$(i)$ $x$ is irrational
$(ii)$ $3x + 2$ is irrational
$(iii)$ $\frac{x}{2}$ is irrational
My working :
$(i) \to (ii) \land (ii \to iii) \land (iii \to i)$
$(i) \to (ii)$:
Proof By contraposition:
x = p/q
y = c/d
3x + 2 = y
3(p/q) + 2 = c/d
3p+2q/q = c / d
c = 3p + 2q
d = q
c/d = c/d
$(ii) \to (iii)$:
Proof by Contraposition:
x/2 = c/d -> 3x+2 = a/b
Now from the previous example we know that 3x+2 = a/b is true.
$(iii) \to (i) $
Proof by contraposition:
x = a/b -> x / 2 = c/d
(a/b) / 2 = c/d
a/2b = c/d
c = a
d = 2b
c / d = c / d