# On the priority of arithmetic operations

Could someone explain the difference between these two problems: 6:2(2 + 1) and 6/2(2 + 1)?

The first one should be read as $$\frac{6}{2(2 + 1)} = \frac{6}{6} = 1$$ while the second is actually $$\left(\frac{6}{2}\right) \times (2 + 1) = 3 \times 3 = 9.$$ Am I right?

First observe $6/2(2+1)$ If you make all brackets you get $(6/2)\cdot (2+1)$ like you wrote it.

Now we observe $6:2(2+1)$ I never saw this notation before but if you read it as a fraction you implicit make brackets around the nominator and denominator. i.e. $(6):(2\cdot (2+1))=(6)/(2\cdot (2+1))$ Now you see that don't have the same brackets and therefore you can get different results.

$6/2(2+1)=9$

$6:2(2+1)=1$

The use of the colon sign to denote a ratio is a distraction. In the first problem you are dividing by everything and in the second example you are not. This can explicitly be seen if double parentheses were used instead of the colon.
First Problem
$6:2(2+1)\rightarrow 6/(2(2+1))=\frac{6}{2(2+1)}$

Second Problem
$6/2(2+1)=6/2*(2+1)=\frac62 *(2+1)$

TLDR - The colon notation is not standard in arithmetic to be used like that so double parentheses should have been used to be more explicit.