Why is any number consisting of all 1’s is a rational number? [on hold]

Is this true? I can’t understand why.



2 Answers


One way to see it is to note x=1.111…x=1.111\dots.

Then 10x=11.111…10x=11.111\dots, so 10x−x=10=9x10x-x=10=9x and x=109x=\frac{10}{9}.

But you first have to accept the existence of such numbers, i.e. accept that they represent a real number. This can cause some troubles when you try to interpret the meaning of 0.999999…0.999999… 🙂

Any real number having an eventually periodic decimal expansion is rational.

Indeed any such number can be written as
x=N+0.a1a2…anb1b2…brb1b2…=N+0.a1a2…an+0.00…0⏟n0s¯b1b2…br=N+A10n+10−n∞∑k=1B10kr,\begin{align}x=N+0.a_1a_2\dots a_nb_1b_2\dots b_rb_1b_2\dots&=N+0.a_1a_2\dots a_n+0.\underbrace{00\dots 0}_{n\,0\mathrm s}\,\overline{b_1b_2\dots b_r}\\
&=N+\frac A{10^n}+10^{-n}\sum_{k=1}^{\infty}B\,10^{kr},
where NN is the integer part and A=a1a2…an,A=a_1a_2\dots a_n, B=b1b2…br\;B=b_1b_2\dots b_r (decimal notation).

Now use the formula for the sum of a geometric series:
x=N+A10n+10−nB10−r1−10−r=N+A10n+B10n+r−10n.x=N+\frac A{10^n}+10^{-n}B\frac{10^{-r}}{1-10^{-r}}=N+\frac A{10^n}+\frac{B}{10^{n+r}-10^n}.