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Can you get a rational nonzero number adding or subtracting 2 irrational numbers? If so, do you have examples?

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- Jul 27th 2005, 08:27 AMnotamathnerdMore Irrational Number ??>
:confused: :confused: Help :confused: :confused:

Can you get a rational nonzero number adding or subtracting 2 irrational numbers? If so, do you have examples? - Jul 27th 2005, 10:17 AMRebesques
Well, if you substract the irrationals $\displaystyle 3+\sqrt{2}$ , $\displaystyle 2+\sqrt{2}$, you 'll get (the very rational) 1 :p

- Jul 27th 2005, 10:24 AMnotamathnerd
you didn't use 2 irrational numbers, 3 is rational

- Jul 27th 2005, 10:34 AMRebesques
Do not judge too fast: :p

$\displaystyle (3+\sqrt{2})-(2+\sqrt{2}) = 1 $

Cheap!!?? :cool: - Jul 27th 2005, 10:50 AMnotamathnerd
but you need to add the irrationals together

- Jul 27th 2005, 11:00 AMRebesquese?
Just did :)

Sorry, but I see kinda stress here...

$\displaystyle 2+\sqrt{2}$ is very irrational, and so is the other fellow. Tell me what you do not understand. - Jul 27th 2005, 12:57 PMnotamathnerd
in the number model 2 + sqrt 2, only sqrt 2 is irrational. 2 as an addend is rational.

- Jul 27th 2005, 01:06 PMRebesques
...So? The number $\displaystyle 2+\sqrt{2}$ is irrational, still. What troubles you? :)

- Jul 27th 2005, 01:13 PMMathGuru
the point is you are simply saying that $\displaystyle \sqrt{2} - \sqrt{2} = 0 $

I don't think that is the point of the question. - Jul 27th 2005, 02:43 PMnotamathnerd
the point of the question is to use two addends, both irrational, and get a sum that is rational.

can it be done? - Jul 27th 2005, 03:32 PMMathGurue^pi*i
I don't know if you can do

irrational + irrational = rational

but you can do

(irrational)^(irrational*i)=rational

for example

$\displaystyle e^{\pi*i} = -1$ - Jul 27th 2005, 11:55 PMRebesquese?
I think I get the idea. The lad asks whether two -irrelevant- irrationals, can add up to a rational. The

**general**answer is**no**. Only with trickery can this happen (like the example with $\displaystyle 2+\sqrt{2}, 3+\sqrt{2}$).

To see this, we remember that a number is rational, if and only if it has a terminating or a periodical decimal expansion. Therefore, if we have two irrationals $\displaystyle A=\alpha.\alpha_0\alpha_1\alpha_2... , B=\beta.\beta_0\beta_1\beta_2... $, then for the sum A+B we have the following possibilities:

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**a)**A+B is non-terminating and unperiodical.

Then, it is irrational.

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**b)**A+B is terminating.

Then, there exists a minimum index k, such that $\displaystyle \alpha_n =-\beta_n, \ \forall \ n\geq k.$ This means that

$\displaystyle A+B=(\alpha+\beta).(\alpha_0+\beta_0)(\alpha_1+ \beta_1)... =

\gamma.\gamma_0\gamma_1\gamma_2...\gamma_{k-1}=C,$

this last number $\displaystyle C$ being rational (as terminating). Then, $\displaystyle A=-B+C$, and so**the two irrationals were related at first hand**!

(**Example**: $\displaystyle A=1+\sqrt{2}=2.141421..., B=1-\sqrt{2}=-0.141421...$. Then $\displaystyle A+B=2.141421... -0.141421... =2$.)

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**c)**A+B is periodical.

Then, there exists a minimum index k and a maximum m, such that the decimal expansion

$\displaystyle A+B= \gamma.\gamma_0\gamma_1\gamma_2...=\gamma.\gamma_0 \gamma_1...\gamma_{k-1}\gamma_k...\gamma_m\gamma_k...\gamma_m...$.

This means that for all naturals t, $\displaystyle \gamma_k=\gamma_{tm-k}$ or $\displaystyle \alpha_k+\beta_k=\alpha_{tm-k}+\beta_{tm-k},$ and so

$\displaystyle \alpha_k -\alpha_{tm-k}=-(\beta_k -\beta_{tm-k})$, for all t. This (again) sais that**A and B are forehand related**, as promised.

(**Example**: Consider the numbers $\displaystyle A=0.1212212221... $ and $\displaystyle B=0.1010010001... $They are unperiodical by construction, and so irrational. If $\displaystyle A=0.\alpha_0\alpha_1...$ and $\displaystyle B=0.\beta_0\beta_1...$, then $\displaystyle \alpha_0=\beta_0=\alpha_2=\beta_2=\alpha_5=\beta_5 ...=1,$ that is they are quite related. And, as if by magic, $\displaystyle A+B= 0.222...$, which is periodical, and so very rational.)

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Sorry for the long answer, hope it was worth it. :eek: - Jul 31st 2005, 12:22 AMRebesquese?
Nobody agrees/disagrees? :) :(

- Jul 31st 2005, 11:55 AMMathGuru
I like the unperiodical by construction trick. But it is still an artificial construction.

So I agree, and I appreciate the trick, and you have answered the question well.