# Prove that concatenation is greater than sum

It appears that the concatenation of two positive integers is strictly greater than their sum.

I can prove this for the case when the order of magnitude of their sum is less than the order of magnitude of the concatenation, however I am struggling with the case when they are equal.

Let $a\ge b\ge 1$ and $10^{n-1}\le a<10^n$ (that is, $a$ has $n$ digits). And let's denote concatenation of $a$ followed by $b$ by $a|b$.

Then the sum $a+b$ is certainly not greater that $2a$ since $b\le a$, and $2a$, in turn is certainly less than $10^n+a$ since $a<10^n$:

$$a+b\le 2a < 10^n+a$$

On the other hand, the concatenation of $b$ followed by $a$ is certainly at least equal to $10^n+a$, which would mean adding $1$ before $a$, since $b\ge1$:

$$b|a=10^nb+a\ge 10^n+a$$

Hence we have $a+b<b|a$.

Now, is it also true that $a|b>a+b$?

Yes, because

$$a|b\ge 10a+1>2a\ge a+b$$

Order of magnitude is equal implies they have same number of digits, say $n$. Concatenation of integers $a,b$ that have n-digits is equal to $10^na+b$ or $10^nb+a$. Both these values are greater than $a+b$.

Hint: If the number $b$ is $k$ digits long, then $a\oplus b = a\left(10^k\right)+b$ ($\oplus$ represents concatenation).