3 \times 10^{17} + 4 \times 10^{16} + 5 \times 10^{15} + 6 \times 10^{14} + 7 \times 10^{13} + 8 \times 10^{12} + 9 \times 10^{11} + 0 \times 10^{10} + 9 \times 10^9 + 8 \times 10^8 + 7 \times 10^7 + 6 \times 10^6 + 5 \times 10^5 + 4 \times 10^4 + 3 \times 10^3 + 2 \times 10^2 + 1 \times 10^1 + 0 \times 10^0 + 5 \times 10^{-1} + 4 \times 10^{-2} + 3 \times 10^{-3} + 2 \times 10^{-4} + 1 \times 10^{-5}$, where $:=$ means "is defined as".

#1: How multiplication works

Say you want to multiply 345 by 678. The "usual method" is:
$\ \ \ \ \ \ 345$
$\ \ \ \ \ \ 678$
$\ \ \ \ \overline{2760}$
$\ \ 2415$
$2070$
$\overline{233910}$

Here's how it actually works:

1) First we write out our numbers explicitly in base-10 notation:

$$345 \times 678 = [(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)] \times [(6 \times 10^2) + (7 \times 10^1) + (8 \times 10^0)]$$

2) Then we distribute:

$$= \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (6 \times 10^2)\right) + \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (7 \times 10^1)\right) + \left(\left[(3 \times 10^2) + (4 \times 10^1) + (5 \times 10^0)\right] \times (8 \times 10^0)\right)$$ $$ = \left(\left[(3 \times 10^2) \times (6 \times 10^2)\right] + \left[(4 \times 10^1) \times (6 \times 10^2)\right] + \left[(5 \times 10^0) \times (6 \times 10^2)\right]\right) + \left(\left[(3 \times 10^2) \times (7 \times 10^1)\right] + \left[(4 \times 10^1) \times (7 \times 10^1)\right] + \left[(5 \times 10^0) \times (7 \times 10^1)\right]\right) + \left(\left[(3 \times 10^2) \times (8 \times 10^0)\right] + \left[(4 \times 10^1) \times (8 \times 10^0)\right] + \left[(5 \times 10^0) \times (8 \times 10^0)\right]\right)$$

3) Now to save space, I'm only going to work on those last 3 terms which will produce the first line of the solution of the "usual method". I will now use the commutativity of addition to reorder the terms:

$$= \left[(5 \times 10^0) \times (8 \times 10^0)\right] + \left[(4 \times 10^1) \times (8 \times 10^0)\right] + \left[(3 \times 10^2) \times (8 \times 10^0)\right] + \cdots$$

4) Now I'll use the commutativity and associativity of multiplication to reorder the factors in each term:

$$= \left[(5 \times 8) \times (10^0 \times 10^0)\right] + \left[(4 \times 8) \times (10^1 \times 10^0)\right] + \left[(3 \times 8) \times (10^2 \times 10^0)\right] + \cdots$$

5) Use our exponent rules:

$$= \left[(5 \times 8) \times 10^0\right] + \left[(4 \times 8) \times 10^1\right] + \left[(3 \times 8) \times 10^2\right] + \cdots$$

6) Here is where our multiplication table comes in:

$$= \left[(40) \times 10^0\right] + \left[(32) \times 10^1\right] + \left[(24) \times 10^2\right] + \cdots$$

7) Recognize that $40$ is shorthand for $(4 \times 10^1) + (0 \times 10^0)$ and so on to get:

$$= \left[(4 \times 10^1 + 0 \times 10^0) \times 10^0\right] + \left[(3 \times 10^1 + 2 \times 10^0) \times 10^1\right] + \left[(2 \times 10^1 + 4 \times 10^0) \times 10^2\right] + \cdots$$

8) From distributivity and our exponent rules:

$$= \left[4 \times 10^1 + 0 \times 10^0\right] + \left[3 \times 10^2 + 2 \times 10^1\right] + \left[2 \times 10^3 + 4 \times 10^2\right] + \cdots$$

9) From commutativity:

$$= \left[0 \times 10^0 + 4 \times 10^1\right] + \left[2 \times 10^1 + 3 \times 10^2\right] + \left[4 \times 10^2 + 2 \times 10^3\right] + \cdots$$

10) From associativity:

$$= (0 \times 10^0) + (4 \times 10^1 + 2 \times 10^1) + (3 \times 10^2 + 4 \times 10^2) + (2 \times 10^3) + \cdots$$

11) From distributivity:

$$= (0) \times 10^0 + (4 +2) \times 10^1 + (3 + 4) \times 10^2 + (2) \times 10^3 + \cdots$$

12) And we finally end up with the first term in the solution of the "usual method" as promised:

$$= (0) \times 10^0 + (6) \times 10^1 + (7) \times 10^2 + (2) \times 10^3 + \cdots = 2760 + \cdots$$

At this point, if I had written out all the other terms instead of just writing "$\cdots$" we could break each term up into base-10 parts and add.

As you can see, the "usual method" has a LOT going on under the surface that (thankfully) you don't normally need to think about.

Answer 3

When You write 12 you actually mean 10+2. Or in general when you represent a number in decimal by writing $a_na_{n-1}a_{n-2} . . . a_2a_1a_0$ we actually mean $ \Sigma_{i=0}^n a_i10^i$. Now, Perhaps It is easy for you to see that you can add term-wise. For example You can verify $a_i 10^i + b_i 10^i = d_i 10^{i+1} + c_i 10^i$ where $c_i = (a_i + b_i)mod 10 ; d_i = Quotient((a_i+b_i)/10)$. Since, above is true for each $i$, your term-wise addition holds and $d_i$ is carried over.