Addition of Irrationals

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Mentallic

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Can such irrationals like [tex]\sqrt{2}, \pi, e[/tex] ever be added/subtracted to another irrational to give a rational result?
This would have to exclude such occurences like: [tex]\sqrt{2}+(1-\sqrt{2})[/tex] and less obvious irrationals that - for their irrational parts - can be expressed as the negative of the irrational it is being summed with (or the positive if being subtracted).
I'd like to avoid logarithms, as they can easily give 2 irrationals to become a rational.
 

arildno

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Well, what you are REALLY asking about, as far as I can get it, is:

Does there exist subsets of the irrationals having the property that adding either two elements yield an irrational?
Answer:
Of course it does, zillions of such, if we accept finite subsets of the irrationals.
Do you want to exclude such sets as well?

Your question is not really interesting.
What would have been interesting from a mathematical point of view, would have been to define a subset of the irrationals by some interesting, or innocuously looking criterion, and then gone on to PROVE that this particular set had the fascinating property that summing to elements never yielded a rational.
 
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Mentallic

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If such irrationals [tex]\sqrt{a},\sqrt{b}[/tex] a,b rational are added together to give a rational, then either of the irrationals must be able to be converted: [tex]\sqrt{b}=c-\sqrt{a}[/tex] where c is also rational.
But I'm more curious in irrationals that don't seem to have any connections, such as [tex]\pi, e[/tex].
yyat thanks, thats what I'm looking for :smile:
 

arildno

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Note that yyat's example is PRECISELY the type of problem that WOULD be very interesting to resolve!

Namely, that we first:
1. Identify a class of irrational numbers (namely those that are integer multiples of either pi or e)

2. Asking the question whether the sum of two elements of this particular subset might have the desired property.

Proving, or disproving it, would be a mathematical achievement.
 

arildno

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Well, how would you, on the GENERAL level, specify the criterion "have no seeming connection"?

All it seems to me is that you want to exclude all classes that are "easy" to prove or disprove, and retain only that bundle of criteria that would make the proof "hard".
 

Mentallic

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wikipedia said:
there is no pair of non-zero integers m and n for which it is known whether mπ + ne is irrational or not
Why are they restricting m and n to being integral? I would consider it difficult to find even for rational m and n and even irrational m and n (excluding variations of [itex]\pi[/itex] and e).

Wikipedia says it isn't known if [tex]2^e[/tex] is irrational or not. Intuition would tell me that a rational to the power of an irrational would result in an irrational, but something this - assumably - simple having yet to be discovered is pretty awesome.
 

Mentallic

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Well, how would you, on the GENERAL level, specify the criterion "have no seeming connection"?

All it seems to me is that you want to exclude all classes that are "easy" to prove or disprove, and retain only that bundle of criteria that would make the proof "hard".
Yes that is exactly it. I want to exclude all the irrationals that can be manipulated to have their irrational part become the negative of the irrational it is being added to.

Say for another e.g. [tex]2\sqrt{2}+(1-\sqrt{8})[/tex]
since [tex]\sqrt{8}[/tex] can be converted into the negative of [tex]2\sqrt{2}[/tex] then this is excluded. [tex]\pi[/tex], [itex]e[/itex] and logs on the other hand have no seeming connection in which they can be manipulated to convert into the negation of the other irrational, so these are the problems I'm interested in.
 

arildno

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please remember that [tex]\sqrt{8}[/tex] IS the negative of of [tex]2\sqrt{2}[/tex]; the "conversion" process you speak of is a switch between different REPRESENTATIONS of a particular number, not some magical conversion of one number into some other.
 

arildno

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I hope you realize that by a mere NEGATIVE criterion for set description, i.e that for any x within it, it is implied that q-x is NOT included in the set (q rational), you have specified a truly gargantuan set!

For example, it includes the tiny, specific set yyat mentioned, and many, many more.

Thus, it is not interesting, because it is too general, there are too few specifics upon the numbers included included for a proof to work upon.
 

Mentallic

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there are too few specifics upon the numbers included included for a proof to work upon.
sorry, what exactly do you mean here? Not enough is known about pi and e to base a proof that (as wikipedia states) [itex]m\pi + ne[/itex] for m, n integral is rational?

Sorry that I had to cut out the obvious ones, I just found them to be too... well... obvious :wink:

please remember that IS the negative of of ; the "conversion" process you speak of is a switch between different REPRESENTATIONS of a particular number, not some magical conversion of one number into some other.
yes to convert the representation is what I was getting at.
Now lets say that there is some integers m and n so that [itex]m\pi+ne[/itex] are rational. This would suggest that e or [itex]\pi[/itex] could be 'converted' into the negation of the other irrational. But until such m and n are found, there is no known representation for e in terms of [itex]\pi[/itex] and vice versa.
 

arildno

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sorry, what exactly do you mean here? Not enough is known about pi and e to base a proof that (as wikipedia states) [itex]m\pi + ne[/itex] for m, n integral is rational?
I meant precisely what I wrote.

You have merely designated a gargantuan set that trivially include subsets we already have no idea about how to prove anything about (for example yyat's set).

What is the point of such a general proposition??
 
This seems to me quite an interesting thing to think about. I think I understand what you are asking...

for example: Is there any n such that [tex]log(n)[/tex] is irrational and [tex]log(n) + \pi = s[/tex] where s is a rational number.

or

Is there any n such that [tex]\sqrt{n}[/tex] is irrational and [tex]\sqrt{n} + e = s[/tex] where s is a rational number.


Are those examples of what you're trying to find?

I'm no mathematician for sure, but it seems like a very valid question to me. There is no apparent simple or "trivial" way of proving this to me.
 
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Is there any n such that [tex]\sqrt{n}[/tex] is irrational and [tex]\sqrt{n} + e = s[/tex] where s is a rational number.
No, this would imply that [tex]e^2-2se+s^2-n=0[/tex] contradicting the known fact that [tex]e[/tex] is transcendental.
 
Sorry yyat, I don't understand. Like I said, I'm not a mathematician so I'm not good at seeing simple proofs such as the one you provided.

I understand that e is proven to be transcendental, but what does that equation you gave mean? Let me see if I understand this correctly:

[tex]e[/tex] is irrational therefore [tex]e^2[/tex] is irrational.
[tex]2se[/tex] is irrational because [tex]e[/tex] is irrational and simply being multiplied by a rational constant.
[tex]n[/tex] may or may not be rational (but most likely it is rational. I guess if we say it's irrational then it's just going in circles, so we can say it's rational?)

So what you're saying is that [tex]e^2 - C*e[/tex] cannot be rational?
 
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Sorry yyat, I don't understand. Like I said, I'm not a mathematician so I'm not good at seeing simple proofs such as the one you provided.

I understand that e is proven to be transcendental, but what does that equation you gave mean? Let me see if I understand this correctly:

[tex]e[/tex] is irrational therefore [tex]e^2[/tex] is irrational.
[tex]2se[/tex] is irrational because [tex]e[/tex] is irrational and simply being multiplied by a rational constant.
[tex]n[/tex] may or may not be rational (but most likely it is rational. I guess if we say it's irrational then it's just going in circles, so we can say it's rational?)

So what you're saying is that [tex]e^2 - C*e[/tex] cannot be rational?
I assumed that [tex]n[/tex] is an integer or a rational number, otherwise one could trivially take [tex]n=(3-e)^2, s=3[/tex].

If [tex]e^2-2se+s^2-n=0[/tex] were true, then e would be the solution of a quadratic equation with rational coefficients. By definition, a transcendental number is not the solution of any (nonzero) polynomial equation with rational coefficients.
 
Oh okay. I was trying to think of a situation like that so we would have to explicitly define n to be rational.

Thanks.
 

Mentallic

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With long numerical calculations to a very high precision, it is still found that [itex]\pi - e[/itex] doesn't seem to be making any patterns in its decimal expansion. From what I understand, for the answer to be rational, there must be a corresponding pattern in both pi and e in some form or another. Since there doesn't seem to be any (and computers have calculated e to 1011 places and pi to 1012 decimal places), what are the chances that [itex]\pi - e[/itex] will end up being rational?
 

Hurkyl

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Mentallic

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P(pi - e is rational) = 0.7762732...(repeating)...
 
Stands for "the probability of even (pi - e is rational) is equal to 0.7752732...(repeating)..., or 7752732/9999999.
 
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With long numerical calculations to a very high precision, it is still found that [itex]\pi - e[/itex] doesn't seem to be making any patterns in its decimal expansion. From what I understand, for the answer to be rational, there must be a corresponding pattern in both pi and e in some form or another. Since there doesn't seem to be any (and computers have calculated e to 1011 places and pi to 1012 decimal places), what are the chances that [itex]\pi - e[/itex] will end up being rational?
If you take any two irrational numbers x,y then x-y will "almost surely" not be a rational number (the rational numbers are set of measure zero in the reals). So the first best guess is that pi-e is irrational and the opposite would be quite a sensation.

A well known related example is the http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant" [Broken] which has not been proven irrational. So far, 14,922,244,771 digits have been calculated, so if it is a rational number it's a very complicated one!
 
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