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If $x$ is irrational then $3x+2$ is irrational.

My Working:

Proof by contraposition:

$3x + 2$ is rational and $x$ is rational

$x = a/b$

$y = c/d$

$3x + 2 = c/d$

$3x = (c/d) - 2$

$x = c-2d/3d$

$a/b = c-2d / 3d$

I am having a hard time understanding as to why do we end up solving for $x$ in this equation to determine weather $3x+2$ is rational or not if $x$ is rational.

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    The phrasing is a bit odd. You don't need to start out assuming that $x$ is rational...just assume that $3x-2$ is rational and then deduce that $x$ is from that (as you do). That contradicts the initial assumption and completes the proof (by contradiction). – 2017-01-06
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    @lulu it should be $3x+2$--:) – 2017-01-06
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    @juniven Very true. – 2017-01-06

2 Answers 2

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What you have is not a proof by contraposition.

If you were actually proving the contrapositive directly, you would not assume that $x$ is rational: you would assume that $3x+2$ is rational and show that this implies that $x$ is rational. Such a proof would start something like this:

If $3x+2$ is rational, there are integers $a$ and $b$ such that $b\ne 0$, and $3x+2=\frac{a}b$.

In order to show that $x$ is rational, you’re going to have to solve for it in terms of $a$ and $b$ in order to express it as a ratio of integers.

Then $x=\frac{a-2b}{3b}$. Clearly $a-2b$ and $3b$ are integers, and $3b\ne 0$ because $b\ne 0$, so $x$ is rational.

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The contrapositive of "If $x$ is irrational then $3x+2$ is irrational" is:

If $3x+2$ is rational then $x$ is rational.

There's no "and" involved in contraposition (contrary to the current phrasing of your question).

So to prove this, you assume that $3x+2$ can be expressed as a fraction of integers, and prove that this implies that $x$ can be expressed as a fraction of integers as well.