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$(i)$ $x$ 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
  • 1
    see this http://math.stackexchange.com/questions/2085544/why-do-i-have-to-solve-for-x-in-this-proof-if-i-choose-not-to-find-a-countere – 2017-01-06
  • 1
    I'm having a really hard time understanding your work, more words would be better. Also, using MathJax is very helpful for formatting, here's a quick tutorial http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-01-06

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