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We can build $\frac{1}{33}$ like this, $.030303$ $\cdots$ ($03$ repeats). $.0303$ $\cdots$ tends to $\frac{1}{33}$.

So,I was wondering this: In the decimal representation, if we start writing the $10$ numerals in such a way that the decimal portion never ends and never repeats; then am I getting closer and closer to some irrational number?

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2 Answers 2

The decimal expansion of a rational number is always repeating (we can view a finite decimal as a repetition of $0$'s)

If $q$ is rational we may write it as an irreducible fraction $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$. Consider the Euclidean division of $a$ by $b:$

At each step, there are only finitely many possible remainders $r\;\;(0\leq r< b)$. Hence, at some point, we must hit a remainder which has previously appeared in the algorithm: the decimals cycle from there i.e. we have a repeating pattern.

Since no rational number can be non-repeating, a non-repeating decimal must be irrational.

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It should be pointed that the repetition is not limited to digits, but to patterns. –  Asaf Karagila Jan 26 '13 at 16:40
    
@AsafKaragila, what do you mean by "patterns"? The continued fraction expansion of the golden ratio is all 1's. –  alancalvitti Jan 29 '13 at 17:44
    
@alan: Where do you see the words "continued fraction" on this page? –  Asaf Karagila Jan 29 '13 at 17:45
    
I'm referring to the word "pattern" in your comment. You said not limited to digits. What's your definition of pattern? Did you mean to say other bases? –  alancalvitti Jan 29 '13 at 17:46
    
@alan: $0.123412341234\ldots$ has a pattern of $1234$. We say that $x\in(0,1)$ has a [decimal] pattern if there are positive integers $k,n$ such that $10^n\cdot x-k=x$. This guarantees that $x$ is rational, even if the decimal expansion is not terminating. You really need to stop assuming that I am saying more than what I write. It's quite annoying at times. –  Asaf Karagila Jan 29 '13 at 18:05

Closer and closer? If you mean by closer and closer as in approximating an irrational, yes. The idea behind closeness is that there is some destination behind where you are going whilst writing the number, but in an irrational number there is no destination, you simply keep writing. If it would be possible, with perfect information, to keep writing out the decimal expansion of an irrational number, making sure there is absolutely no repetitiveness or patterns, then you would be getting arbitrarily close to the irrational number, (in terms of $\epsilon$ close). An irrational number has a non-terminating, non-repeating decimal expansion. So there is no real idea of closeness here, unless you are talking about distance, ($\epsilon$ close), in which case yes you are getting closer.

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Answers found. Thank you for the help. –  ReekMaths Mar 12 '13 at 18:51

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