Thrifty numbers Part 2

One example is sufficient to support the fact of minimum of 2 distinct integers:

$\boxed{0} + \boxed{0} = \boxed{0}$

$\boxed{0} - \boxed{0} = \boxed{0}$

$\boxed{0} \times \boxed{0} = \boxed{0}$

$\boxed{0} \div \boxed{1} = \boxed{0}$

Simplest set applied. The last row is flexible!

Answer: $\boxed{2}$

$\large{ \boxed{0} \, + \, \boxed{1} \, = \, \boxed{1} \\ \boxed{1} \, - \, \boxed{0} \, = \, \boxed{1} \\ \boxed{0} \, \times \, \boxed{1} \, = \, \boxed{0} \\ \boxed{0} \, \div \, \boxed{1} \, = \, \boxed{0} \\ }$ It is clearly seen, that I have used only 2 distinct integers.

Assume that there exists a single integer n that could satisfy every box. Take a look at row 4. $\frac{n}{n}$ = n. You cannot divide by zero, so the integer that satisfies this line cannot be zero. We can remove the n factor from the numerator of denominator of $\frac{n}{n}$ , giving us $\frac{1}{1}$ = n, or just 1 = n. Therefore, if such a number n exists, it must be 1. However, if you replace n for 1 in n + n = n, the first line, this is a contradiction because 1 + 1 does not equal 1. Therefore, no such integer exists in which n could satisfy every box.

It can be proven that there exists a combination of 2 integers that can satisfy the equations by providing an example: 0 + 0 = 0; 0 - 0 = 0; 0 * 0 = 0; 1 / 1 = 1; Therefore, the minimum number of distinct integers is 2.

What I did:

1 + 1 = 2

2 - 1 = 1

1 x 1 = 1

1 / 1 = 1

"Given that the integers 0,1,2,…,9"

Zero to nine is …….integers. Ten integers, if zero is included. But the problem allows all NINE integers. I guess zero doesn't count? Sort of a no account number. Please excuse the poor excuse for an attempt at mathematical humor.