Prove that $r+x$ is irrational

Question

If $r$ is rational ($r$$\ne$$0$) and $x$ is irrational, prove that $r+x$ is irrational.

Assume that $r+x$ is rational. Then $r+x=(\frac{p}{q})$, where $p$ , $q$ are $\in$ $\mathbb{Z}$, and $p$ and $q$ are in lowest terms. Then we have $x=(\frac{p}{q})-r$= $(\frac{p-rq}{q})$. Since $p-rq$ is $\in$ $\mathbb{Z}$, $x$ is rational. This is a contradiction since $x$ was assumed to be irrational. Therefore, $r+x$ is irrational. QED

I wanted to try proving by contradiction. Just wanted to know if everything looked okay.

Well, mostly good. I don't see why $p-rq\in \mathbb Z$ (though you only need it to be rational). — lulu, Sep 03 '17 at 21:46
You idea is correct just use the fact that $rq$ is rational and not an integer...+1 for your efforts though — Marios Gretsas, Sep 03 '17 at 21:46
The proof is basically correct (modulo the two above comments), though it is unnecessary to assume that $p/q$ is in lowest terms, since you don't actually use that hypothesis anywhere. Since the fact that you can make that assumption depends on the fundamental theorem of arithmetic, which is somewhat deep, you may not want to mention it. — Xander Henderson, Sep 03 '17 at 21:58

hamam_Abdallah · Answer 1 · 2017-09-04T15:41:19.903

6

hint

Observe that if $x+r $ is rational, then

$$(x+r)+(-r)=x $$ as a sum of two rationals will be rational. $(\Bbb Q $ is a field $) $.

edited Sep 04 '17 at 15:41

answered Sep 03 '17 at 23:38

hamam_Abdallah

62,951

The downvote is underserved, the argument is excellent. – Sep 04 '17 at 15:58

score 4 · Answer 2 · 2017-09-04T16:10:55.563

4

Lemma: the difference of two rationals is rational.

Indeed, $$\frac ab-\frac cd=\frac{ad-bc}{bd}\in\mathbb Q.$$

Main theorem:

If $x+r$ is rational, so is $(x+r)-r=x+(r-r)=x$.

Note that this proof requires associativity of the addition of reals.

edited Sep 04 '17 at 16:10

answered Sep 04 '17 at 16:02

I was wondering how necessary it is to provide the note that the proof requires associativity of the addition of reals? Does it make the proof more "complete?" – Skm Sep 05 '17 at 22:31
@S.K.M.: yes it does. – Sep 06 '17 at 10:09

score 3 · Answer 3 · answered Sep 04 '17 at 00:29

The comments essentially yield the full solution but I'll provide a quick proof for you.

Assume that $x$ is irrational and that $r$ is a rational number such that $x+r=s$ is rational. Then we have a contradiction when we observe $x= s-r$ since $x$ is assumed irrational and $s-r$ is rational (difference between rational numbers is rational).

score 0 · Answer 4 · answered Sep 04 '17 at 15:48

$r$ rational means $r=\frac ab$.

$x$ irrational means that it can be expressed as a ratio between integers.

If by contradiction $r+x$ was rational, then $r+x=\frac cd$ thus $x=\frac cd-r=\frac cd-\frac ab\in\Bbb Q$ thus $x$ would be rational, contradiction.

score 0 · Answer 5 · answered Jan 28 '24 at 07:46

0

Let $s=r+x$, be a rational, then we have $s-r=x$ which is a contradiction as by closure of addition of rational numbers, the LHS is rational while $x$ is irrational.

answered Jan 28 '24 at 07:46

Daniel Aricatt

13
4

Prove that $r+x$ is irrational

5 Answers5