Because the number 8 is not change, but 9 is split up, so the digit number of base₉ is the remander after divde by 9, up there is the remainder after divide by 9, etc.

So we have:

\(A\left(base_{10}\right)=\left(a\equiv x\left(mod9\right)\right)\overline{10...00}+\left(a-\left(9\overline{x0...00}\right)\equiv y\left(mod9\right)\right)\overline{10...10}+...+\left(a-\left(9\overline{x0...00}-\left(...-\left(\overline{z0...00}\right)\overline{10...00}\right)...\right)\equiv t\left(mod9\right)\right)\left(base_9\right)\)

Use this formula, we get:

\(22\left(base_{10}\right)=\left(22\equiv2\left(mod9\right)\right)10+\left(22-2\times9\right)\equiv4\left(mod9\right)\left(base_9\right)\)

So when write the number 22 in Base₉, we get 24.