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

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?

• What's $(\pi+1)-\pi$? Commented Apr 7, 2015 at 2:16
• 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]" Commented Apr 7, 2015 at 2:20
• I think it's still an open question whether or not $\pi - e$ is rational or irrational.
– user14972
Commented Apr 7, 2015 at 11:21
• 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 $e$). Commented Apr 7, 2015 at 11:55
• 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. Commented Apr 7, 2015 at 12:57

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

• @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 $a$. Commented Apr 7, 2015 at 10:11
• @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. Commented Apr 7, 2015 at 10:55
• @user2813274 if for irrational numbers a and b, a-b=q a rational number, then we have a=b+q.
– ant
Commented Apr 7, 2015 at 12:59
• Are there two irrational square roots that differ by a rational number? Certainly. Consider $\sqrt{(\sqrt{2}+1)^2}$ and $\sqrt{2}$. Commented Apr 7, 2015 at 15:44
• @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. Commented Apr 7, 2015 at 16:15

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.
• 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"? Commented Apr 7, 2015 at 9:02
• "no cool examples like $\pi^{7/5}−\sqrt{131}$ can possibly work" - you've never heard that $e^\pi - \pi = 20$? Commented Apr 7, 2015 at 12:15
• @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. Commented Apr 7, 2015 at 15:54
• @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). Commented Apr 7, 2015 at 16:06
• @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) Commented Apr 8, 2015 at 16:08

$\hspace{20mm} \sqrt{2}-\sqrt{2}=0$

• Yeah exactly. Another example of this: PI and PI+1 are both irrational :) Commented Apr 9, 2015 at 6:24

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

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

• 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 $\mathbb R \setminus\mathbb Q$) Commented Apr 7, 2015 at 4:50
• @Quincunx - you mean irrational numbers, not imaginary numbers Commented Apr 7, 2015 at 8:05
• @Belgi Yes, that's correct. Silly me. Commented Apr 7, 2015 at 21:24

One of many: $\sqrt{2}+(5-\sqrt{2})=5$.

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

• That's an interesting way of doing it Commented Apr 8, 2015 at 14:38

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.

• $1/\phi = \phi - 1$ isn't that special. For example, $1/\sqrt{2}=\sqrt{2}/2$ which is similar. This is going to be the case with all roots $\xi$ of polynomials irreducible over $\mathbb{Q}[X]$: in that case $\mathbb{Q}[\xi]=\mathbb{Q}(\xi)$ meaning that $1/\xi$ has a simple expression as a polynomial of $\xi$.
– cody
Commented Apr 8, 2015 at 21:27
• What does that squiggly E mean?
– Neil
Commented Apr 10, 2015 at 1:03
• It's the greek letter xi, a standard letter for dealing with algebraic irrational numbers.
– cody
Commented Apr 10, 2015 at 2:21
• @Neil; what do you mean here by pure irrational numbers? one with without an additive rational component?
– user231063
Commented May 24, 2017 at 3:44

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

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.

• +1 I think yours is the best answer. sin^2(x) -1 = -cos^2(x).
– Neil
Commented Apr 8, 2015 at 2:08
• Please learn to type in LaTeX format. I've cleaned up the mess for you this time. Commented Apr 8, 2015 at 16:10

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.

• See the other cases, it clearly does not cover the whole question Commented Apr 9, 2015 at 17:52
• 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. Commented Apr 14, 2019 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. Commented Apr 15, 2019 at 19:56

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 $(a-b)\in \Bbb Q$.

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.

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

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.

• Every irrational is the sum of an irrational and a non-zero rational: For example, $\sqrt{2}=(\sqrt{2}-1)+1)$. Commented Feb 6 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. Commented Feb 6 at 23:29

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: