I am not sure how to show this. This is what I tried.

Proposition: If x is rational number and y is an irrational number then x+y is an irrational number.

Logically speaking this is

P=x is rational

Q= Y is irrational

R= x+y is irrational

$P\wedge Q \rightarrow R$

Explanation: What I did

$x=\frac{p}{q}$ P and Q integer

Y cannot be written as the a rational by definition

So therefore


Will not be a rational number as there cannot be a common denominator.

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    $\begingroup$ This doesn't do the trick as you claim what is to be shown. Assume that $x+y$ is rational show and that this implies that $y$ is rational. $\endgroup$ – Stefan Mesken Jun 17 '14 at 16:59
  • $\begingroup$ you mean I have to do the contra postive (If x+y is rational then either x is irrational or y is rational) ? $\endgroup$ – Fernando Martinez Jun 17 '14 at 17:04
  • $\begingroup$ No, note that $\sqrt 2 - \sqrt 2 = 0$ is rational, but neither is $\sqrt 2$, nor $-\sqrt 2$.Instead show: "If $x+y$ is rational and $x$ is rational, then $y$ is rational." $\endgroup$ – Stefan Mesken Jun 17 '14 at 17:10

$\color{green}{x=\frac{a}{b}}$ (since $x$ is rational), for $a, b \in \mathbb{Z}$.

Assume, for a contradiction, that $x+y$ is rational.

Then $x+y=\frac{p}{q}$, for $p, q \in \mathbb{Z}$.

Re-arranging: $$\color{red}y=\frac{p}{q}-\color{green}x=\frac{p}{q}-\color{green}{\frac{a}{b}}=\color{red}{\frac{pb-aq}{qb}}\left(=\frac{p'}{q'}\right),$$ which is rational (it's the quotient of two integers)-- a contradiction (since $y$ was assumed to be irrational).



PS: the following is not really necessary, but it's a useful question:

Why are $pb-aq$ and $qb$ integers? I'll leave you to answer that.

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Hint : Assume that x+y is a rational. What can you say about y=(x+y)-x ?

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  • $\begingroup$ Well If x is rational plus y is irrational minus x being rational y would become irrational? $\endgroup$ – Fernando Martinez Jun 17 '14 at 17:01
  • $\begingroup$ It's a proof by contradiction. Assume that the conclusion of the proposition is false : x+y is rational. Now show that would imply a contradiction with the hypothesis : "y is irrational". Indeed, y=(x+y)-x is the difference between two rational numbers, so it is a rational number. $\endgroup$ – Patissot Jun 17 '14 at 17:12

A slight generalization reveals the group-theoretical essence of the matter:

Lemma $ $ Let $\,Q\,$ be a nonempty subset of the additive group $\,\Bbb R\,$ (or any abelian group), and let $\,\overline Q\,$ be the complement of $\,Q\,\,$in $\,\Bbb R.\,$ Then $\,Q + \overline Q\,\subseteq\, \overline Q\iff Q - Q \subseteq Q\iff Q\,$ is a subgroup of $\,\Bbb R.\,$

Thus $\,\Bbb Q + \overline{\Bbb Q}\,\subseteq\, \overline{ \Bbb Q},\,$ i.e. rational + irrational = irrational, is simply a special case of the above, since $\,\Bbb Q\,$ is an additive subgroup of $\,\Bbb R,\,$ being closed under subtraction (so the Subgoup Test applies).

Therefore the above composition law is simply a complementary view of the Subgroup Test. See this answer for the simple proof and further examples, e.g. integer + noninteger = noninteger, and algebraic $*$ transcendental = transcendental if $\ne 0,\,$ etc.

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  • $\begingroup$ Doubt: Must the group be abelian for the lemma to hold? $\endgroup$ – Anant Jun 18 '14 at 4:37
  • $\begingroup$ @Anant Look at the linked proof: does it require commutativity? $\endgroup$ – Gone Jun 18 '14 at 12:51
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    $\begingroup$ $g_1 - g_2 = \bar{g} \iff g_1 - g_2 + g_2 = \bar{g} + g_2 \iff g_1 = \bar{g} + g_2 \overset{\text{commutativity}}{=} g_2 + \bar{g}$. So, without commutativity, the theorem holds only for $\bar{G} + G = \bar{G}$ and not $G + \bar{G} = \bar{G}$? $\endgroup$ – Anant Jun 18 '14 at 13:40
  • $\begingroup$ @Anant But notice that the Subgroup Test can be applied in two ways, yielding both forms $$\begin{eqnarray} G - G = G &\iff& \bar G + G = \bar G\\ -G + G = G &\iff& G + \bar G = \bar G \end{eqnarray}\qquad $$ $\endgroup$ – Gone Jun 18 '14 at 14:29
  • $\begingroup$ Hmm. I am not able to think of a reason :( $\endgroup$ – Anant Jun 18 '14 at 16:57

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