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I had a little back and forth with my logic professor earlier today about proving a number is irrational. I proposed that 1 + an irrational number is always irrational, thus if I could prove that 1 + irrational number is irrational, then it stood to reason that was also proving that the number in question was irrational.

Eg. $\sqrt2 + 1$ can be expressed as a continuous fraction, and through looking at the fraction, it can be assumed $\sqrt2 + 1$ is irrational. I suggested that because of this, $\sqrt2$ is also irrational.

My professor said this is not always true, but I can't think of an example that suggests this.

If $x+1$ is irrational, is $x$ always irrational?

Actually, a better question is: if $x$ is irrational, is $x+n$ irrational, provided $n$ is a rational number?

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    $\begingroup$ I think the problem your professor is pointing out is, "How do you prove that 1+x is irrational?" Knowing that 2<sup>1/2</sup> + 1 is irrational is enough to prove the irrationality of all numbers of the form 2<sup>1/2</sup> + x, x rational, but what about the rest of them? While it is true that x+r irrational implies x is irrational when r is irrational, you don't have another way of proving that x+r is irrational in the first place. $\endgroup$
    – chepner
    Oct 28, 2013 at 16:33
  • $\begingroup$ possible duplicate of The sum of irrationals is irrational? $\endgroup$
    – user17762
    Oct 28, 2013 at 17:34
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    $\begingroup$ @chepner Why are you trying to use HTML tags when you can use $\LaTeX$? $\endgroup$
    – Bakuriu
    Oct 28, 2013 at 18:53
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    $\begingroup$ Too much time spend on stackexchange.com, where the need for $\LaTeX$ is minimal, I suppose. I forgot I could :) $\endgroup$
    – chepner
    Oct 28, 2013 at 19:09
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    $\begingroup$ @Dason Actually, a better question is: if x is irrational, is x+n irrational, provided n is a rational number? Directly from the question. $\endgroup$
    – Cruncher
    Oct 28, 2013 at 20:04

10 Answers 10

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Suppose $x$ is irrational and $x+\dfrac pq=\dfrac mn$ then, $x=\dfrac mn-\dfrac pq=\dfrac{mq-np}{nq}$ so, $x$ would then be rational. :)

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    $\begingroup$ You could delete "$x$ is irrational and" and thus avoid any need to use contradiction. $\endgroup$
    – Henry
    Oct 28, 2013 at 23:50
  • $\begingroup$ @PraphullaKoushik How you know such m/n exist that gives x+(p/q) = m/n? Please give any example. Because I am afraid, there is possibility that x+(p/q) can not be expressed by any m/n. $\endgroup$ Oct 29, 2013 at 12:07
  • $\begingroup$ @PraphullaKoushik it would seem to me that it would be impossible to add a rational (p / q) to irrational (x) and achieve a rational (m / q) (I am no expert, this is a question?) $\endgroup$
    – major-mann
    Oct 29, 2013 at 12:08
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    $\begingroup$ @the two above: he is stating that for $a + b = c$, if $b$ and $c$ are rational, they can be expressed as a fraction of integers, and if that is true, then $a$ must also be expressed as a fraction, thus $a$ must be rational. He is proving that an irrational + rational must be irrational, because a rational - rational must be rational. $\endgroup$
    – gator
    Oct 29, 2013 at 12:15
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Look at the contrapositive: If $x$ is rational, then $x+n$ is rational. Clearly this is a true statement.

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    $\begingroup$ This is the converse, not the contrapositive. $\endgroup$ Oct 28, 2013 at 18:03
  • $\begingroup$ It's essentially the contrapositive of his first question in the last paragraph. Of course it's the converse of the second. I meant to answer the first, but didn't notice that he didn't change the 1 to an $n$ there until now. $\endgroup$
    – Casteels
    Oct 28, 2013 at 18:53
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    $\begingroup$ Huh? "Provided $n$ is a rational number " are the last few words in the question. $\endgroup$
    – Casteels
    Oct 29, 2013 at 3:18
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    $\begingroup$ Re: Cameron Buie's upvoted comment: Up until the very last sentence, riista is asking whether or not the following statement is true: "if $x+1$ is irrational, then $x$ is irrational". I answered with the contrapositive to this statement (with $1$ replaced by $n$). $\endgroup$
    – Casteels
    Oct 29, 2013 at 16:42
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    $\begingroup$ Am I going crazy? For the third time: There are two questions above. The first, taking up 95% of the post is if the following is true "If $x+1$ is irrational, then $x$ is irrational". The second (what he calls "better") question is if, for a rational $n$, this is true: "If $x$ is irrational, then $x+n$ is irrational". I was answering the first, but accidentally replaced 1 with $n$. $\endgroup$
    – Casteels
    Oct 29, 2013 at 20:58
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A proof in the style of "mathematics made difficult": Note that a number $r$ is rational if and only if $\mathbb Q(r) = \mathbb Q$. Now it is easy to see that $\mathbb Q(\gamma) = \mathbb Q(r+\gamma)$ for all rational $r$ and arbitrary $\gamma$. So if $r+\gamma$ is irrational, then $\gamma$ is also.

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    $\begingroup$ So what is $\mathbb Q(r)$? Apparently not the tail probability of the standard normal distribution. Is it $\mathbb Q$ union {$r$}? $\endgroup$
    – LarsH
    Oct 28, 2013 at 15:40
  • $\begingroup$ @LarsH Field extension. $\endgroup$ Oct 28, 2013 at 17:52
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    $\begingroup$ I'm torn between my admiration of the simplicity of this and my admiration of any reference to "Mathematics made Difficult". The number of upvotes resulting should be inferred, assuming the Axiom of Choice and the continued existence of bricks. $\endgroup$
    – Tynam
    Oct 28, 2013 at 18:09
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In fact, for any rational number $ r $ it is true that the irrationality of $ x+r $ implies the irrationality of $ x $. This is due to the fact that the rationals are closed under addition. Assume that $ x+r $ is irrational and (for contradiction) that $ x $ is rational, by the fact that the rationals are closed under addition ($\mathbb {Q}$ is a field) you get that $ x+r $ is rational. Contradiction.

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Note that the sum of two rationals is always rational, and that if $n$ is rational then $-n$ is rational. Now suppose that $x$ is any number and $n$ is rational.

Suppose $x+n$ is rational, then $(x+n)+(-n)$ is rational. Therefore $x+(n+(-n))$ is rational. Therefore $x+0$ is rational, and finally $x$ is rational.

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    $\begingroup$ This is how I would do it, though I would simply say "If $\frac{a}{b}$ is rational and $x+\frac{a}{b}$ is rational, then $x+\frac{a}{b}-\frac{a}{b}$ is rational so $x$ is rational". $\endgroup$
    – Henry
    Oct 28, 2013 at 23:54
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Yes, it is true that $x+1$ being irrational implies $x$ is irrational. Given that $x+1$ is irrational, assume $x=\frac ab$ with $a,b$ integers. Then $x+1=\frac {a+b}b$ would be rational as well.

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  • $\begingroup$ sir, can you elaborate this part for me i can't get it = Then $x+1=\frac{a+b}{b}$ $\endgroup$
    – Deiknymi
    Oct 28, 2013 at 13:55
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    $\begingroup$ $x+1 = \frac{a}b + 1 = \frac{a}b + \frac{b}b = \frac{a + b}{b}$ $\endgroup$
    – filmor
    Oct 28, 2013 at 14:50
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The rational numbers are closed under addition and subtraction. Let $w$ be any irrational number and $r$ be a rational number. Since $$ (r + w) - r = w$$ and $w$ is irrational, one of the subtrahends here is irrational. Since $r$ is rational, the irrational quantity must by $r + w$.

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This follows essentially from the fact that as additive abelian groups $\mathbb Q$ is a normal subgroup of $\mathbb R$, which is true since both groups are abelian and the latter contains the former. Hence we have a quotient homomorphism $\varphi: \mathbb R \rightarrow \mathbb R / \mathbb Q$. $r \in \mathbb R$ is rational iff $\varphi (r) = 0_{\mathbb R / \mathbb Q}$, where $0_{\mathbb R / \mathbb Q}$ is the identity element of $\mathbb R / \mathbb Q$.

Consider then for $x \in \mathbb R - \mathbb Q$, $r \in \mathbb Q$, that \begin{align*} \varphi (x+r)&= \varphi(x) + \varphi(r) \\ &= \varphi(x) + 0_{\mathbb R / \mathbb Q} \\ &= \varphi(x) \\ & \ne 0_{\mathbb R / \mathbb Q} \text{ by assumption that }x \not\in \mathbb Q, \end{align*}

so $x+r \not \in \mathbb Q$.

Of course, the above is just reinterpreting the above elementary proofs in a more general context, but this lets us apply the same line of reasoning to a wide variety of things including modular arithmetic, rotations/reflections of geometric objects, Rubik's cube moves, matrix multiplication, permutations of sets of numbers, exotic differentiable structures on spheres, ...

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Let x be rational and y be irrational. So let us assume that x + y is rational.
(x+y) - x will also be a rational since rational - rational is always rational.
therefore y is also rational. but that is a contradiction. Hence proved!

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  • $\begingroup$ is it true that $irrational^{another irrational}$ is irrational? $\endgroup$
    – shma2001
    Dec 11, 2014 at 15:47
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    $\begingroup$ @user156035 no. for example, $e$ and $\ln(2)$ are irrational, but $e^{\ln(2)} = 2$, which is rational. $\endgroup$ Dec 12, 2014 at 17:18
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Lemma:

If $a$ is rational, and $b$ is rational, then $a+b$ is rational.

Proof:

Let $a=p/q$, and $b=r/s$, where $p,q,r,s$ are integers. Then, $a+b=(ps+qr)/(qs)\blacksquare$

Now, let $a$ be rational, so that $-a$ is rational. If $a+x$ is rational, then $x=(a+x)+(-a)$ is rational, by the lemma. The desired result is the contrapositive: if $x$ is irrational, then $a+x$ cannot be rational, for then $x$ would be rational, which is a contradiction; hence, if $x$ is irrational, then $a+x$ is irrational.

This is a proof by contrapositive: when you prove an implication $P\implies Q$ by first proving that $\neg Q\implies \neg P$. Here, we assume that $a$ is rational. Then, we prove that $$ \text{$a+x$ is rational} \implies \text{$x$ is rational} $$ Then, to prove the implication $$ \text{$x$ is irrational} \implies \text{$a+x$ is irrational} $$ we assume, for the sake of contradiction, that if $x$ is rational, then $a+x$ can be rational. This would imply that $x$ is irrational, which is a contradiction. Hence, $a+x$ is irrational.

Proofs by contrapositive are similar in nature to proofs by contradiction, except that the proof of $\neg Q\implies\neg P$ is a direct proof. This means that any intermediate results deduced using this direct proof can be used in other proofs. This is why, generally speaking, mathematicians favour contraposition over contradiction. See here for further discussion.

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