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[tex]\frac{1}{2^3} + \frac{1}{2^9} + \frac{1}{2^{27}} + \frac{1}{2^{81}} + \cdots[/tex]

i have no idea where to begin. thanks in advance.

note: this is not a homewrok problem.

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- Thread starter murshid_islam
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- #1

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[tex]\frac{1}{2^3} + \frac{1}{2^9} + \frac{1}{2^{27}} + \frac{1}{2^{81}} + \cdots[/tex]

i have no idea where to begin. thanks in advance.

note: this is not a homewrok problem.

- #2

StatusX

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What's true about the decimal expansion of rational numbers?

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Is that fact unique to a

- #5

HallsofIvy

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That doesn't necessarily help you FIND the sum, but your problem is to prove the number is irrational. What happens if you think of this as a number written in binary?

- #6

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it becomes 0.001000001000000000000000001.......That doesn't necessarily help you FIND the sum, but your problem is to prove the number is irrational. What happens if you think of this as a number written in binary?

clearly it doesn't terminate and no pattern repeats. but does this fact of irrational numbers hold true for any base?

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Gib Z

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Fine recipe for proving that real numbers represented by sums of murshid's kind are irrational or even trancendent is http://mathworld.wolfram.com/LiouvillesApproximationTheorem.html" [Broken]

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and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.

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CRGreathouse

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and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.

The most common form I see is expressing it as a/b in lowest terms and then showing that a/b = c/d with 0 < c < a.

- #11

HallsofIvy

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and then prove that if that was the case, then

x<a<x+1

where x is some integer and therefore there is contradiction; an integer cannot lie between two integers.

The most common form I see is expressing it as a/b in lowest terms and then showing that a/b = c/d with 0 < c < a.

But, in this particular case that would be much more difficult that just observing that the number, in base 2, is neither a terminating nor a repeating "decimal".

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Werg22,CRGreathouse:

Of course,"*Reductio ad absurdum*" is ,pretty often, a logical way of organization of the proofs.Unfortunately,as HallsofIvy indicates, it doesn't tell anything on how to construct proofs from case to case!

It may appear you two sound like suggesting there is elementar proof in this case,something like [itex]\sqrt{2}[/itex] is irrational?

Please,show your proofs ,I'm not aware this can be done at elementar level.

Proofs that real numbers :

[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(n!)}}[/tex]

Or

[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(m^n)}};m\geq 2[/tex]

are irrational and transcendent can be constructed by employment of joined consequences of*Dirichlet's Theorem * and mentined above *Liouville's Theorem*.I don't think I can classify Lioville's Tm by elementar label...

Of course,"

It may appear you two sound like suggesting there is elementar proof in this case,something like [itex]\sqrt{2}[/itex] is irrational?

Please,show your proofs ,I'm not aware this can be done at elementar level.

Proofs that real numbers :

[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(n!)}}[/tex]

Or

[tex]\sum_{n=1}^{\infty}\frac{1}{2^{(m^n)}};m\geq 2[/tex]

are irrational and transcendent can be constructed by employment of joined consequences of

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- #13

HallsofIvy

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What do you mean by "elementary proof"?

Euclid's proof seems pretty elementary to me: If [itex]\sqrt{2}[/itex] is rational then [itex]\sqrt{2}= \frac{m}{n}[/itex], reduced to lowest terms. Then [itex]2= \frac{m^2}{n^2}[/itex] so [itex]2n^2= m^2[/itex]. But the square of any odd number is odd ((2n+1)^{2}= 4n^{2}+ 4n+ 1= 2(2n^{2}+ 2n)+ 1). Since m^{2} is even, m must be even: m= 2k for some integer k. Then [itex]2n^2= (2k)^2= 4k^2[/itex] and so [itex]n^2= 2k^2[/itex]- that is, n is also even, contradicting the fact that m and n are relatively prime.

A direct proof, somewhat**less** "elementary" is this: [itex]\sqrt{2}[/itex] obviously satisfies the equation x^{2}- 2= 0. By the "rational root theorem", any rational root of that equation must have numerator that evenly divides the constant term, 2, and denominator that evenly divides the leading coefficient 1. The only possible rational roots, then, are 2 and -2, neither of which satisfies the equation. x^{2}- 2= 0 has no rational roots so [itex]\sqrt{2}[/itex].

That, of course, has little to do with the original question, but I like to show off!

Euclid's proof seems pretty elementary to me: If [itex]\sqrt{2}[/itex] is rational then [itex]\sqrt{2}= \frac{m}{n}[/itex], reduced to lowest terms. Then [itex]2= \frac{m^2}{n^2}[/itex] so [itex]2n^2= m^2[/itex]. But the square of any odd number is odd ((2n+1)

A direct proof, somewhat

That, of course, has little to do with the original question, but I like to show off!

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- #14

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The one that doesn't rely on theorems from math analysis (complex analysis in particular.)What do you mean by "elementary proof"?

Read my post more carefully.I'm not questioning if the irrationality of [itex]\sqrt{2}[/itex] can be prooven by elementar math methods,but if OP's problem can be solved in similar way.

I think it can't (but you can never know for sure ).

If you can solve problems ,and prove irrationality (let alone transcendence) of the numbers represented by the sums given above,in "Euclid's" fashion ,post your solution.We would like to know.

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