Question

© Question attached :-

Answer 1

http://gyazo.com/69a1dc285f9d89f9e211e5166b33bd3d

Answer 2

Since irrational numbers cannot be represented as fractions, I think the key to the proof has something to do with proving that there is no way to represent that number as a fraction. Also I assume you need to represent that number as an infinite sum. I'll keep thinking about it, but maybe that will help you somehow.

Answer 3

I had an interesting thought. There was a way that you could convert repeating decimals to fractions in which you multiple the decimal by 10^(however many digits repeat). You then subtract the decimal from this number getting rid of the repeating decimals and converting to a fraction. I"ll do 1/3 as an example.\[(1/3)=.333333...\]\[10*(1/3)=3.33333333333...\]subtracting these two equations:\[(10-1)*(1/3)=3 \implies (1/3)=3/9=1/3\] So even if you are handed a long decimal, as long as it repeats at some point, you can convert it to a fraction. Not sure if this will be useful or not but I'll keep thinking.

Answer 4

Hmm, you can write the given number as
\[\begin{align*}\frac{10}{100}+\frac{100}{100000}+\frac{1000}{1000000000}+\cdots&=\frac{10^1}{10^2}+\frac{10^2}{10^5}+\frac{10^3}{10^9}+\frac{10^4}{10^{14}}+\frac{10^5}{10^{20}}+\cdots\\\\
&=10^{-1}+10^{-3}+10^{-6}+10^{-10}+10^{-15}+\cdots\end{align*}\]
I'm not sure if there's a distinct pattern here...

Answer 5

I mean to say that there is a pattern, but I don't think there's a simple, concise way to describe it with an infinite series.

Answer 6

Ah, my mistake. There is an \(n\)-th term formula for the exponents. Each successive exponent is the sum of triangular numbers.

Answer 7

So could you write that as 10^-(arithmetic sum) or something? I'm still not sure that would be useful.. lol

Answer 8

So as a series, we have
\[\text{given number}=\Huge\sum_{n=1}^\infty \left(\frac{1}{10}\right)^{\frac{n(n+1)}{2}}\]

Answer 9

So I would argue that the number is actually rational, unless there's some detail I'm missing...

Answer 10

Well they can calculate pi using infinite series so I don't think that implies rationality.

Answer 11

Oh right, for some reason I thought we were dealing with a finite sum of rational numbers, not infinite :P. Not the first time I've made this mistake.

In any case, I suppose this doesn't really prove anything about the number's ir/rationality. Just a nice succinct way of expressing it

Answer 12

Since there are different number of zeros in between any two consecutive ones, we can never find a repeating pattern....Thus, the number is irrational.

Answer 13

Yeah. But now make that into a proof.. lol.

Answer 14

I think it would start something like assume the given number = a/b. Then do who knows what, proving that a/b can't exist, then say and therefore the number is irrational. XD

Answer 15

I kind of regret not having had a proof/mathematical reasoning class at times like this. Any of you proof lovers have any input?

Answer 16

This expansion has a rule. The rule is there is one more 0 after each successive 1. With this rule, we could continue writing the decimal expansion. But though it has a rule, we cannot find any repeating patterns. This number, therefore, cannot be the expansion of a fraction.
Hence, its partial sum do not converge to any rational number.