An irrational number is a real number that can't be written as a fraction: since all rational numbers ([tex] \mathbb{Q} [/tex]) are real numbers ([tex] \mathbb{R} [/tex]), the irrational numbers are all elements in [tex] \mathbb{R} [/tex] that don't belong to [tex] \mathbb{Q} [/tex].
The rational numbers are closed with respect to sum and multiplication, which means that the sum/product of two rational numbers is a rational number.
Note that, of all the options, only [tex] \sqrt{3} [/tex] is irrational, because the other options are
[tex] 3 = \dfrac{3}{1},\quad 2 =\dfrac{2}{1},\quad 0.\overline{23} = \dfrac{23}{99} [/tex]
so they all can be written as fraction. This means that
[tex] -\dfrac{5}{9} \times 3,\quad -\dfrac{5}{9} \times 2,\quad -\dfrac{5}{9} \times\dfrac{23}{99}[/tex]
are all rational numbers, because they are the multiplication of two rational numbers.
So, your only hope to get an irrational number is to multiply
[tex]-\dfrac{5}{9}\times \sqrt{3} [/tex]
Expression 2 works exactly in the same way, because everything we said about the product is true for the sum as well.
