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Why irrational means having an infinite decimal expansion?

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Do you mean "infinite without a repeating pattern"? –  naslundx Apr 19 at 14:28
    
@naslundxs I mean that an irrational number have an end. You count and count then you get to an end, you know all the digits of that number. –  user144243 Apr 19 at 14:31

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If the decimal was finite, then it can be represented as $0.d_1 d_2 d_3 \dots d_n$ where $0 \leq d_i < 10$ are the digits of the decimal expansion and $n$ is finite. This can be represented as a fraction, namely $\frac{d_1d_2d_3 \dots d_n}{10^n}$ ($d_1d_2d_3 \dots d_n$ is a number in base $10$, not the product of all $d_i$)

If the decimal was infinite but repeating, say it is $0.\overline{b_1b_2b_3 \dots b_n}$. Then it can be represented as a fraction, namely $\frac{b_1b_2b_3 \dots b_n}{10^n - 1}$

Since they can be represented as fractions of integers, they are not irrational.

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Thanks Michael . –  user144243 Apr 19 at 14:42

Your claim in the title is different from what you have in the question.

The following claim below is true. $$\text{Irrational }\implies \text{infinite decimal expansion}$$ This is equivalent to the contrapositive, i.e., $$\text{Not infinite decimal expansion } \implies \text{not irrational}$$ which (after getting rid of double negative) is same as $$\text{Finite decimal expansion } \implies \text{rational}$$ I assume you can prove this statement.

Let the finite decimal expanion have $k$ digits after the decimal point, i.e., the number is $a + 0.a_1 a_2 \ldots a_k$, where $a$ is an integer and $a_i \in \{0,1,\ldots,9\}$. This can be written as $$a + \dfrac{a_1 a_2 \ldots a_k}{10^k} = \dfrac{10^ka + a_1a_2\ldots a_k}{10^k}$$


However, note that any number with infinite decimal expansion need not be irrational. For instance, the number $0.\bar{3}$ is the infinite decimal expansion for the rational number $1/3$.

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What you are saying is correct. And as Git Gud says you can look at what happens when the decimal expansion is finite. suppose then the number in decimal looks like $a_ma_{m-1}\dots a_1.d_1d_2\dots d_n$ Then it is equivalent to $a_ma_{m-1}\dots a_1+\frac{d_1d_2\dots d_n}{10^n}$ which is rational.

However not all numbers with infinite expansion are irrational. look at $.\overline9$ which is $1$.

see here also.

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