How addition and multiplication works 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: 


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

*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?
 A: 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.
A: 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. 
