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Are there any irrational numbers that have a difference of a rational number?

For example, if you take $\pi - e$, it looks like it will be irrational ($0.423310\ldots$) - however, are there any irrational numbers where this won't be the case?

Edit to keep up with the answers:

Cases where it won't be the case:

  • $yX - y(X + n)$, where $X$ is irrational, or equivalent have been covered

  • $e^{\pi i} = -1$ has been covered

  • the golden ratio ($\phi$) has been covered

Are there any other cases?

$e^\pi - \pi$ comes close, but not quite - are there any cases such as this where the result is a (proper) rational number?

Bart Michels
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user2813274
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    What's $(\pi+1)-\pi$? – Steven Stadnicki Apr 07 '15 at 02:16
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    A related thing to note: there exist irrational numbers $x,y$ such that $x^y$ is rational, for example $e, \ln (2)$ are both irrational yet $e^{\ln(2)} = 2$ is rational. Properties like this can be stated mathematically that "The irrational numbers are not closed under [addition/multiplication/exponentiation]" – JMoravitz Apr 07 '15 at 02:20
  • I see. In which what others have pointed out is just how things are by definition. And they have not given x == y solutions... If you only go by definition, you'll know the answer yourself. – 0fnt Apr 07 '15 at 04:28
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    I think it's still an open question whether or not $\pi - e$ is rational or irrational. –  Apr 07 '15 at 11:21
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    I am surprised no one points out the simplest example $\phi - \frac{1}{\phi} = 1$ where $\phi$ is one of the most famous irrational numbers (probably after $\sqrt{2}$, $\pi$ and $e$). – achille hui Apr 07 '15 at 11:55
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    Perhaps it should be noted that the last sentence in your question needs a good bit of cleanup. To prove that a statement is true in general, it must be true in general. That is why several people, including myself, simply provided an example instead of a proof. One counterexample to a claim is enough to disprove it. – Daniel W. Farlow Apr 07 '15 at 12:57
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    More seriously: The numbers pi and e show up all over the place in mathematics. It's fair to say that they are more interesting than, say, the square root of 147 divided by 12. Maybe a sufficiently ingenious person could come up with something interesting about any number. Like the anecdote abut Ramanujan, that one day a friend came to see him and commented that the number of his taxicab, 1729, was not an interesting number. To which Ramanujan instantly replied, "No, that's a very interesting number. It's the smallest number that can be expressed as the sum of two cubes in two different ways." – Jay Apr 07 '15 at 16:37
  • I'm trying to figure out why the comment by @achillehui has any upvotes. It's obviously false. – DWin Feb 21 '24 at 20:23

14 Answers14

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If $q$ is a rational number, and a is an irrational number, then $q+a$ is irrational, $(q+a)-a=q$.

mich95
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    @user2813274: then this answer gives you plenty of examples to answer your question. $a$ and $a+q$ are irrational (which you don't dispute), and their difference is rational. You can fill in any rational value for $q$ and any irrational value for $a$. – Steve Jessop Apr 07 '15 at 10:11
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    @user2813274 Think one step further: any two numbers that confirm to your question, can be written in the form given in this answer! So, mich95 has not only given you a way to construct those two numbers, but has also given the equation that all your 'candidates' must confirm to. – Sanchises Apr 07 '15 at 10:55
  • @sanchises is there any proof that this is the only form in which it will be true? it's not really two irrational numbers, rather than the same one minus itself... – user2813274 Apr 07 '15 at 12:35
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    @user2813274 if for irrational numbers a and b, a-b=q a rational number, then we have a=b+q. – ant Apr 07 '15 at 12:59
  • @ant and.. any cases where q isn't zero? can you prove if any such cases exist/do not exist? (PHI has been included in Neil's answer already) – user2813274 Apr 07 '15 at 14:17
  • @user2813274 ant's comment makes no assumptions on $q=0$ or $q\neq 0$, just that $q$ is rational as you wanted it to be. Maybe easier: say, we want two rational numbers which differ by an integer. We 'magically' pick two rational numbers $a$ and $b$, that differ by an integer $q$; you can write this as $a-b=q$. Now, we simply rewrite this, and you get $a=b+q$ – Sanchises Apr 07 '15 at 14:41
  • @sanchises I know there is no assumption on if q=0 or not, however I am asking because the q=0 scenario has been covered to where it's beating a dead horse by now.. so unless it's something different (q != 0, which I would be interested in), then it's been covered - I would avoid the phrasing "which differ by an integer" as I would be interested in case that result in .1 difference of .5555(repeating) for example, hence rational numbers, and why I used the word difference in the question. – user2813274 Apr 07 '15 at 14:42
  • @user2813274 If you agree this holds for $q\neq 0$, then why is this not an example? Or do you want two irrational numbers that can be written without using addition or substraction? In that case, you should rephrase your question, because mathematics do not require an irrational number to be written in any specific way. – Sanchises Apr 07 '15 at 14:47
  • @sanchises I was considering rewriting the question, however I felt like it would invalidate most of the answers already posted, which are numerous and technically correct - I was hoping that the repeats would just wind down and more answers like Jyrki's and Neil's would come out at this point (In hindsight, I should of started out with the "trivial case" in the answer removed already..) but regardless, I am still interested in the "Proof" part of the answer - can you guarantee that there aren't any other combinations of irrational numbers? or that they are all of form X? – user2813274 Apr 07 '15 at 14:53
  • @user2813274 They can all be written, per definition, of the above form - it's basically what you're asking, written in symbols. The answer does not state that there are no other ways of writing it; compare $4+1$ which one could also write as $5$. Since there are many, many ways to come to an irrational number (really, almost all numbers are irrational!), you should really specify! For example, are there two irrational square roots which differ by a rational number? I'm afraid I don't know the answer to that though :) – Sanchises Apr 07 '15 at 15:36
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    Are there two irrational square roots that differ by a rational number? Certainly. Consider $\sqrt{(\sqrt{2}+1)^2}$ and $\sqrt{2}$. – apnorton Apr 07 '15 at 15:44
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    @sanchises: No there isn't. If $\sqrt p+q = \sqrt r$ with positive rational $p,q,r$, then square both sides and rearrange to get $\sqrt p = \frac{r-p-q^2}{2q}$, so $\sqrt p$ wasn't irrational. – hmakholm left over Monica Apr 07 '15 at 16:15
  • @sanchises If I were to word it as "irrational square roots", then that would exclude the transcendentals (even my own example with Pi and e), however I admit I don't have the best wording either..

    to:anorton that can be re-written as 3+2(sqrt(2)) I believe?

     – user2813274 Apr 07 '15 at 17:42
  • @user2813274 There are uncountable many transcendental numbers, of which many are not 'named'. So again, what is your criterion? All transcendental numbers on the Wikipedia list? Because $\pi+1$ is also transcendental, but apparently you don't want numbers like that. However, I'm afraid there is no mathematical class of 'named numbers'; $\pi$ is just 'special' because we use it a lot, but mathematically no different from $\pi+1$. I don't think you're gonna get a better answer than this. Also, you mean $\sqrt{2}+1$ according to my 'calculations' (i.e., putting it into MATLAB's symbolic engine). – Sanchises Apr 07 '15 at 18:19
  • @sanchises I have edited the question - that is indeed a class of answer, however there are others (also in the question), and I am asking if there are any more - and yes, I seem to have made a typo somewhere when I put it in, it does refactor into 1+sqrt(2) – user2813274 Apr 07 '15 at 18:22
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This is a surprisingly tricky question, if you discount the answers already given. The set of irrational numbers can be further split into

  • algebraic numbers (zeros of polynomials with integer coefficients, such as $\sqrt{7}$, $\root3\of2$ that are zeros of $x^2-7$ and $x^3-2$ respectively), and
  • transcendental numbers - the rest of them. Famous known transcendentals include $e,\pi$, $\log n$ for an integer $n>1$. In general it is very difficult to prove that a number given by some formula is transcendental. The odds are in favor of a number being transcendental unless it is "obviously" algebraic (such as $\sin(\pi/4)=\sqrt2/2$ that happens to be algebraic). That is, unless the formula only involves rationals and root extractions.

What can be said in general is the following

  • The difference between two algebraic numbers is irrational, unless it is of the type described in other answers. The methods needed to identify, when this may be the case involve the theory of field extensions. Algebraic number theory in particular. See questions carrying that tag.
  • The difference between an algebraic and a transcendental is ALWAYS transcendental, hence also irrational. So no cool examples like $\pi^{7/5}-\sqrt{131}$ can possibly work. Such a difference is automatically irrational.
  • The difference between two transcendentals? Who knows? I'm not aware of any non-trivial examples.
Jyrki Lahtonen
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    Great answer, but I'm not sure I get "the difference between two algebraic numbers is irrational, unless it is of the type described in other answers". Is it always obvious when two algebraic numbers are "of the type described in other answers"? – dumbledad Apr 07 '15 at 09:02
  • @dumbledad: All I'm saying that in the case of two irrational algebraic numbers we kinda have an algorithm telling us when the difference might be rational. When that is the case, it will always be of the type $a=\root3\of{7+\sqrt2}$, $b=\root3\of{7+\sqrt{2}}-1$. Differences like $\sqrt3-\sqrt{113}$ and $\sqrt3-\root3\of{5+\sqrt{23}}$ are easy to handle (and they are irrational). – Jyrki Lahtonen Apr 07 '15 at 09:32
  • Ah, I see, thanks. I was troubled that given two arbitrary algebraic equations proving whether any of the roots of one differ by an integer from a root of the other sounds hard since the algebraic equations themselves may be hard to solve. – dumbledad Apr 07 '15 at 09:40
  • @dumbledad. They may be hard to solve. But we can still usually settle questions like whether the difference is rational by studying at the field extensions generated by the respective roots. I am possibly painting too simple a picture. We can have cases like $\theta$ is the real root $x^5+x^2+1=0$, and then use $\theta+1$ as the other. If the other algebraic were $\theta^4+\theta^3$ instead, we could easily tell that the difference is irrational. – Jyrki Lahtonen Apr 07 '15 at 09:48
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    "no cool examples like $\pi^{7/5}−\sqrt{131}$ can possibly work" - you've never heard that $e^\pi - \pi = 20$? – user253751 Apr 07 '15 at 12:15
  • @immibis $$e^\pi-\pi\approx 19.9990999791894757672664429847?$$ Just a silly approximation. On the other hand $$e^{\pi\sqrt{163}}\approx262537412640768743.9999999999992500725972$$ Much closer to an integer, but no cigar. The latter is actually cool, because it has an explanation coming from the theory of ellipctic curves and algebraic number fields. And, unfortunately but not entirely unexpectedly, I didn't first realize that your point was to link to XKCD (+1) :-) – Jyrki Lahtonen Apr 07 '15 at 12:20
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    @dumbledad: If you have two algebraic numbers plus their (monic) minimal polynomials, then it is easy to check if one of them is a translation of the other. Just subtract the second coefficients and divide by the degree; since those coefficients are minus the sum of the roots, what you get is the only possible horizontal translation distance. Then it is trivial to see if the other coefficients match that. – hmakholm left over Monica Apr 07 '15 at 15:54
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    @immibis: $e^\pi-\pi$ would not be one of those cool examples, but belongs in the last bullet instead. (Except this particular difference is known to be transcendental, as shown by Nesterenko). – hmakholm left over Monica Apr 07 '15 at 16:06
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    @JyrkiLahtonen: Could you elaborate on what do you mean by "of the type described in other answers"? Just cases where the second number is of the form "first number + rational number"? What about, for example, $\sqrt{3+\sqrt{8}}-\sqrt{2}$? – Aleksi Torhamo Apr 08 '15 at 14:51
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    @Aleksi: Undoubtedly you know that $\sqrt{3+\sqrt8}=1+\sqrt2$, so this is of that type. Anyway, we have methods (from the theory of field extensions) to first observe that the field $\Bbb{Q}(\sqrt{3+\sqrt{8}})$, though pretending to be a quartic extension of the rationals, is actually a quadratic extension. This is because if we try polynomials with two zeros among $\pm\sqrt{3\pm\sqrt8}$ we find one with rational coefficients. From that point on we can use the method based on minimal polynomials described in Henning Makholm's comment. – Jyrki Lahtonen Apr 08 '15 at 15:19
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    @JyrkiLahtonen: I probably should've emphasized the word "form", or something. It is of that type, but the point I was trying to make (in my head, anyway...) was, isn't every instance of that type by definition? Ie. if the difference of two irrational numbers is rational, one is always the other plus the rational difference. So isn't the sentence essentially just "The difference between two algebraic numbers is irrational, unless the difference between them is rational"? (I don't have enough math background to follow the thing about field extensions, unfortunately) – Aleksi Torhamo Apr 08 '15 at 16:08
  • @Aleksi. Understood. The comments under mich95's answer explain that all the examples are of this type. Basically I wanted to point out that A) with algebraic numbers we can more or less algorithmically (I haven't thought everything through) decide, whether their difference is rational or not. B) the difference between an algebraic and a transcendental is immediate, C) Can't say anything about two transcendentals :-). This answer was in response to a meta thread bemoaning (in a not entirely constructive way) the "trivial" nature of the early answers. – Jyrki Lahtonen Apr 08 '15 at 16:29
  • @Jyrki: Ah, okay, thanks! :) (Also, I hadn't expanded those comments; I'm a bit amused that the exact same example was used there, too :-) – Aleksi Torhamo Apr 08 '15 at 16:57
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$\hspace{20mm} \sqrt{2}-\sqrt{2}=0$

Nicolas Bourbaki
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Let $x = \sqrt{2}$ and $y = \sqrt{2} - 1$. Clearly, $x,y \in \mathbb{I}$. Now,

$$x - y = \sqrt{2}-\sqrt{2} + 1 = 1 \in \mathbb{Q}.$$

Decaf-Math
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    It's strange notation to call the imaginary numbers $\mathbb I$, because they are not closed under addition, multiplication, etc. In other words, they don't form something like a group or a ring or a field. The more standard notation is to call them $\mathbb R - \mathbb Q$ (IE $\mathbb R \setminus\mathbb Q$) – Justin Apr 07 '15 at 04:50
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    @Quincunx - you mean irrational numbers, not imaginary numbers – Belgi Apr 07 '15 at 08:05
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    @Belgi Yes, that's correct. Silly me. – Justin Apr 07 '15 at 21:24
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One of many: $\sqrt{2}+(5-\sqrt{2})=5$.

Daniel W. Farlow
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How about $\frac{log_{2}(3)}{log_{2}(\frac{3}{2})}$ and $\frac{log_{2}(\frac{3}{4})}{log_{2}(\frac{3}{2})}$, then $\frac{log_{2}(3)}{log_{2}(\frac{3}{2})} + \frac{log_{2}(\frac{3}{4})}{log_{2}(\frac{3}{2})} = \frac{log_{2}(\frac{9}{4})}{log_{2}(\frac{3}{2})} = 2 \frac{log_{2}(\frac{3}{2})}{log_{2}(\frac{3}{2})} = 2.$

sqtrat
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For Transcendental numbers I don't know if there is one that will work. But I can tell you why it won't for regular irrational numbers like square root 2 or cube root 3.

There isn't a way that you can add pure irrational numbers and end up with another pure irrational number at the end. for example: (x)^1/a + (y)^1/b won't give you (z)^1/c where those are all integers and result in an irrational number.

unless it happens to be the same irrational number in which case it ends up being 2 times itself.

The only thing that I can think of that would be close to your intention would be the golden ratio (PHI): (1-(5)^.5)/2

This is because of a very interesting property! :) 1/PHI = PHI-1

On further consideration the transcendental one ultimately has a problem, once you can figure out that you can get 1 transcendental number from another by simple arithmetic then you realize that one of the numbers will have to be simpler than the other so you would write the more complex one in terms of the simpler one.

Consider the area of a 4d hypersphere, which is a 4 dimensional ball. It has an area of ((pi^2)r^4)/2 where r is the radius. Now also say that you sum up all the numbers from 1/1 + 1/2^2 + 1/3^2 +...1/n^2 and you aren't 100% sure that you know what that is equal to, but you have a really good decimal value and it seems like a new and interesting irrational/transcendental number

Now you start playing around with the value of the radius in your hypersphere and all of a sudden when you put in r as (1/3)^.25 you realize that the numbers come out the same! and that adding up all the squares of the reciprocals of every number to infinity you end up with pi^2/6. So what happened is now you rewrote your transcendental value in terms of another transcendental value

What I'm trying to say is that if someone had a really useful and interesting transcendental number that they called ku because they think it is cool, but then someone figures out that ku is just pi+777 then people would switch to writing pi+777 instead of writing ku.

Neil
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$e^{\pi i}=-1$

Not exactly what you asked for, but manipulation of irrationals and imaginaries CAN give a rational number. (Okay, you said irrational, not imaginary. But ...)

$\sqrt8-2\sqrt2=0$

Okay, maybe ending up with a difference of zero is a boring special case.

Others have noted the obvious category of cases, $(\pi+1) - \pi = 1$, etc. That seems pretty boring, but what's the difference between such a "boring" case and an "interesting" case? Intuitively, we want a case where $x-y=1$ and where we can't see $x$ or part of $x$ in $y$ or vice versa. i.e. we want two irrationals, $x$ and $y$, and an integer $n$, such that $x-y=n$ but where $x$ is not written as "$y+n$". Like $\sqrt p-\sqrt q=1$. That particular "format" doesn't work, but maybe there's some other format?

Oh, duh, there's an obvious set of cases from trig. For example:

$$\sin(\pi/2) - \sin(\pi/4) * \cos(\pi/4) = 1/2$$

Well, maybe that doesn't count because $\sin(\pi/2)$ is not irrational, even though I used a non-algebraic function to get there.

Well, $\sin^2(x)+\cos^2(x)=1$, which is not a difference but a sum, but if you want to pick nits, I could say it's minus the negative.

Marc van Leeuwen
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Jay
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Are there any irrational numbers that have a difference of a rational number?

Turn the question around:

If I have an irrational number $X$, can I add a rational number $q$ to it? Will the result be an irrational number?

And I think the answer is, obviously, yes in nearly all cases.

You've already ruled this out as a covered case:

  • $yX - y(X + n)$, where $X$ is irrational, or equivalent have been covered

but I think it really answers the whole question.

Ross Presser
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  • See the other cases, it clearly does not cover the whole question – user2813274 Apr 09 '15 at 17:52
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    No need for "nearly": adding a rational A to an irrational B can only result in an irrational C. Otherwise (i.e. if C could end up being rational), the difference between two rationals (C - A) would be irrational (B), which is clearly not possible because the field of rationals $\mathbb{Q}$ is closed with respect to sum. – polettix Apr 14 '19 at 21:54
  • @polettix Thanks. I've struck out the word "nearly". After reviewing this though, and reading the other answers, I'm not sure my answer is really relevant. – Ross Presser Apr 15 '19 at 19:56
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I do not understand what you mean by 'this won't be the case' but if you ask if there are many such pairs of irrational numbers $a,b$ that their difference is rational, then yes—for each irrational $a$ there is a countably infinite set of irrational numbers $B\subset (\Bbb R\setminus\Bbb Q)$ such that each $b\in B$ satisfies $(a-b)\in \Bbb Q$.

CiaPan
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I would like to give another approach.

Let $x$ be an irrational number, and consider it's decimal expansion, $a_0.a_1a_2a_3......$. Then let $y=b_0.b_1b_2b_3......$and for each $i\in \mathbb N - \{0\},$ let $a_i= b_i$. Then $x-y = a_0 - b_0,$ which are both integers, and are clearly rational.

A. Thomas Yerger
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Difference of any two same irrational numbers will give $0$ which is a rational number.

Soham
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A special case of the problem posed: Let z be any irrational that is not the sum of an irrational plus a nonzero rational. Let r be any nonzero rational. Then $\sqrt{2}$ cannot be equal to z+r. For if $\sqrt{2}$=z+r then $z^2+2rz+r^2-2=0$ (1). The two solutions of (1) are z1=-r-$\sqrt{2}$ , z2=-r+$\sqrt{2}$ both of which do not conform with our prerequisition about z.

Kostas
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  • Every irrational is the sum of an irrational and a non-zero rational: For example, $\sqrt{2}=(\sqrt{2}-1)+1)$. – Hanul Jeon Feb 06 '24 at 23:28
  • The issue in your "proof" is that you assumed that there is $z$ that is not the sum of an irrational and a non-zero rational. There is no such $z$, so the remaining part of your argument is meaningless while it can be correct. – Hanul Jeon Feb 06 '24 at 23:29
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It is it always the case, that:

The sum of an 'rational number': "$x;\,x\in \mathbb{Q}$", and an 'irrational number': "$y;\,y\in \mathbb{IR}$", i.e, "$m=(x+y)$", is an 'irrational number': "$m;\,m \in \mathbb{IR}$"?

$$\forall(x\in \mathbb{Q})\forall(y \in \mathbb{IR}):\,(x + y) \in \mathbb{IR}$$

Thus, in these cases, $\text{the difference;}\quad \text{"}m-y\text{"}\,$, between the 'irrational valued sum', "$m";\,\text{where}\, m=x+y$, and its 'irrational component', "$y$", is 'the sum's rational component ': "$x$", $$\text{where,}$$ $$[m-y=x+y-y=x]\rightarrow [m-y=x]\,\rightarrow \,(m-y)\in\mathbb{Q}\,\,\text{, because, }\,x\in \mathbb{Q}$$

$$\text{Where, the difference between two irrationals,}\,(m,y)\in\mathbb{IR};\quad\text{that is: '(m - y)'};\,(m-y)\in\mathbb{Q},\,\,\text{ is a rational number.}$$

It may not always be the case that the difference between two irrational numbers is rational.

Nonetheless, the sum $x+y$ of an irrational number,$y$ and a rational number, $x$ is always irrational, as far as I know.

Thus, for all such cases of the form adduced above, the difference between two irrational numbers, $'m-y'; \,(m,y)\in \mathbb{IR} $, where '$m=x+y';x \in \mathbb{Q}$, will be rational, as its just $x$.

Remember thought that, at most one of $$(\,(x-y)\,,(x+y)\,) \in \mathbb{Q}$$

As its always the case that, for two irrationals, $(x,y)$ either the difference betwixt them $x-y$, or their sum $x+y$, must be irrational valued. On pain of contradiction. And these are not mutually exclusive alternatives. Both could be irrational.

At least one of $$(\,(x-y)\,,(x+y)\,) \in \mathbb{IR}\,\,\,\text{where} \, (x,y)\in \mathbb{IR}$$;

For example,

$$x=2\pi \in\mathbb{IR},\, y=\pi\in \mathbb{IR}$$.

$$x+y=2\pi +\pi=3\pi\in \mathbb{IR}$$

$$x-y=2\pi-\pi=\pi\in \mathbb{IR}$$ .

$x=2\pi\in\mathbb{IR}$, as $x$ is a rational multiple of an irrational number, $\pi$.

If, on the contrary, $x=2\pi$ were rational, then $\pi$, an irrational number, $\pi=\frac{2\pi}{2}=\frac{x}{2}$, would be a rational fraction, of a rational number, $x=2\pi$, and thus a number of form;

$$\pi=\frac{2\pi}{2}=\frac{x}{2}=\frac{1}{2}\times\frac{p}{q}=\frac{p}{2q}\,\text{as}\,\,x= \frac{p}{q}\,\text{where}\,(p,q) \in \mathbb{Z}\,\text{as}\, x \in \mathbb{Q}$$

$$\text{but because}\, q \in \mathbb{Z}\, \rightarrow t=2q \in \mathbb{Z}$$

Thus, $$\pi= \frac{p}{2q}= \frac{p}{t}$$

$\pi$ would be a rational form:

$$ \text{as}\, (p,t)\in \mathbb{Z} \, \rightarrow\, \pi =\frac{p}{t}\in \mathbb{Q} $$

that is, a fraction, with integer denominator and numerator, and thus a rational number, a contradiction:

William Balthes
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