# More Irrational Number ??>

• July 27th 2005, 09:27 AM
notamathnerd
More 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?
• July 27th 2005, 11:17 AM
Rebesques
Well, if you substract the irrationals $3+\sqrt{2}$ , $2+\sqrt{2}$, you 'll get (the very rational) 1 :p
• July 27th 2005, 11:24 AM
notamathnerd
you didn't use 2 irrational numbers, 3 is rational
• July 27th 2005, 11:34 AM
Rebesques
Do not judge too fast: :p

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

Cheap!!?? :cool:
• July 27th 2005, 11:50 AM
notamathnerd
but you need to add the irrationals together
• July 27th 2005, 12:00 PM
Rebesques
e?
Just did :)

Sorry, but I see kinda stress here...

$2+\sqrt{2}$ is very irrational, and so is the other fellow. Tell me what you do not understand.
• July 27th 2005, 01:57 PM
notamathnerd
in the number model 2 + sqrt 2, only sqrt 2 is irrational. 2 as an addend is rational.
• July 27th 2005, 02:06 PM
Rebesques
...So? The number $2+\sqrt{2}$ is irrational, still. What troubles you? :)
• July 27th 2005, 02:13 PM
MathGuru
the point is you are simply saying that $\sqrt{2} - \sqrt{2} = 0$
I don't think that is the point of the question.
• July 27th 2005, 03:43 PM
notamathnerd
the point of the question is to use two addends, both irrational, and get a sum that is rational.
can it be done?
• July 27th 2005, 04:32 PM
MathGuru
e^pi*i
I don't know if you can do

irrational + irrational = rational

but you can do

(irrational)^(irrational*i)=rational

for example

$e^{\pi*i} = -1$
• July 28th 2005, 12:55 AM
Rebesques
e?
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 $2+\sqrt{2}, 3+\sqrt{2}$).

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 $A=\alpha.\alpha_0\alpha_1\alpha_2... , B=\beta.\beta_0\beta_1\beta_2...$, 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 $\alpha_n =-\beta_n, \ \forall \ n\geq k.$ This means that

$A+B=(\alpha+\beta).(\alpha_0+\beta_0)(\alpha_1+ \beta_1)... =
\gamma.\gamma_0\gamma_1\gamma_2...\gamma_{k-1}=C,$

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

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

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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

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

This means that for all naturals t, $\gamma_k=\gamma_{tm-k}$ or $\alpha_k+\beta_k=\alpha_{tm-k}+\beta_{tm-k},$ and so
$\alpha_k -\alpha_{tm-k}=-(\beta_k -\beta_{tm-k})$, for all t. This (again) sais that A and B are forehand related, as promised.

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

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Sorry for the long answer, hope it was worth it. :eek:
• July 31st 2005, 01:22 AM
Rebesques
e?
Nobody agrees/disagrees? :) :(
• July 31st 2005, 12:55 PM
MathGuru
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.