$$\frac{6}{30}$$  

$$\frac{5}{13}$$  

$$\frac{1}{5}$$  

$$\frac{5}{30}$$  

Are always bigger than prime numbers

Have 2 pairs of factors

Have more than 2 factors

Don't have factors

9

6

3

8

$$6^4$$  

$$4^6$$  

$$4^2$$  

4x6

n(A)

∀

(A)

A'

A' = { 1,2,3,4,5}

A' = { 4,5}

A' = { 1,2,3,}

A' = { }

Add up the digits. The sum must be even and divisible by 3.

The last digit must be a 3.

The digits must add up to 9.

Add up the digits. The sum must be divisible by 3.

It is divisible by 2 (even) and 5 (ends in a 0 or 5)

It's divisible by 2 (even) and 3 (sum of digits is a multiple of 3)

The sum of the digits is 12, a multiple of 6

The sum of the digits is even

$$\frac{2}{4}=\frac{2}{6}$$  because their numerators are equal

$$\frac{2}{4}<\frac{2}{6}$$  because sixths is bigger than fourths

$$\frac{2}{4}>\frac{2}{6}$$  because fourths are bigger than sixths